Alternating Current

Resonance frequency

$ \begin{aligned} & f_0=\frac{1}{2 \pi \sqrt{L C}} \\ & =\frac{1}{2 \pi \sqrt{1 \times 10^{-7} \times 10^{-3}}} \\ & =15.9 \mathrm{kHz} \end{aligned} $

Alternating current,

$ i=i_{\text {peak }} \sin (\omega t) $

where $\omega=2 \pi f=2 \pi \times 60=120 \pi \mathrm{rad} / \mathrm{s}$

$ \begin{aligned} & \Rightarrow \quad i=5 \sin (120 \pi t) \\ & \Rightarrow \quad 5=5 \sin (120 \pi t) \\ & \Rightarrow \quad \sin (120 \pi t)=1=\sin \left(\frac{\pi}{2}\right) \\ & \therefore \quad 120 \pi t=\frac{\pi}{2} \\ & \Rightarrow \quad t=\frac{1}{240} \mathrm{~s} \end{aligned} $

Angular frequency of $A C$ source, $\omega=2 \times 10^3 \mathrm{rad} / \mathrm{s}$

Inductive reactance, $X_L=\omega L=\left(2 \times 10^3\right) \times\left(10 \times 10^{-3}\right)=20 \Omega$

Capacitive reactance, $X_C=\frac{1}{\omega C}=\frac{1}{\left(2 \times 10^3\right) \times\left(25 \times 10^{-6}\right)}=20 \Omega$

Impedance $Z$ for circuit $=R\left(\because X_L=X_C\right.$ and circuit is at resonance $)$

∴ current amplitude, $i_0=\frac{V_0}{Z}=\frac{V_0}{R}=\frac{220}{100}=2.2 \mathrm{~A}$

To determine the current in the circuit and the phase angle between the current and voltage, follow these steps:

Components and their Reactance:

Inductive reactance, $ X_L = 45 \, \Omega $

Capacitive reactance, $ X_C = 25 \, \Omega $

Resistance, $ R = 20 \, \Omega $

Calculate Current (I):

The current through the series circuit can be calculated using the formula:

$ I = \frac{V}{\sqrt{(X_L - X_C)^2 + R^2}} $

Substituting the given values:

$ I = \frac{220}{\sqrt{(45 - 25)^2 + 20^2}} = \frac{220}{\sqrt{20^2 + 20^2}} $

Simplifying further:

$ I = \frac{220}{\sqrt{400 + 400}} = \frac{220}{\sqrt{800}} = \frac{220}{2\sqrt{200}} = \frac{220}{2\sqrt{100 \times 2}} = \frac{220}{20\sqrt{2}} = \frac{11}{\sqrt{2}} $

Therefore, the current is approximately:

$ I = 7.779 \, \text{A} \, (\approx 7.8 \, \text{A}) $

Calculate Phase Angle ($\phi$):

The phase angle can be calculated using the tangent of the angle:

$ \tan \phi = \frac{X_L - X_C}{R} = \frac{45 - 25}{20} = \frac{20}{20} = 1 $

Hence, the phase angle $\phi$ is:

$ \phi = 45^{\circ} $

Thus, the current in the circuit is approximately 7.8 A, and the phase angle is $45^{\circ}$.

$V_L$ leads current $I$ by $\frac{\pi}{2}$

$\begin{aligned} & \therefore V_L=V_0 \sin \left(\omega t+\frac{\pi}{2}\right) \quad\left(\because I=I_0 \sin \omega t\right) \\ & V_0=I_0 X_L \\ & \Rightarrow V_L=I_0 X_L \cos (\omega t)=I_0 \omega L \cos (\omega t) \end{aligned}$

Let's analyze the problem step by step. We are given a step-up transformer with the following specifications:

Primary Voltage (V_p) = $220 \mathrm{~V}$
Secondary Voltage (V_s) = $11000 \mathrm{~V}$
Power (P) = $88$ watt

The power input to the transformer is equal to the power output, assuming there is no loss in the transformer.

Hence,

$P_{primary} = P_{secondary} = P = 88 \mathrm{~W}$

The power equations can be written as:

$P_{primary} = V_p I_p$

$P_{secondary} = V_s I_s$

Given the power in the secondary winding is $88 \mathrm{~W}$ and the secondary voltage is $11000 \mathrm{~V}$, we can find the secondary current $I_s$ using the formula:

$P_{secondary} = V_s I_s$

Rearranging to solve for $I_s$ gives:

$I_s = \frac{P}{V_s}$

Substituting the given values:

$I_s = \frac{88 \mathrm{~W}}{11000 \mathrm{~V}}$

Simplifying this expression:

$I_s = \frac{88}{11000} \mathrm{~A}$

$I_s = 0.008 \mathrm{~A}$

$I_s = 8 \mathrm{~mA}$

Therefore, the current in the secondary circuit, ignoring the power loss in the transformer, is 8 mA.

