Question 1: Palette 1/89

DPT MCQ

The time period $T$ of an alternating quantity is related to its frequency $v$ by the formula $T = \frac{1}{v}$. The time taken to reach from zero to the positive peak value is one-fourth of the time period ($t = \frac{T}{4}$), and the time taken for the change from positive peak to negative peak is half of the time period ($t' = \frac{T}{2}$).

If the frequency of an alternating current is 50 Hz then the time taken for the change from zero to positive peak value and positive peak value to negative peak value of current are respectively

A

1/200 sec, 1/100 sec

B

1/100 sec, 1/200 sec

C

200 sec, 100 sec

D

1/50 sec, 1/100 sec

Question 2: Palette 2/89

DPT MCQ

The general equation of an alternating current is expressed as $i = i_{0} \sin \omega t = i_{0} \sin 2\pi v t$, where $i_{0}$ is the peak value of current and $v$ is the frequency. The peak value $i_{0}$ is related to the root mean square value $i_{rms}$ by the relation $i_{0} = \sqrt{2} i_{rms}$.

What will be the equation of ac of frequency 75 Hz if its r.m.s. value is 20 A

A

$i = 20 \sin 150 \pi t$

B

$i = 20\sqrt{2} \sin(150 \pi t)$

C

$i = \frac{20}{\sqrt{2}} \sin(150 \pi t)$

D

$i = 20\sqrt{2} \sin(75\pi t)$

Question 3: Palette 3/89

DPT MCQ

The alternating voltage can be represented as $V = V_{0} \sin \omega t = V_{0} \sin (\frac{2\pi}{T}) t$. To find the time $t$ when the voltage reaches a fraction of its peak value, substitute the value into the equation and solve for $t$ using inverse trigonometric relations.

At what time (From zero) the alternating voltage becomes $\frac{1}{\sqrt{2}}$ times of it's peak value. Where T is the periodic time

A

$\frac{T}{2}$ sec

B

$\frac{T}{4}$ sec

C

$\frac{T}{8}$ sec

D

$\frac{T}{12}$ sec

Question 4: Palette 4/89

DPT MCQ

The instantaneous alternating e.m.f. expressed as a sine function is given by $E = E_{0} \sin \omega t = E_{0} \sin 2\pi v t$, where $E_{0}$ is the peak value, $v$ is the frequency, and $t$ is the time.

The peak value of an alternating e.m.f. E is given by $E = E_{0} \cos \omega t$ is 10 volts and its frequency is 50 Hz. At time $t = \frac{1}{600}$ sec, the instantaneous e.m.f. is

A

10 V

B

$5\sqrt{3}$ V

C

5 V

D

1 V

Question 5: Palette 5/89

DPT MCQ

The instantaneous value of an alternating current is given by the equation $i = i_{0} \sin (\omega t + \phi_{0})$, where $i_{0}$ is the peak current, $\omega$ is the angular frequency, $t$ is the time, and $\phi_{0}$ is the initial phase angle. To find the current at a specific time, substitute the given time value into the equation.

The instantaneous value of current in an ac circuit is $i = 2 \sin (100\pi t + \pi/3)$ A. The current at the beginning ($t = 0$) will be

A

$2\sqrt{3}$ A

B

$\sqrt{3}$ A

C

$\frac{\sqrt{3}}{2}$ A

D

Zero

Question 6: Palette 6/89

DPT MCQ

The alternating voltage can be simplified using trigonometric identities such as $\sin 2\theta = 2 \sin \theta \cos \theta$ to express it in the standard form $V = V_{0} \sin \omega t$, where $V_{0}$ is the peak voltage and $\omega = 2\pi v$ gives the frequency.

The voltage of an ac source varies with time according to the equation $V = 100 \sin(100\pi t) \cos(100\pi t)$ where $t$ is in seconds and $V$ is in volts. Then

A

The peak voltage of the source is 100 volts

B

The peak voltage of the source is 50 volts

C

The peak voltage of the source is $100/\sqrt{2}$ volts

D

The frequency of the source is 50 Hz

Question 7: Palette 7/89

DPT MCQ

The time difference corresponding to a phase difference $\phi$ for an alternating current of frequency $v$ (or time period $T$) is given by the formula $T.D. = \frac{T}{2\pi} \times \phi = \frac{1}{2\pi v} \times \phi$.

If the frequency of ac is 60 Hz the time difference corresponding to a phase difference of $60^{\circ}$ is

A

60 sec

B

1 sec

C

$\frac{1}{60}$ sec

D

$\frac{1}{360}$ sec

Question 8: Palette 8/89

DPT MCQ

The average power dissipated in an alternating current circuit is given by the formula $P = \frac{1}{2} V_{0} i_{0} \cos \phi$, where $V_{0}$ is the peak voltage, $i_{0}$ is the peak current, and $\phi$ is the phase difference between voltage and current.

In an ac circuit, V and i are given by $V = 100 \sin(100 t)$ volt, and $i = 100 \sin(100 t + \frac{\pi}{3})$ mA. The power dissipated in circuit is

A

$10^{4}$ watt

B

10 watt

C

2.5 watt

D

5 watt

Question 9: Palette 9/89

DPT MCQ

When an alternating current and a direct current are supplied together, the total current is a mixture of a dc component and an alternating current. The r.m.s. value of such a combined current is calculated using the relation $i_{rms} = \sqrt{(dc)^{2} + (r.m.s. value of ac)^{2}}$.

In a circuit an alternating current and a direct current are supplied together. The expression of the instantaneous current is given as $i = 3 + 6 \sin \omega t$. Then the r.m.s. value of the current is

A

3 A

B

6 A

C

$3\sqrt{2}$ A

D

$3\sqrt{3}$ A

Question 10: Palette 10/89

DPT MCQ

The r.m.s. value of a complex alternating quantity containing multiple harmonic components is equal to the square root of the sum of the squares of the r.m.s. values of its individual components, given by $E_{rms} = \sqrt{(\frac{a}{\sqrt{2}})^{2} + (\frac{b}{\sqrt{2}})^{2}} = \frac{1}{\sqrt{2}}(\sqrt{a^{2} + b^{2}})$.

The r.m.s. value of the alternating e.m.f. $E = (8 \sin \omega t + 6 \sin 2\omega t)$ V is

A

7.05 V

B

14.14 V

C

10 V

D

20 V

Question 11: Palette 11/89

DPT MCQ

The phase difference between alternating voltage and current is given by $\Delta \phi = \phi_{2} - \phi_{1}$ relative to current, where positive or negative signs determine whether voltage leads or lags the current.

Voltage and current in an ac circuit are given by $V = 5 \sin(100\pi t - \frac{\pi}{6})$ and $i = 4 \sin(100\pi t + \frac{\pi}{6})$

A

Voltage leads the current by $30^{\circ}$

B

Current leads the voltage by $30^{\circ}$

C

Current leads the voltage by $60^{\circ}$

D

Voltage leads the current by $60^{\circ}$

Question 12: Palette 12/89

DPT MCQ

The root mean square (r.m.s.) value of the wattless current is given by the formula $i_{WL(rms)} = i_{rms} \sin \phi = \frac{i_{0}}{\sqrt{2}} \sin \phi$, where $i_{0}$ is the peak current and $\phi$ is the phase difference between the alternating current and potential difference.

The instantaneous values of current and potential difference in an alternating circuit are $i = \sin \omega t$ and $E = 100 \cos \omega t$ respectively. r.m.s. value of wattless current (in amp) in the circuit is

A

1

B

$\frac{1}{\sqrt{2}}$

C

100

D

Zero

Question 13: Palette 13/89

DPT MCQ

The wattless current in an alternating current circuit is given by $i_{WL} = i_{rms} \sin \phi$, where $i_{rms}$ is the root mean square current and $\phi$ is the phase angle. The power factor of the circuit is defined as the cosine of the phase angle, given by $p.f. = \cos \phi$.