The correct option is Option A: 8 mA.

The given equation for the amplitude of the charge oscillating in a circuit is:

$Q = Q_0 e^{- \frac{R t}{2 L} }$

We need to find the time, $t$, at which charge amplitude decreases to $0.50 Q_0$. So, we set $Q = 0.50 Q_0$:

$0.50 Q_0 = Q_0 e^{- \frac{R t}{2 L} }$

Divide both sides by $Q_0$:

$0.50 = e^{- \frac{R t}{2 L} }$

Take the natural logarithm on both sides to solve for $t$:

$\ln(0.50) = - \frac{R t}{2 L}$

Recall that $\ln(0.50) = - \ln(2)$. Substituting the given value $\ln(2) = 0.693$, we get:

$-0.693 = - \frac{R t}{2 L}$

Remove the negative signs from both sides:

$0.693 = \frac{R t}{2 L}$

Now, solve for $t$:

$t = \frac{2 L \cdot 0.693}{R}$

Substituting the given values $R = 1.5 \Omega$ and $L = 12 \mathrm{~mH} = 12 \times 10^{-3} \mathrm{~H}$, we have:

$t = \frac{2 \times 12 \times 10^{-3} \cdot 0.693}{1.5}$

Calculate the numerator and denominator:

$t = \frac{16.632 \times 10^{-3}}{1.5}$

Finally, compute the value of $t$:

$t = 11.09 \times 10^{-3} \mathrm{~s} = 11.09 \mathrm{~ms}$

Therefore, the time at which the charge amplitude decreases to $0.50 Q_0$ is nearly 11.09 ms. Thus, the correct answer is:

Option B: 11.09 ms

An ideal transformer is a device that transfers electrical energy from one circuit to another through inductively coupled conductors—the transformer's coils. The primary coil (with $N_P$ turns) is connected to the input voltage ($V_P$), and the secondary coil (with $N_S$ turns) delivers the output voltage ($V_S$).

The relationship between the input (primary) voltage and output (secondary) voltage in a transformer is determined by the ratio of the number of turns in the primary coil to the number of turns in the secondary coil, expressed by the formula:

In this case, the given turns ratio is $\frac{N_P}{N_S} = \frac{1}{2}$. To find the correct expression for $\frac{V_S}{V_P}$, we need to invert the given ratio to reflect the relationship with respect to the voltages. This means:

Therefore, substituting into the voltage ratio equation:

This indicates that the secondary voltage ($V_S$) is twice the primary voltage ($V_P$). If we were to express the ratio $V_S:V_P$ based on this, it would be:

Option B $2: 1$

This is the correct answer as it represents that, in accordance with the transformer's turns ratio, the voltage in the secondary coil is twice that in the primary coil.

Capacitive Reactance

$\begin{aligned} X_C =\frac{1}{\omega C}=\frac{1}{2 \pi f C}=\frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} \\ =\frac{1000}{3.14}\end{aligned}$

$ \begin{aligned} & V_{\mathrm{rms}}=210 \mathrm{~V} \\ & i_{\mathrm{rms}}=\frac{V_{\mathrm{rms}}}{X_C}=\frac{210}{X_C} \\ & \text { Peak current }=\sqrt{2} i_{\mathrm{ms}}=\sqrt{2} \times \frac{210}{1000} \times 3.14=0.932 \\ & \qquad =0.93 \mathrm{~A} \end{aligned} $

$\tan \phi=\frac{\mathrm{X}_{\mathrm{L}}}{\mathrm{R}}$

$\begin{aligned} & \mathrm{X}_{\mathrm{L}}=\omega \mathrm{l}=100 \pi \times \frac{1}{\pi}=100 \Omega \\ & \mathrm{R}=100 \Omega \\ & \tan \phi=\frac{100}{100}=1 \\ & \phi=45^{\circ} \end{aligned}$

$\begin{aligned} & Z_1=\sqrt{X_L^2+R^2} \quad \quad X_L=O \text { (D.C. circuit) } \\\\ & Z_1=10 \Omega \\\\ & Z_2=\sqrt{X_C^2+R^2} \\\\ & X_C=\frac{1}{2 \pi \times 50 \times \frac{10^3}{\pi} \times 10^{-6}}=10 \Omega \\\\ & Z_2=\sqrt{(10)^2+(10)^2} \\\\ & =10 \sqrt{2} \\\\ & Z_1 < Z_2\end{aligned}$

Power dissipated is maximum of purely resistive circuit.

As frequency is very high

$\mathrm{X_C \approx 0}$

$\mathrm{X_L\to\alpha}$

Effective circuit will be

Effective impedance of circuit will be = 3$\Omega$

The goal is to find the frequency at which resonance occurs in a series LCR circuit. To achieve resonance, the inductive reactance ($X_L$) should equal the capacitive reactance ($X_C$).

Given are:

• Inductance, $L = 10 ~\mathrm{mH} = 10 \times 10^{-3} ~\mathrm{H}$
• Capacitance, $C = 1 ~\mu\mathrm{F} = 1 \times 10^{-6} ~\mathrm{F}$
• Resistance, $R = 100 ~\Omega$ (which does not directly impact the resonance frequency)

Resonance occurs when the angular frequency ($\omega$) meets the condition $\omega L = \frac{1}{\omega C}$, leading to the formula for calculating the resonance frequency ($f$) as $f = \frac{\omega}{2\pi} = \frac{1}{2\pi\sqrt{LC}}$.

Substituting the given values:

$f = \frac{1}{2\pi\sqrt{10 \times 10^{-3} \cdot 1 \times 10^{-6}}} = 1.59 \mathrm{kHz}$

Thus, the frequency at which resonance occurs in this LCR circuit is 1.59 kHz.

To find the current in the primary winding of an ideal step-down transformer, we first understand that an ideal transformer's power input (primary side) equals its power output (secondary side). This is because an ideal transformer is assumed to have 100% efficiency, meaning there are no losses in the transformation process.

The power of the lamp (which is connected to the secondary side of the transformer) is given as $60\,\mathrm{W}$, and it operates at $12\,\mathrm{V}$. The voltage on the primary side of the transformer is $220\,\mathrm{V}$.

Using the power formula $P = IV$, where $P$ is the power in watts, $I$ is the current in amperes, and $V$ is the voltage in volts, we equate the power on both sides of the transformer because of its ideal nature, i.e., $P_{\text{primary}} = P_{\text{secondary}}$.

Hence, $60\,\mathrm{W} = 220\,\mathrm{V} \times I_{\text{primary}}$.

Solving for $I_{\text{primary}}$ gives:

$I_{\text{primary}} = \frac{60\,\mathrm{W}}{220\,\mathrm{V}} = 0.2727... \approx 0.27\,\mathrm{A}$.

Therefore, the current in the primary winding of the transformer is approximately $0.27\,\mathrm{A}$.

When an AC source is connected to a capacitor, the current (I) depends on the capacitive reactance (X_C), where $X_{C} = \frac{1}{\omega C}$, with $\omega$ representing the angular frequency $\omega = 2\pi f$, and C denoting the capacitance.

The formula for capacitive reactance shows that it is inversely proportional to the frequency (f) of the operation.

Thus, if the frequency decreases, the capacitive reactance increases because $X_{C} = \frac{1}{\omega C}$ and $\omega$ is proportional to frequency.

As capacitive reactance increases, the displacement current (I), which flows through the capacitor, decreases according to the relationship $I = \frac{V}{X_{C}}$, with V being the voltage across the capacitor.

So, among the given choices, the correct interpretation is that the displacement current decreases with a decrease in operating frequency.

The formula for calculating the magnetic energy stored in an inductor is given by:

$E = \frac{1}{2} L I^2$

where:

• $E$ is the energy stored (in joules, J),
• $L$ is the inductance of the inductor (in henrys, H),
• $I$ is the current flowing through the inductor (in amperes, A).

Plugging the given values into the formula:

$E = \frac{1}{2} \times 4 \mu H \times (2 A)^2$

$E =\frac{1}{2} \times 4 \times 10^{-6} H \times 4 A^2$

$E = 2 \times 10^{-6} H \cdot 4$

$E = 8 \times 10^{-6} J$

Hence, the energy stored in the inductor is $8 \mu J$.

To find the net impedance, Z, of the circuit, we need to calculate the inductive reactance (XL), the capacitive reactance (XC), and use the resistance (R) in the circuit. Since R is given as 10 $ \Omega $, we calculate the reactances as follows:

Inductive reactance, $ X_L = 2\pi fL $, where $ f $ is the frequency and $ L $ is the inductance. With $ f = 50 $ Hz and $ L $ given as $ \frac{50}{\pi} $ mH, or $ \frac{50 \times 10^{-3}}{\pi} $ H, we find $ X_L = 2 \pi \times 50 \times \frac{50 \times 10^{-3}}{\pi} = 5 $ $ \Omega $.