The r.m.s. current in an ac circuit is 2 A. If the wattless current be $\sqrt{3} A$, what is the power factor

A

$\frac{1}{\sqrt{3}}$

B

$\frac{1}{\sqrt{2}}$

C

$\frac{1}{2}$

D

$\frac{1}{3}$

Question 14: Palette 14/89

DPT MCQ

The concept involved is the relationship between power, r.m.s. voltage, r.m.s. current, power factor, and peak voltage in an alternating current circuit. The formula for power dissipated is $P = V_{rms} i_{rms} \cos\phi$, and the relation between peak voltage and r.m.s. voltage is $V_{0} = \sqrt{2} V_{rms}$.

The r.m.s. value of alternating current in a circuit is 4 A and power factor is 0.5. If the power dissipated in the circuit is 100W, then the peak value of voltage in the circuit is

A

50 volt

B

100 volt

C

35 volt

D

70 volt

Question 15: Palette 15/89

DPT MCQ

The concept involved is the relationship between impedance, resistance, and phase angle in an alternating current circuit. The formula relating them is $\cos\phi = \frac{R}{Z}$, where $Z$ is the impedance, $R$ is the resistance, and $\phi$ is the phase angle between the current and the e.m.f.

the impedance of an ac circuit is $200~\Omega$ and the phase angle between current and e.m.f is $60^{\circ}$. What is the resistance of the circuit

A

$50~\Omega$

B

$300~\Omega$

C

$100~\Omega$

D

$100\sqrt{3}~\Omega$

Question 16: Palette 16/89

DPT MCQ

The concept involved is the calculation of root mean square (r.m.s.) current in a half-wave rectified alternating current circuit. The formula for the r.m.s. value of current in a half-wave rectifier is $i_{rms} = \frac{i_{0}}{2} = \frac{V_{0}}{2R}$, where $V_{0}$ is the peak voltage and $R$ is the resistance.

An ac voltage source of $E = 150~sin~100~t$ is used to run a device which offers a resistance of $20~\Omega$ and restricts the flow of current in one direction only. The r.m.s. value of current in the circuit will be

A

1.58 A

B

0.98 A

C

3.75 A

D

2.38 A

Question 17: Palette 17/89

DPT MCQ

The concept involved is the determination of frequency from the time period and the phase relationship between two sinusoidal waveforms. The frequency is given by $v = \frac{1}{T}$, where $T$ is the time period, and the phase relationship is determined by the horizontal shift between the waveforms expressed in radians.

From the image given below, two sinusoidal voltages of the same frequency are shown in the diagram. What is the frequency, and the phase relationship between the voltages

A

$0.4~Hz, -\frac{\pi}{4}$

B

$2.5~Hz, +\frac{\pi}{2}$

C

$2.5~Hz, -\frac{\pi}{2}$

D

$2.5~Hz, -\frac{\pi}{4}$

Question 18: Palette 18/89

DPT MCQ

The concepts involved are peak value, r.m.s. value, average value, and frequency of an alternating current, calculated using the formulas $i_{0} = \frac{V_{0}}{R}$, $i_{rms} = \frac{i_{0}}{\sqrt{2}}$, $i_{av} = \frac{2}{\pi}i_{0}$, and $v = \frac{\omega}{2\pi}$.

In a resistive circuit $R = 10~\Omega$ and applied alternating voltage $V = 100~sin~100~\pi~t$, find the peak current, r.m.s. current, average current, and frequency respectively

A

$10~A,~5\sqrt{2}~A,~6.37~A,~50~Hz$

B

$5~A,~10~A,~3.14~A,~100~Hz$

C

$10\sqrt{2}~A,~5~A,~5~A,~50~Hz$

D

$10~A,~10~A,~6.37~A,~25~Hz$

Question 19: Palette 19/89

DPT MCQ

The concepts involved are time period, power factor, power dissipation, and time difference in a purely resistive alternating current circuit, calculated using the formulas $T = \frac{1}{v}$, $p.f. = \cos\phi = 1$, $P = \frac{1}{2}V_{0}i_{0}\cos\phi$, and $T.D. = \frac{T}{2\pi}\times\phi$.

In a resistive circuit $R = 10~\Omega$ and applied alternating voltage $V = 100~sin~100~\pi~t$, find the time period, power factor, power dissipated, and time difference respectively

A

$0.02~sec,~1,~500~W,~0$

B

$0.05~sec,~0,~250~W,~T/2$

C

$0.02~sec,~0,~0~W,~T/4$

D

$0.01~sec,~1,~500~W,~T/2$

Question 20: Palette 20/89

DPT MCQ

The concepts involved are frequency and time period in an alternating current circuit, calculated using the formulas $v = \frac{\omega}{2\pi}$ and $T = \frac{1}{v}$.

In a purely inductive circuit if $L = \frac{100}{\pi} \times 10^{-3}~H$ and applied alternating voltage is given by $V = 100~sin~100~\pi~t$, find the frequency and time period respectively

A

$50~Hz,~0.02~sec$

B

$100~Hz,~0.01~sec$

C

$25~Hz,~0.04~sec$

D

$50~Hz,~0.05~sec$

Question 21: Palette 21/89

DPT MCQ

The concept involved is the calculation of root mean square (r.m.s.) current in a purely capacitive alternating current circuit, where the ammeter reads the r.m.s. value of current given by the formula $i_{rms} = \frac{V_{rms}}{X_{C}} = V_{rms} \times \omega \times C$.

An alternating voltage $E = 200\sqrt{2}~sin(100~t)$ is connected to a 1 microfaracd capacitor through an ac ammeter. The reading of the ammeter shall be

A

10 mA

B

40 mA

C

80 mA

D

20 mA

Question 22: Palette 22/89

DPT MCQ

In an alternating current circuit containing resistance and reactance, the impedance $Z$ is related to the peak voltage $V_0$ and peak current $i_0$ by $Z = \frac{V_0}{i_0}$. The relationship between impedance, resistance, and reactance is given by $Z^{2} = R^{2} + X_{L}^{2}$.

The current and voltage in an ac circuit are respectively given by i = sin 314 t and e = 200 sin (314 t + pi / 3) from the image given below. If the resistance is $100$ ohm, then the reactance of the circuit is

A

$100 / \sqrt{3}$ ohm

B

$200$ ohm

C

$100 \sqrt{3}$ ohm

D

$200 \sqrt{3}$ ohm

Question 23: Palette 23/89

DPT MCQ

For an alternating current circuit containing a resistor (bulb) and an inductor in series, the total applied voltage is related to the voltage across the resistor and inductor by the equation $V = \sqrt{V_{R}^{2} + V_{L}^{2}}$. The inductive reactance is given by $X_{L} = 2\pi\nu L$, and the voltage across the inductor is $V_{L} = i X_{L}$.

A bulb of $60$ volt and $10$ watt is connected with $100$ volt of ac source with an inductance coil in series. If bulb illuminates with it's full intensity then value of inductance of coil is from the image given below

A

$1.28$ H

B

$2.15$ H

C

$3.27$ H

D

$3.89$ H

Question 24: Palette 24/89

DPT MCQ

For a solenoid acting as an RL circuit, the resistance $R$ is determined under DC conditions by $R = \frac{V}{I_{dc}}$, and the impedance $Z$ under AC conditions is given by $Z = \frac{V}{I_{ac}}$. The relationship between impedance, resistance, and inductive reactance is $Z^{2} = R^{2} + X_{L}^{2}$, where $X_{L} = 2\pi f L$.

When $100$ volt dc is applied across a solenoid, a current of $1.0$ amp flows in it. When $100$ volt ac is applied across the same coil, the current drops to $0.5$ amp. If the frequency of ac source is $50$ Hz the impedance and inductance of the solenoid are from the image gven below

A

$200$ ohms and $0.5$ henry

B

$100$ ohms and $0.86$ henry

C

$200$ ohms and $1.0$ henry

D

$100$ ohms and $0.93$ henry

Question 25: Palette 25/89

DPT MCQ

In an RLC series circuit, the current is maximum at resonance where the inductive reactance equals the capacitive reactance ($X_{L} = X_{C}$). The angular frequency at resonance is given by the formula $\omega_{0} = \frac{1}{\sqrt{LC}}$.