Capacitive reactance, $ X_C = \frac{1}{2\pi fC} $, where $ C $ is the capacitance. Given $ C = \frac{10^3}{\pi} $ μF, or $ \frac{10^{-3}}{\pi} $ F, results in $ X_C = \frac{1}{2 \pi \times 50 \times \frac{10^{-3}}{\pi}} = 10 \Omega $.

With these values, the net impedance $ Z $ of the circuit can be calculated using the formula:

$ Z = \sqrt{R^2 + (X_L - X_C)^2} $

Substituting $ R = 10 \Omega $, $ X_L = 5 \Omega $, and $ X_C = 10 \Omega $, we find:

$ Z = \sqrt{10^2 + (5 - 10)^2} = \sqrt{100 + 25} = \sqrt{125} = 5\sqrt{5} \Omega $

Therefore, the correct option is $ 5\sqrt{5} \Omega $.

$V = {V_0}\sin \omega t$

$V = {q \over C}$

$q = CV\sin \omega t$

${{dq} \over {dt}} = CV{d \over {dt}}\sin \omega t$

$i = CV\omega \cos \omega t$

$i = CV\omega \sin \left( {\omega t + {\pi \over 2}} \right)$

In a.c circuit current lead by voltage by a phase ${\pi \over 2}$

Thus statement I is correct while II is incorrect.

We know, for A.C. source

${X_L} = \omega L$

$ = 2\pi f(L)$

$ = 100\pi (2 \times {10^{ - 3}})$

$ = 0.2\pi \,\Omega = 0.628\,\Omega $

For D.C. source

The inductor behaves as a closed circuit offering no resistance at all (at steady state) as $\omega = 0$ (For D.C.)

$\therefore$ ${X_L} = 0\,\Omega $

${P_{avg}} = {V_{rms}} \times {I_{rms}} \times \cos \phi $

Here, $\phi = 0^\circ $ as current and voltage are in same phase in a resistor.

$ \Rightarrow 100 = 200 \times {I_{rms}} \Rightarrow {I_{rms}} = {1 \over 2} \Rightarrow {{{I_0}} \over {\sqrt 2 }} = {1 \over 2} \Rightarrow {I_0} = {1 \over {\sqrt 2 }} = 0.707\,A$

We know,

RMS value of A.C. ${E_{rms}} = {{{E_0}} \over {\sqrt 2 }}$

${E_0} = \sqrt 2 {E_{rms}}$

$\omega L = {1 \over {\omega C}}$

$\omega = {1 \over {\sqrt {LC} }} = {1 \over {\sqrt {10 \times 10 \times {{10}^{ - 6}}} }}$

$\omega = 100$

$\omega = 2\pi f \Rightarrow f = {\omega \over {2\pi }}$

${v_0} = {f_0} = {{100} \over {2\pi }} = {{50} \over \pi }$ Hz, $\omega = 100$

$v = f = {{100} \over {2\pi }} = {{50} \over \pi }$

$\text { Given, turn ratio of a transformer, } \frac{N_1}{N_2}=\frac{50}{1}$

$\Rightarrow \quad N_1=50 N_2$

$\text { Since, } \quad \frac{N_1}{N_2}=\frac{V_1}{V_2}$

$\begin{array}{rlrl} & 50=\frac{120}{V_2} \quad\left[V_1=120 \mathrm{~V} \text { (given) }\right] \\ \Rightarrow & V_2=\frac{12}{5} \mathrm{~V} \end{array}$

Output power at secondary coil,

$\begin{aligned} & P_s=\frac{V_2^2}{R_2} \quad \quad \quad\left[R_2=1 \Omega \text { (given) }\right] \\ & =\frac{\left(\frac{12}{5}\right)^2}{1}=\frac{144}{25}=5.76 \mathrm{~W} \end{aligned}$

Given the values :

$ L = 25 \, \text{mH} = 25 \times 10^{-3} \, \text{H} $

$ C = 10 \, \mu\text{F} = 10^{-5} \, \text{F} $

The time period $ T $ for an LC oscillation can be calculated using the formula :

$ T = 2\pi \sqrt{LC} $

Substituting the given values :

$ T = 2\pi \sqrt{25 \times 10^{-3} \times 10^{-5}} = \pi \times 10^{-3} \, \text{s} = \pi \, \text{ms} $

For the current in the circuit to be at its maximum, it occurs at :

$ t = \frac{T}{4} = \frac{\pi}{4} \, \text{ms} $

Therefore, the time after which the current in the circuit will be maximum is $\frac{\pi}{4} \, \text{ms}$.