In an ac circuit, containing an inductance and a capacitor in series, the current is found to be maximum when the value of inductance is $0.5$ henry and a capacitance of $8 \ \mu F$. The angular frequency of the input ac voltage must be equal to from the image gven below

A

$500$ rad/sec

B

$5 \times 10^{4}$ rad/sec

C

$4000$ rad/sec

D

$5000$ rad/sec

Question 26: Palette 26/89

DPT MCQ

In an RL series circuit, the total impedance is determined by the combination of resistance $R$ and inductive reactance $X_{L}$, given by the formula $Z = \sqrt{R^{2} + X_{L}^{2}}$, where $X_{L} = 2\pi\nu L$.

A resistance of $40$ ohm and an inductance of $95.5$ millihenry are connected in series in a $50$ cycles/second ac circuit. The impedance of this combination is very nearly from the image gven below

A

$30$ ohm

B

$400$ ohm

C

$50$ ohm

D

$60$ ohm

Question 27: Palette 27/89

DPT MCQ

In an RC series circuit, the total impedance is $Z = \sqrt{R^{2} + X_{C}^{2}}$ where $X_{C} = \frac{1}{2\pi\nu C}$. The power factor is given by $\cos\phi = \frac{R}{Z}$ and the average power dissipated is $P = \frac{V_{rms}^{2} \cos\phi}{Z}$.

A $\frac{2.5}{\pi} \ \mu F$ capacitor and $3000$-ohm resistance are joined in series to an ac source of $200$ volt and $50 \text{ sec}^{-1}$ frequency. The power factor of the circuit and the power dissipated in it will respectively be from the image gven below

A

$0.6, 0.06 \text{ W}$

B

$0.06, 0.6 \text{ W}$

C

$0.6, 4.8 \text{ W}$

D

$4.8, 0.6 \text{ W}$

Question 28: Palette 28/89

DPT MCQ

For an RLC series circuit, the impedance is minimum at resonance where inductive reactance equals capacitive reactance ($X_{L} = X_{C}$). The formula for inductance at minimum impedance is $L = \frac{1}{4\pi^{2}\nu^{2}C}$.

A telephone wire of length $200 \ km$ has a capacitance of $0.014 \ \mu F$ per $km$. If it carries an ac of frequency $5 \ kHz$, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum from the image gven below

A

$0.35 \ mH$

B

$35 \ mH$

C

$3.5 \ mH$

D

Zero

Question 29: Palette 29/89

DPT MCQ

In an AC circuit containing a resistor $R$ and an inductor $L$ connected in series, the inductive reactance is given by $X_L = 2 \pi \nu L$. The total impedance $Z$ of the $RL$ series circuit is calculated using the formula $Z = \sqrt{R^2 + X_L^2}$.

A resistance of $40$ ohm and an inductance of $95.5$ millihenry are connected in series in a $50$ cycles/second ac circuit. The impedance of this combination is very nearly

A

$30$ ohm

B

$60$ ohm

C

$40$ ohm

D

$50$ ohm

Question 30: Palette 30/89

DPT MCQ

In an RLC series circuit, the current is maximum at resonance where the inductive reactance equals the capacitive reactance ($X_{L} = X_{C}$). The angular frequency at resonance is given by the formula $\omega_{0} = \frac{1}{\sqrt{LC}}$.

In an ac circuit, containing an inductance and a capacitor in series, the current is found to be maximum when the value of inductance is $0.5$ henry and a capacitance of $8 \ \mu F$. The angular frequency of the input ac voltage must be equal to from the image gven below

A

$500$ rad/sec

B

$5 \times 10^{4}$ rad/sec

C

$4000$ rad/sec

D

$5000$ rad/sec

Question 31: Palette 31/89

DPT MCQ

For the impedance of a series RLC circuit to be minimum, the circuit must be at resonance, which means that the inductive reactance equals the capacitive reactance, given by the condition $X_L = X_C$, or $2 \pi \nu L = \frac{1}{2 \pi \nu C}$.

A telephone wire of length 200 km has a capacitance of 0.014 microfarad per km. If it carries an ac of frequency 5 kHz, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum

A

3.5 mH

B

0.35 mH

C

35 mH

D

Zero

Question 32: Palette 32/89

DPT MCQ

In an RC series circuit, the total impedance is $Z = \sqrt{R^{2} + X_{C}^{2}}$ where $X_{C} = \frac{1}{2\pi\nu C}$. The power factor is given by $\cos\phi = \frac{R}{Z}$ and the average power dissipated is $P = \frac{V_{rms}^{2} \cos\phi}{Z}$.

A $\frac{2.5}{\pi} \ \mu F$ capacitor and $3000$-ohm resistance are joined in series to an ac source of $200$ volt and $50 \text{ sec}^{-1}$ frequency. The power factor of the circuit and the power dissipated in it will respectively be from the image gven below

A

$0.6, 0.06 \text{ W}$

B

$0.06, 0.6 \text{ W}$

C

$0.6, 4.8 \text{ W}$

D

$4.8, 0.6 \text{ W}$

Question 33: Palette 33/89

DPT MCQ

For an RLC series circuit, the impedance is minimum at resonance where inductive reactance equals capacitive reactance ($X_{L} = X_{C}$). The formula for inductance at minimum impedance is $L = \frac{1}{4\pi^{2}\nu^{2}C}$.

A telephone wire of length $200 \ km$ has a capacitance of $0.014 \ \mu F$ per $km$. If it carries an ac of frequency $5 \ kHz$, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum from the image gven below

A

$0.35 \ mH$

B

$35 \ mH$

C

$3.5 \ mH$

D

Zero

Question 34: Palette 34/89

DPT MCQ

The instantaneous value of an alternating current expressed as a sinusoidal function of time is given by $I = I_{0} \sin(\omega t)$, where $\omega = \frac{2\pi}{T}$. Substituting $t = \frac{T}{8}$ gives the value at that specific instant.

Find out instantaneous current value for $I = I_{0} \sin(\omega t)$ at $t = \frac{T}{8}$

A

$\frac{I_{0}}{2}$

B

$\frac{I_{0}}{\sqrt{2}}$

C

$\sqrt{2} I_{0}$

D

$I_{0}$

Question 35: Palette 35/89

DPT MCQ

The instantaneous value of an alternating voltage is given by the general sinusoidal expression $V = V_{0} \sin(\omega t)$, where $V_{0}$ is the peak voltage and $\omega$ is the angular frequency. Substituting the given time $t$ into the equation yields the voltage at that specific instant.

Find out instantaneous voltage for $V = 200 \sin (400 \pi t)$ at $t = \frac{1}{800}$ sec

A

$100$ V

B

$200$ V

C

$200\sqrt{2}$ V

D

$100\sqrt{2}$ V

Question 36: Palette 36/89

DPT MCQ

The instantaneous value of an alternating current with a phase lag is given by the general formula $I = I_{0} \sin(\omega t - \phi)$, where the angular frequency $\omega$ is related to the time period $T$ by $\omega = \frac{2\pi}{T}$. Substituting the given time $t = \frac{T}{2}$ determines the value at that specific instant.

Find out instantaneous value for $I = I_{0} \sin\left(\omega t - \frac{\pi}{6}\right)$ at $t = \frac{T}{2}$

A

$\frac{I_{0}}{2}$

B

$I_{0}$

C

$\frac{I_{0}}{\sqrt{2}}$

D

$0$

Question 37: Palette 37/89

DPT MCQ

For a sinusoidal alternating current given by $I = I_{0} \sin(\omega t)$, the phase angle corresponding to a current of $\frac{I_{0}}{\sqrt{2}}$ starting from zero is $\frac{\pi}{4}$. The time taken is related to angular frequency $\omega = 2\pi f$ and phase difference by $\Delta t = \frac{T}{2\pi} \times \phi = \frac{1}{8f}$.

Find the time taken by current to reach $\frac{I_{0}}{\sqrt{2}}$ if the frequency of current is $50$ Hz

A

$1.5$ ms

B

$2.5$ ms

C

$5.0$ ms

D

$10.0$ ms

Question 38: Palette 38/89

DPT MCQ

The average value or mean value of an alternating voltage over a positive half cycle is given by the formula $V_{avg} = \frac{2 V_{0}}{\pi} = \frac{2 \sqrt{2} V_{rms}}{\pi}$, where $V_{rms}$ is the root mean square value and $V_{0}$ is the peak value of the voltage.

If an alternating current main supply is given to be $220$ V. What would be the average e.m.f. during a positive half cycle?

A

$155$ V

B

$198$ V

C

$220$ V

D

$311$ V

Question 39: Palette 39/89

DPT MCQ

The root mean square (RMS) value of a complex alternating current is defined as the square root of the mean of the squared instantaneous current values over a full cycle, given by $I_{rms} = \sqrt{\langle I^{2} \rangle}$, where the mean square value is calculated by taking the average of the squared function components over time.

Find RMS value of $I = 3 + 4 \sin\left(\omega t + \frac{\pi}{3}\right)$

A

$\sqrt{15}$

B

$\sqrt{17}$

C

$\sqrt{19}$

D

$\sqrt{21}$

Question 40: Palette 40/89

DPT MCQ

The root mean square (RMS) value of a combined or sinusoidal alternating current is determined by taking the square root of the average of the squared function over a complete cycle, expressed as $I_{rms} = \sqrt{\langle I^{2} \rangle}$, using the property that the time average of $\sin^{2}(\omega t)$ or $\cos^{2}(\omega t)$ over a full cycle is $\frac{1}{2}$ and that of simple sine or cosine terms is zero.

Find out RMS value of following currents (i) $I = a + b \sin(\omega t)$ (ii) $I = a \sin(\omega t) + b \cos(\omega t)$

A

(i) $\sqrt{a^{2} + \frac{b^{2}}{2}}$, (ii) $\sqrt{\frac{a^{2}}{2} + \frac{b^{2}}{2}}$

B

(i) $\sqrt{a^{2} + b^{2}}$, (ii) $\sqrt{a^{2} + \frac{b^{2}}{2}}$

C

(i) $\sqrt{\frac{a^{2}}{2} + b^{2}}$, (ii) $\sqrt{a^{2} + b^{2}}$

D

(i) $a + \frac{b}{\sqrt{2}}$, (ii) $\frac{a + b}{\sqrt{2}}$

Question 41: Palette 41/89

DPT MCQ

The general equation of an alternating current is given by $i = i_{0} \sin \omega t$, where $i_{0}$ is the peak value and $\omega$ is the angular frequency ($\omega = 2\pi v$). The root mean square (r.m.s.) value of current is related to the peak value by the formula $i_{rms} = \frac{i_{0}}{\sqrt{2}}$.

The equation of an alternating current is $i = 50\sqrt{2} \sin 400\pi t$ ampere then the frequency and the root mean square of the current are respectively

A

400 $\pi$ Hz, $50\sqrt{2}$ amp

B

50 Hz, 200 amp

C

200 Hz, 50 amp

D

200 Hz, $50\sqrt{2}$ amp

Question 42: Palette 42/89

DPT MCQ

For an alternating current circuit containing a resistor (bulb) and an inductor in series, the total applied voltage is related to the voltage across the resistor and inductor by the equation $V = \sqrt{V_{R}^{2} + V_{L}^{2}}$. The inductive reactance is given by $X_{L} = 2\pi\nu L$, and the voltage across the inductor is $V_{L} = i X_{L}$.

A bulb of $60$ volt and $10$ watt is connected with $100$ volt of ac source with an inductance coil in series. If bulb illuminates with it's full intensity then value of inductance of coil is from the image given below

A

$1.28$ H

B

$2.15$ H

C

$3.27$ H

D

$3.89$ H

Question 43: Palette 43/89

DPT MCQ

When a DC source is applied, the inductive reactance is zero, so the resistance is given by $R = \frac{V}{I_{dc}}$. When an AC source is applied, the impedance of the circuit is given by $Z = \frac{V}{I_{ac}}$, and the relationship between impedance, resistance, and inductive reactance is $Z = \sqrt{R^2 + X_L^2}$, where $X_L = 2 \pi \nu L$.

When 100 volt dc is applied across a solenoid, a current of 1.0 amp flows in it. When 100 volt ac is applied across the same coil, the current drops to 0.5 amp. If the frequency of ac source is 50 Hz the impedance and inductance of the solenoid are

A

200 ohms and 0.5 henry

B

100 ohms and 0.86 henry

C

200 ohms and 1.0 henry

D

100 ohms and 0.93 henry

Question 44: Palette 44/89

DPT MCQ

When the current in a series LC circuit is maximum, the circuit is in a state of electrical resonance. The condition for resonance is that the inductive reactance equals the capacitive reactance, leading to the resonance angular frequency formula given by $\omega_0 = \frac{1}{\sqrt{LC}}$.

In an ac circuit, containing an inductance and a capacitor in series, the current is found to be maximum when the value of inductance is 0.5 henry and a capacitance of 8 micro farad. The angular frequency of the input ac voltage must be equal to

A

500 rad/sec

B

$5 \times 10^4$ rad/sec

C

4000 rad/sec

D

5000 rad/sec

Question 45: Palette 45/89

DPT MCQ

In an AC circuit containing a resistor $R$ and an inductor $L$ connected in series, the inductive reactance is given by $X_L = 2 \pi \nu L$. The total impedance $Z$ of the $RL$ series circuit is calculated using the formula $Z = \sqrt{R^2 + X_L^2}$.

A resistance of $40$ ohm and an inductance of $95.5$ millihenry are connected in series in a $50$ cycles/second ac circuit. The impedance of this combination is very nearly

A

$30$ ohm

B

$60$ ohm

C

$40$ ohm

D

$50$ ohm

Question 46: Palette 46/89

DPT MCQ

In an RLC series circuit, the current is maximum at resonance where the inductive reactance equals the capacitive reactance ($X_{L} = X_{C}$). The angular frequency at resonance is given by the formula $\omega_{0} = \frac{1}{\sqrt{LC}}$.

In an ac circuit, containing an inductance and a capacitor in series, the current is found to be maximum when the value of inductance is $0.5$ henry and a capacitance of $8 \ \mu F$. The angular frequency of the input ac voltage must be equal to from the image gven below

A

$500$ rad/sec

B

$5 \times 10^{4}$ rad/sec

C

$4000$ rad/sec

D

$5000$ rad/sec

Question 47: Palette 47/89

DPT MCQ

In an AC circuit containing a resistor $R$ and a capacitor $C$ in series, the total impedance is given by $Z = \sqrt{R^2 + X_C^2}$, where the capacitive reactance is $X_C = \frac{1}{2 \pi \nu C}$. The power factor of the circuit is given by $\cos\phi = \frac{R}{Z}$ and the power dissipated is $P = \frac{V_{rms}^2 \cos\phi}{Z}$.

$\frac{2.5}{\pi} \mu F$ capacitor and $3000$-ohm resistance are joined in series to an ac source of $200$ volt and $50 \text{ sec}^{-1}$ frequency. The power factor of the circuit and the power dissipated in it will respectively

A

$0.6, 0.06 W$

B

$0.06, 0.6 W$

C

$0.6, 4.8 W$

D

$4.8, 0.6 W$

Question 48: Palette 48/89

DPT MCQ

For the impedance of a series RLC circuit to be minimum, the circuit must be at resonance, which means that the inductive reactance equals the capacitive reactance, given by the condition $X_L = X_C$, or $2 \pi \nu L = \frac{1}{2 \pi \nu C}$.

A telephone wire of length 200 km has a capacitance of 0.014 microfarad per km. If it carries an ac of frequency 5 kHz, what should be the value of an inductor required to be connected in series so that the impedance of the circuit is minimum

A

3.5 mH

B

0.35 mH

C

35 mH

D

Zero

Question 49: Palette 49/89

DPT MCQ

In an RC series circuit, the total impedance is $Z = \sqrt{R^{2} + X_{C}^{2}}$ where $X_{C} = \frac{1}{2\pi\nu C}$. The power factor is given by $\cos\phi = \frac{R}{Z}$ and the average power dissipated is $P = \frac{V_{rms}^{2} \cos\phi}{Z}$.

A $\frac{2.5}{\pi} \ \mu F$ capacitor and $3000$-ohm resistance are joined in series to an ac source of $200$ volt and $50 \text{ sec}^{-1}$ frequency. The power factor of the circuit and the power dissipated in it will respectively be from the image gven below

A

$0.6, 0.06 \text{ W}$

B

$0.06, 0.6 \text{ W}$

C

$0.6, 4.8 \text{ W}$

D

$4.8, 0.6 \text{ W}$

Question 50: Palette 50/89

DPT MCQ

The time period of an alternating quantity is given by $T = \frac{2\pi}{\omega}$, where $\omega$ is the angular frequency. The phase difference between voltage and current given in the form $V = V_{0} \sin(\omega t + \phi_{1})$ and $i = i_{0} \sin(\omega t + \phi_{2})$ is calculated as $\Delta\phi = \vert{}\phi_{1} - \phi_{2}\vert{}$.

The current and voltage in a circuit are given by $i = 3.5 \sin (628 t + 30^{\circ})$ A and $V = 28 \sin (628 t - 30^{\circ})$ volt. Find (a) time period of current (b) phase difference between voltage and current.

A

$T = 0.02$ s, $\Delta\phi = 30^{\circ}$

B

$T = 0.01$ s, $\Delta\phi = 60^{\circ}$

C

$T = 0.01$ s, $\Delta\phi = 30^{\circ}$

D

$T = 0.02$ s, $\Delta\phi = 60^{\circ}$

Question 51: Palette 51/89

DPT MCQ

The resultant of two sinusoidal alternating voltages of the same frequency with a phase difference is determined using vector addition (phasor addition). For two vectors of equal magnitude $V_{0}$ separated by an angle $\theta$, the resultant amplitude is given by $R = \sqrt{V_{1}^{2} + V_{2}^{2} + 2 V_{1} V_{2} \cos\theta}$, and the phase is shifted by half of the phase angle when magnitudes are equal.

From the image given below, if $V_{1} = V_{0} \sin(\omega t)$ and $V_{2} = V_{0} \sin\left(\omega t + \frac{\pi}{3}\right)$, then find the resultant of the two voltages.

A

$V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

B

$\sqrt{2} V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

C

$\sqrt{3} V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

D

$2 V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

Question 52: Palette 52/89

DPT MCQ

The resultant of two sinusoidal alternating voltages of the same frequency with a phase difference is determined using vector addition (phasor addition). For two vectors of equal magnitude $V_{0}$ separated by an angle $\theta$, the resultant amplitude is given by $R = \sqrt{V_{1}^{2} + V_{2}^{2} + 2 V_{1} V_{2} \cos\theta}$, and the phase is shifted by half of the phase angle when magnitudes are equal.

From the image given below, if $V_{1} = V_{0} \sin(\omega t)$ and $V_{2} = V_{0} \sin\left(\omega t + \frac{\pi}{3}\right)$, then find the resultant of the two voltages.

A

$V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

B

$\sqrt{2} V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

C

$\sqrt{3} V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

D

$2 V_{0} \sin\left(\omega t + \frac{\pi}{6}\right)$

Question 53: Palette 53/89

DPT MCQ

An ac ammeter measures the root mean square (rms) value of alternating current. The peak current $I_{0}$ is given by Ohm's law $I_{0} = \frac{V_{0}}{R}$, and the rms current is related to the peak current by $I_{rms} = \frac{I_{0}}{\sqrt{2}}$.

A voltage $V = 60 \sin(\pi t)$ volt is applied across a $20~\Omega$ resistor. What will an ac ammeter in series with the resistor read?

A

$1.5~A$

B

$2.12~A$

C

$3.0~A$

D

$4.24~A$

Question 54: Palette 54/89

DPT MCQ

In a pure inductive circuit connected to an alternating source $E = E_{0} \sin(\omega t)$, the current lags behind the emf by a phase angle of $\pi/2$. The peak current is given by $I_{0} = \frac{E_{0}}{X_{L}}$ where the inductive reactance is $X_{L} = \omega L$, resulting in the current equation $I = I_{0} \sin(\omega t - \frac{\pi}{2})$.

An alternating current source $E = 100 \sin(1000t)$ volt is connected through an inductor of $10~\mu H$, then write down the equation of current.

A

$10000 \sin(1000t + \frac{\pi}{2})~\text{A}$

B

$1000 \sin(1000t - \frac{\pi}{2})~\text{A}$

C

$10000 \sin(1000t - \frac{\pi}{2})~\text{A}$

D

$100 \sin(1000t - \frac{\pi}{2})~\text{A}$

Question 55: Palette 55/89

DPT MCQ

An alternating current ammeter measures the root mean square (rms) value of current. The rms current in a pure inductive circuit is given by $I_{rms} = \frac{E_{rms}}{X_{L}}$, where the inductive reactance is $X_{L} = \omega L$.

An alternating voltage $E = 200 \sqrt{2} \sin(100t)$ volt is applied to a $2~H$ inductor through an a.c. ammeter. What will be the reading of the ammeter?

A

$1.41~A$

B

$1.0~A$

C

$0.7~A$

D

$2.0~A$

Question 56: Palette 56/89

DPT MCQ

The capacitive reactance $X_{C}$ offered by a capacitor of capacitance $C$ connected to an alternating voltage source of frequency $f$ (or angular frequency $\omega = 2\pi f$) is given by the formula $X_{C} = \frac{1}{\omega C} = \frac{1}{2\pi f C}$.

A capacitor of $50$ pF is connected to an a.c. source of frequency $1$ kHz. Calculate its reactance.

A

$\frac{10^6}{\pi} \Omega$

B

$\frac{10^7}{\pi} \Omega$

C

$\frac{10^5}{\pi} \Omega$

D

$\frac{10^8}{\pi} \Omega$

Question 57: Palette 57/89

DPT MCQ

The inductive reactance $X_{L}$ in an alternating current circuit connected to an inductor of inductance $L$ is given by $X_{L} = \omega L$, and according to Ohm's law, $X_{L} = \frac{V_{rms}}{I_{rms}}$, where $V_{rms}$ is the root-mean-square voltage and $I_{rms}$ is the root-mean-square current.

In given circuit applied voltage $V = 50\sqrt{2} \sin (100\pi t)$ volt and ammeter reading is $2$ A then calculate value of $L$ from the image given below

A

$\frac{1}{2\pi}$ H

B

$\frac{1}{4\pi}$ H

C

$\frac{1}{\pi}$ H

D

$\frac{1}{3\pi}$ H

Question 58: Palette 58/89

DPT MCQ

The impedance of a pure inductive circuit is given by the inductive reactance $Z = X_{L} = \omega L = 2\pi f L$. The root-mean-square current is given by $I_{rms} = \frac{V_{rms}}{Z}$, the peak current is $I_{0} = \sqrt{2} I_{rms}$, and in a pure inductor circuit, the alternating current lags behind the applied alternating voltage by a phase angle of $\frac{\pi}{2}$.

An inductor of $\frac{10}{\pi}$ mH is connected across an ac-source of $50$ V, $50$ Hz. Find out impedance, root mean square current, peak current, and the equation of current.

A

$Z = 1 \Omega, I_{rms} = 50$ A, $I_{0} = 50\sqrt{2}$ A, $I = 50\sqrt{2}\sin(100\pi t - \frac{\pi}{2})$

B

$Z = 2 \Omega, I_{rms} = 25$ A, $I_{0} = 25\sqrt{2}$ A, $I = 25\sqrt{2}\sin(100\pi t - \frac{\pi}{2})$

C

$Z = 1 \Omega, I_{rms} = 10$ A, $I_{0} = 10\sqrt{2}$ A, $I = 10\sqrt{2}\sin(100\pi t - \frac{\pi}{2})$

D

$Z = 0.5 \Omega, I_{rms} = 100$ A, $I_{0} = 100\sqrt{2}$ A, $I = 100\sqrt{2}\sin(100\pi t - \frac{\pi}{2})$

Question 59: Palette 59/89

DPT MCQ

The impedance $Z$ of an inductor of inductance $L$ connected to an alternating source of angular frequency $\omega$ is given by $Z = \omega L$, so that the slope of the $Z$ versus $\omega$ graph represents the inductance $L$, and the ratio of inductances for the two inductors is given by $\frac{L_{1}}{L_{2}} = \frac{\tan \theta_{1}}{\tan \theta_{2}}$.

Across two different inductors same alternating source is connected. For this $Z$ versus $\omega$ graph is given. Find out the ratio $\frac{L_{1}}{L_{2}}$ from the image given below

A

$3$

B

$\sqrt{3}$

C

$\frac{1}{\sqrt{3}}$

D

$\frac{1}{3}$

Question 60: Palette 60/89

DPT MCQ

The impedance $Z$ of an alternating current circuit containing a resistor of resistance $R$ and a capacitor of capacitive reactance $X_{C}$ connected in series is given by the formula $Z = \sqrt{R^2 + X_{C}^2}$.

Calculate the impedance of the circuit shown from the image given below

A

$30 \Omega$

B

$40 \Omega$

C

$50 \Omega$

D

$70 \Omega$

Question 61: Palette 61/89

DPT MCQ

The resistance of the coil is determined using Ohm's law with direct current as $R = \frac{V}{I}$, and the impedance under alternating current is given by $Z = \frac{V}{I}$. The inductive reactance is then calculated using the relation $X_{L} = \sqrt{Z^2 - R^2}$.

When $10$ V, dc is applied across a coil current through it is $2.5$ A, if $10$ V, $50$ Hz A.C. is applied current reduces to $2$ A. Calculate reactance of the coil.

A

$5 \Omega$

B

$4 \Omega$

C

$3 \Omega$

D

$2 \Omega$

Question 62: Palette 62/89

DPT MCQ

The impedance of a device where current is in phase with voltage is pure resistance $R = \frac{V}{I}$, and the impedance of a device where current leads voltage by $\frac{\pi}{2}$ radians is pure capacitive reactance $X_{C} = \frac{V}{I}$. When connected in series, the impedance is given by $Z = \sqrt{R^2 + X_{C}^2}$ and the current is $I_{rms} = \frac{V_{rms}}{Z}$.

When an alternating voltage of $220$ V is applied across a device X, a current of $0.5$ A flows through the circuit and is in phase with the applied voltage. When the same voltage is applied across another device Y, the same current again flows through the circuit but it leads the applied voltage by $\frac{\pi}{2}$ radians. Calculate the current flowing in the circuit when same voltage is applied across the series combination of X and Y.

A

$0.25$ A

B

$0.35$ A

C

$0.50$ A

D

$0.70$ A

Question 63: Palette 63/89

DPT MCQ

For an $RL$ series circuit connected to an alternating voltage source, the total voltage is given by $V = \sqrt{V_{R}^2 + V_{L}^2}$, where $V_{R}$ is the voltage across the resistor (bulb) and $V_{L}$ is the voltage across the inductor. The required inductance is determined using $L = \frac{X_{L}}{\omega}$ where $X_{L} = \frac{V_{L}}{I}$.

Find out the required inductance to put in series of bulb ($10$ W, $60$ V) to run it safely across an alternating supply of $100$ V, $60$ Hz from the image given below

A

$\frac{2}{\pi}$ H

B

$\frac{3}{\pi}$ H

C

$\frac{4}{\pi}$ H

D

$\frac{5}{\pi}$ H

Question 64: Palette 64/89

DPT MCQ

For an $RL$ series circuit connected to an alternating voltage source, the total voltage is given by $V = \sqrt{V_{R}^2 + V_{L}^2}$, where $V_{R}$ is the voltage across the resistor (bulb) and $V_{L}$ is the voltage across the inductor. The required inductance is determined using $L = \frac{X_{L}}{\omega}$ where $X_{L} = \frac{V_{L}}{I}$.

Find out the required inductance to put in series of bulb ($10$ W, $60$ V) to run it safely across an alternating supply of $100$ V, $60$ Hz from the image given below

A

$\frac{2}{\pi}$ H

B

$\frac{3}{\pi}$ H

C

$\frac{4}{\pi}$ H

D

$\frac{5}{\pi}$ H

Question 65: Palette 65/89

DPT MCQ

An AC ammeter measures the root mean square (rms) value of current, which is given by $I_{rms} = \frac{V_{rms}}{R}$, where $V_{rms} = \frac{V_{0}}{\sqrt{2}}$ and $V_{0}$ is the peak voltage.

A voltage $V = 60 \sin \pi t$ volt is applied across a $20 \Omega$ resistor. What will an ac ammeter in series with the resistor read?

A

$1.5 \sqrt{2} A$

B

$2.12 A$

C

$3.0 A$

D

$1.0 A$

Question 66: Palette 66/89

DPT MCQ

In a pure inductive circuit, the alternating current lags behind the applied alternating emf by a phase angle of $\frac{\pi}{2}$, and the peak current is given by $I_{0} = \frac{E_{0}}{X_{L}}$ where $X_{L} = \omega L$.

An alternating current source $E = 100 \sin(1000t)$ volt is connected through an inductor of $10 \, \mu\text{H}$ then write down the equation of current.

A

$10000 \sin(1000t - \frac{\pi}{2}) \, \text{A}$

B

$5000 \sin(1000t - \frac{\pi}{2}) \, \text{A}$

C

$1000 \sin(1000t - \frac{\pi}{2}) \, \text{A}$

D

$10000 \cos(1000t) \, \text{A}$

Question 67: Palette 67/89

DPT MCQ

An AC ammeter measures the root mean square (rms) value of current, which for a pure inductive circuit is given by $I_{rms} = \frac{E_{rms}}{X_{L}}$, where $X_{L} = \omega L$ and $E_{rms} = \frac{E_{0}}{\sqrt{2}}$.

An alternating voltage $E = 200 \sqrt{2} \sin(100t)$ volt is applied to a $2 \, \text{H}$ inductor through an ac ammeter. What will be the reading of the ammeter?

A

$2.0 \, \text{A}$

B

$0.5 \, \text{A}$

C

$1.0 \, \text{A}$

D

$1.41 \, \text{A}$

Question 68: Palette 68/89

DPT MCQ

The capacitive reactance is given by $X_{C} = \frac{1}{2 \pi f C}$, the rms current is $I_{rms} = \frac{V_{rms}}{X_{C}}$, and the peak current is $I_{0} = \sqrt{2} I_{rms}$. When the frequency is doubled, capacitive reactance becomes half, and the current in the circuit is doubled.

A $15.0 \, \mu\text{F}$ capacitor is connected to a $220 \, \text{V}, 50 \, \text{Hz}$ source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current?

A

$X_{C} = 212.2 \, \Omega$, $I_{rms} = 1.04 \, \text{A}$, $I_{0} = 1.47 \, \text{A}$; reactance becomes half and current is doubled

B

$X_{C} = 106.1 \, \Omega$, $I_{rms} = 2.08 \, \text{A}$, $I_{0} = 2.94 \, \text{A}$; reactance becomes double and current is halved

C

$X_{C} = 424.4 \, \Omega$, $I_{rms} = 0.52 \, \text{A}$, $I_{0} = 0.74 \, \text{A}$; reactance remains same and current remains same

D

$X_{C} = 212.2 \, \Omega$, $I_{rms} = 1.04 \, \text{A}$, $I_{0} = 1.47 \, \text{A}$; reactance becomes double and current is doubled

Question 69: Palette 69/89

DPT MCQ

The root mean square current in a purely capacitive circuit is given by $I_{rms} = \frac{V_{rms}}{X_{C}}$, where the capacitive reactance $X_{C}$ is calculated as $X_{C} = \frac{1}{2 \pi f C}$.

A $60 \, \mu\text{F}$ capacitor is connected to a $110 \, \text{V}, 60 \, \text{Hz}$ a.c. supply. Determine the rms value of the current in the circuit.

A

$1.25 \, \text{A}$

B

$2.49 \, \text{A}$

C

$3.10 \, \text{A}$

D

$4.20 \, \text{A}$

Question 70: Palette 70/89

DPT MCQ

The opposition to current offered by a resistor is independent of frequency ($R$), by an inductor is directly proportional to frequency ($X_{L} = 2\pi f L$), and by a capacitor is inversely proportional to frequency ($X_{C} = \frac{1}{2\pi f C}$).

The given graphs (a) and (b) represent the variation of the opposition offered by the circuit element to the flow of alternating current, with frequency of the applied emf from the image given below. Identify the circuit element corresponding to each graph.

A

(a) Resistor, (b) Inductor

B

(a) Inductor, (b) Capacitor

C

(a) Capacitor, (b) Resistor

D

(a) Resistor, (b) Capacitor

Question 71: Palette 71/89

DPT MCQ

In an R-L series circuit, the total impedance is $Z = \frac{V}{I}$, the resistance is $R = Z \cos\phi$, and the inductive reactance is $X_{L} = Z \sin\phi$, where $\phi$ is the phase angle by which the voltage leads the current.

When a series combination of inductance and resistance are connected with a $10 \, \text{V}, 50 \, \text{Hz}$ a.c. source, a current of $1 \, \text{A}$ flows in the circuit. The voltage leads the current by a phase angle of $\frac{\pi}{3}$ radian. Calculate the values of resistance and inductive reactance.

A

$R = 5 \, \Omega$, $X_{L} = 8.66 \, \Omega$

B

$R = 10 \, \Omega$, $X_{L} = 5 \, \Omega$

C

$R = 8.66 \, \Omega$, $X_{L} = 5 \, \Omega$

D

$R = 5 \, \Omega$, $X_{L} = 5 \, \Omega$

Question 72: Palette 72/89

DPT MCQ

In a series R-L circuit connected to an alternating voltage source $V = V_0 \sin(\omega t)$, the phase difference $\phi$ between the alternating voltage and current is given by $\tan\phi = \frac{X_L}{R} = \frac{\omega L}{R}$, where $X_L = \omega L$ is the inductive reactance, $R$ is the resistance, and $\omega$ is the angular frequency.

From the image given below, the current in the shown circuit is found to be $i = 4\sin(314t - \frac{\pi}{4})$ A. Find the value of inductance.

A

$0.5$ H

B

$0.78$ H

C

$1.0$ H

D

$0.25$ H

Question 73: Palette 73/89

DPT MCQ

When a coil (having resistance $R$ and inductance $L$) is connected to a DC source, the inductive reactance is zero because frequency $f = 0$, so only resistance limits the current ($R = V/I_{dc}$). When connected to an AC source, the opposition is given by impedance $Z = \sqrt{R^2 + X_L^2} = V/I_{ac}$, where inductive reactance $X_L = \omega L$.

A current of $4$ A flows in a coil when connected to a $12$ V dc source. If the same coil is connected to a $12$ V, $50$ rad/s a.c. source, a current of $2.4$ A flows in the circuit. Determine the inductance of the coil.

A

$40$ mH

B

$60$ mH

C

$80$ mH

D

$100$ mH

Question 74: Palette 74/89

DPT MCQ

For a DC source, the coil offers only ohmic resistance $R = V / I_{dc}$. For an AC source, the coil offers impedance $Z = V / I_{ac}$, where $Z = \sqrt{R^2 + X_L^2}$ and $X_L = \omega L$.

A current of 4A flows in a coil when connected to a 12V dc source. If the same coil is connected to a 12V, 50 rad/s a.c. source, a current of 2.4A flows in the circuit. Determine the inductance of the coil.

A

0.04 H

B

0.08 H

C

0.12 H

D

0.16 H

Question 75: Palette 75/89

DPT MCQ

The impedance $Z$ of a series RL circuit is given by $Z = \sqrt{R^2 + X_L^2}$, where the inductive reactance $X_L = \omega L$. The angular frequency $\omega$ is obtained from the voltage equation $E = E_0 \sin(\omega t)$.

An alternating voltage E=200 sin(300t) volt is applied across a series combination of R=10 $\Omega$ and inductance of 800mH. Calculate the impedance of the circuit.

A

200.5 $\Omega$

B

220.2 $\Omega$

C

240.2 $\Omega$

D

260.5 $\Omega$

Question 76: Palette 76/89

DPT MCQ

The impedance of the coil is $Z = \sqrt{R^2 + X_L^2}$. The RMS current is $I_{rms} = V_{rms} / Z$, and the maximum current is $I_0 = \sqrt{2} I_{rms}$. The phase angle $\phi$ is given by $\tan \phi = X_L / R$, and the time lag is $\Delta t = \phi / \omega = \phi / (2 \pi f)$.

A coil of reactance 100 $\Omega$ and resistance 100 $\Omega$ is connected to a 240 V, 50 Hz a.c. supply.
(a) What is the maximum current in the coil?
(b) What is the time lag between the voltage maximum and the current maximum?

A

(a) 2.4 A, (b) 2.5 ms

B

(a) 3.4 A, (b) 3.5 ms

C

(a) 2.4 A, (b) 5.0 ms

D

(a) 1.2 A, (b) 2.5 ms

Question 77: Palette 77/89

DPT MCQ

The phase angle $\phi$ in an RL circuit is given by $\tan \phi = X_L / R$. The inductive reactance is $X_L = 2 \pi f L$, from which the self-inductance $L$ can be calculated.

An inductance has a resistance of 100 $\Omega$. When a.c. signal of frequency 1000 Hz is applied to the coil, the applied voltage leads the current by 45°. Calculate the self inductance of the coil.

A

10.5 mH

B

15.9 mH

C

20.2 mH

D

25.4 mH

Question 78: Palette 78/89

DPT MCQ

For a series a.c. circuit, the phase angle between current and source voltage is given by $\tan\phi = \frac{X_L - X_C}{R}$.
Inductive reactance is $X_L = \omega L$ and capacitive reactance is $X_C = \frac{1}{\omega C}$.
For a purely reactive circuit, $R = 0$ and the phase difference is $\frac{\pi}{2}$.
If $|X_L - X_C| = R$, then $\phi = \frac{\pi}{4}$.

Match the following circuit components connected across an a.c. source of angular frequency $\omega = 200$ rad/s with the phase difference between current and source voltage from the image given below.
Column I:
(A) Series combination of $10 \Omega$ and $500 \mu F$
(B) Pure inductor of $5 H$
(C) Pure capacitor of $500 \mu F$
(D) Series combination of $3 \mu F$ and $4 H$
(E) Series combination of $5 H$ and $1 k\Omega$
Column II:
(p) $\frac{\pi}{2}$
(q) $\frac{\pi}{6}$
(r) $\frac{\pi}{4}$
(s) $\frac{\pi}{3}$
(t) None of the above

A

A-p, B-q, C-r, D-s, E-t

B

A-r, B-p, C-p, D-p, E-r

C

A-q, B-r, C-s, D-p, E-r

D

A-s, B-p, C-p, D-r, E-q

Question 79: Palette 79/89

DPT MCQ

In a series LCR circuit the applied voltage is the phasor sum of the component voltages: $V = \sqrt{V_R^2 + (V_L - V_C)^2}$.

For the given LCR-series circuit, find out applied source voltage of ac from the image given below (with $V_R = 40$ V, $V_L = 50$ V, $V_C = 20$ V).

A

40 V

B

50 V

C

70 V

D

110 V

Question 80: Palette 80/89

DPT MCQ

When $V_L = V_C$ the series LCR circuit is at resonance, so the entire applied voltage appears across R and $I = \frac{V}{R}$.

A capacitor, a resistor and a inductor are connected in series to an ac-source of 110 V and frequency 60 Hz. Find reading of voltmeter $V_3$ and ammeter in the given LCR-series circuit from the image given below (with $V_1 = 300$ V, $V_2 = 300$ V, $R = 220 \Omega$).

A

$V_3 = 220$ V, $I = 0.25$ A

B

$V_3 = 0$ V, $I = 1$ A

C

$V_3 = 110$ V, $I = 1$ A

D

$V_3 = 110$ V, $I = 0.5$ A

Question 81: Palette 81/89

DPT MCQ

Impedance of a series LCR circuit: $Z = \sqrt{R^2 + (X_L - X_C)^2}$.

Find out the impedance of given circuit from the image given below (with $R = 4 \Omega$, $X_L = 9 \Omega$, $X_C = 6 \Omega$).

A

3 $\Omega$

B

4 $\Omega$

C

7 $\Omega$

D

5 $\Omega$

Question 82: Palette 82/89

DPT MCQ

For a series R-C circuit $Z = \sqrt{R^2 + X_C^2}$; peak current $I_0 = \frac{V_0}{Z}$; an ac ammeter reads the rms value $I_{rms} = \frac{I_0}{\sqrt{2}}$; and $V_R = I_{rms} R$, $V_C = I_{rms} X_C$.

Find out reading of A.C. ammeter and also calculate the potential difference across resistance and capacitor from the image given below (with $E = 100 \sin \omega t$ volt, $R = 10 \Omega$, $X_C = 10 \Omega$).

A

Ammeter = 10 A, $V_R = V_C = 100$ V

B

Ammeter = 5 A, $V_R = V_C = 50$ V

C

Ammeter = 5 A, $V_R = V_C = 25$ V

D

Ammeter = 7.07 A, $V_R = V_C = 70.7$ V

Question 83: Palette 83/89

DPT MCQ

$\omega = 2\pi f$, $X_L = \omega L$, $X_C = \frac{1}{\omega C}$, $Z = \sqrt{R^2 + (X_L - X_C)^2}$ and $I_{rms} = \frac{E_{rms}}{Z}$.

In LCR circuit with an AC source $R = 300 \Omega$, $C = 20 \mu F$, $L = 1.0$ H, $E_{rms} = 50$ V and $f = \frac{50}{\pi}$ Hz. Find RMS current in the circuit.

A

0.5 A

B

1.0 A

C

0.1 A

D

0.2 A

Question 84: Palette 84/89

DPT MCQ

AC ammeters and voltmeters always read RMS values. The equivalent resistance of the main branch determines the RMS current, and the voltmeter reads the RMS potential difference across the specific component.

In given circuit find the reading of ammeter and voltmeter from the image given below (with $E = 50 \sin 100\pi t$ volt, and resistors $10 \Omega$, $20 \Omega$, $10 \Omega$).

A

Ammeter = 2.5 A, Voltmeter = 25 V

B

Ammeter = 5.0 A, Voltmeter = 50 V

C

Ammeter = 1.25 A, Voltmeter = 12.5 V

D

Ammeter = 3.5 A, Voltmeter = 35 V

Question 85: Palette 85/89

DPT MCQ

In a parallel AC circuit, the voltage across each element is the same and equal to the source voltage $V = V_0 \sin \omega t$. The current through a resistor is $i_R = \frac{V}{R}$, through an inductor is $i_L = \frac{1}{L} \int V dt$, and through a capacitor is $i_C = C \frac{dV}{dt}$.

For the circuit shown in figure, write down the instantaneous current through each element from the image given below (with $V = V_0 \sin \omega t$ applied across parallel R, L, C).

A

$i_R = \frac{V_0}{R} \sin \omega t$, $i_L = -\frac{V_0}{\omega L} \cos \omega t$, $i_C = V_0 \omega C \cos \omega t$

B

$i_R = \frac{V_0}{R} \cos \omega t$, $i_L = \frac{V_0}{\omega L} \sin \omega t$, $i_C = -V_0 \omega C \sin \omega t$

C

$i_R = \frac{V_0}{R} \sin \omega t$, $i_L = \frac{V_0}{\omega L} \cos \omega t$, $i_C = V_0 \omega C \sin \omega t$

D

$i_R = \frac{V_0}{R} \sin \omega t$, $i_L = -\frac{V_0}{\omega L} \sin \omega t$, $i_C = V_0 \omega C \cos \omega t$

Question 86: Palette 86/89

DPT MCQ

At resonance in a series LCR circuit, the inductive reactance equals the capacitive reactance ($X_L = X_C$), and the angular frequency is $\omega_r = \frac{1}{\sqrt{LC}}$. The impedance is minimum and equal to $R$, so $I_{rms} = \frac{V_{rms}}{R}$ and the amplitude of current is $I_0 = \sqrt{2} I_{rms}$.

A variable frequency 230V alternating voltage source is connected across a series combination of $L = 5.0$ H, $C = 80 \mu$F and $R = 40 \Omega$. Calculate (a) The angular frequency of the source at resonance. (b) Amplitude of current at resonance frequency.

A

(a) 50 rad/s, (b) 8.13 A

B

(a) 100 rad/s, (b) 5.75 A

C

(a) 50 rad/s, (b) 5.75 A

D

(a) 100 rad/s, (b) 8.13 A

Question 87: Palette 87/89

DPT MCQ

At resonance in a series LCR circuit, inductive reactance equals capacitive reactance: $X_L = X_C \implies \omega_0 L = \frac{1}{\omega_0 C}$.
Impedance is minimum: $Z = R$, and the circuit behaves purely resistive, so phase difference $\phi = 0$.

A coil of resistance R and inductance L is connected in series with a capacitor C and complete combination is connected to a.c. voltage. Circuit resonates when angular frequency of supply is $\omega = \omega_0$. Find out relation between $\omega_0$, L and C and phase difference between V and I at resonance.

A

$\omega_0 = \frac{1}{LC}$, $\phi = 45°$

B

$\omega_0 = \sqrt{LC}$, $\phi = 90°$

C

$\omega_0 = \frac{1}{\sqrt{LC}}$, $\phi = 0°$

D

$\omega_0 = \frac{L}{C}$, $\phi = 60°$

Question 88: Palette 88/89

DPT MCQ

At half power frequencies, current is $\frac{I_{max}}{\sqrt{2}}$, so impedance $Z = \sqrt{2} R$.
This gives $|X_L - X_C| = R$ and $\tan\phi = \frac{X_L - X_C}{R} = \pm 1$.

Find the phase difference between voltage and current in series LCR circuit at half power frequencies.

A

$\frac{\pi}{6}$

B

$\frac{\pi}{3}$

C

$\frac{\pi}{4}$

D

$\frac{\pi}{2}$

Question 89: Palette 89/89

DPT MCQ

Current is maximum at resonance: $\omega_r = \frac{1}{\sqrt{LC}}$ and $f_r = \frac{1}{2\pi\sqrt{LC}}$.
Quality factor: $Q = \frac{\omega_r L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}$.

A series LCR circuit with $L = 0.12$ H, $C = 480$ nF, $R = 23 \Omega$ is connected to a 230 V variable frequency supply. Find –
(a) Source frequency for which current is maximum.
(b) Q–factor of the given circuit.

A

(a) $6.63 \times 10^2$ Hz, (b) Q = 21.7

B

(a) $3.31 \times 10^2$ Hz, (b) Q = 10.85

C

(a) $6.63 \times 10^2$ Hz, (b) Q = 10.85

D

(a) $1.33 \times 10^3$ Hz, (b) Q = 43.4