Given, in $L C R$ series circuit,

Source voltage (resultant voltage) $V=120 \mathrm{~V}$

Voltage across inductor, $V_L=50 \mathrm{~V}$

Voltage across resistor, $V_R=40 \mathrm{~V}$

In $L$-$C$-$R$ series circuit,

$\begin{aligned} V & =\sqrt{V_R^2+\left(V_L-V_C\right)^2} \\ \text{or} \quad 120 & =\sqrt{40^2+\left(50-V_C\right)^2} \\ 120^2 & =40^2+\left(50-V_C\right)^2 \end{aligned}$

$\begin{aligned} & \Rightarrow \quad\left(50-V_C\right)^2=12800 \\ & \Rightarrow \quad 50-V_C=80 \sqrt{2} \Rightarrow V_C=10(5-8 \sqrt{2}) \end{aligned}$

We know that, for series R-C circuit

$V^2=V_C^2+V_R^2$

$\begin{aligned} 100 & =64+V_R^2 \\\\ \Rightarrow & V_R^2=36 \Rightarrow V_R=\sqrt{36}=6 \mathrm{~V} \end{aligned}$

$\begin{aligned} & \text { Also, } \tan \phi=\frac{V_C}{V_R} \Rightarrow \tan \phi=\frac{8}{6} \\\\ & \tan \phi=\frac{4}{3} \\\\ & \therefore \phi=\tan ^{-1}\left(\frac{4}{3}\right) \end{aligned}$

Rise of current in $L-R$ circuit is given by

$I=I_0\left(1-e^{-t / \tau}\right)$

where, $\quad I_0=\frac{E}{R}=\frac{5}{5}=1 \mathrm{~A}$

Now, $\tau=\frac{L}{R}=\frac{10}{5}=2 \mathrm{~s}$

After $2 \mathrm{~s}$, i.e. at $t=2 \mathrm{~s}$

Rise of current, $I=\left(1-e^{-1}\right) \mathrm{A}$

$\therefore \tan \phi=\frac{\omega L-\frac{1}{\omega C}}{R}$

$\phi$ being the angle by which the current leads the voltage.

Given, $\phi=45^{\circ}$

$\begin{aligned} \therefore \quad & \tan 45^{\circ} =\frac{\omega L-\frac{1}{\omega C}}{R} \Rightarrow 1=\frac{\omega L-\frac{1}{\omega C}}{R} \\ & R =\omega L-\frac{1}{\omega C} \Rightarrow \omega C=\frac{1}{\omega L-R} \\ \Rightarrow \quad & C =\frac{1}{\omega(\omega L-R)}=\frac{1}{2 \pi f(2 \pi f L-R)} \end{aligned}$

For circuit A

Impedence, $Z_A=\sqrt{R^2+\frac{1}{\omega^2 c^2}}$

Current in circuit $I_R^A=\frac{V}{\sqrt{R^2+\frac{1}{\omega^2 c^2}}}$ ..... (i)

Potential difference across $C$,

$V_C^A=I_R^A \times \frac{1}{\omega c}=\frac{V}{\sqrt{(R \omega c)^2+1}}$

For circuit B

Impedence, $Z_B=\sqrt{R^2+\frac{1}{(4 \omega)^2 c^2}}$

Current in circuit, $I_R^B=\frac{V}{\sqrt{R^2+\frac{1}{(4 \omega c)^2}}}$

Potential difference across $C$,

$V_C^B=I_R^B \times \frac{1}{4 \omega c}=\frac{V}{\sqrt{(4 R \omega c)^2+1}}$ ..... (ii)

Hence from above Eqs. (i) and (ii), we conclude that

$I_R^B> I_R^A \text { and } V_C^A> V_C^B .$

The average value of alternating emf $E=E_0 \sin \omega t$ over a positive half cycle is

$E_{\text {av }} \text { half cycle }=\frac{2 E_0}{\pi}=0.6369$

Percentage of $E_{\mathrm{av}}=0.6369 \times 100=63.69 \%$

Also rms value of emf is given as

$E_{\mathrm{rms}}=\frac{E_0}{\sqrt{2}}=0.7072$

Percentage

$\begin{aligned} E_{\mathrm{rms}} & =0.7072 \times 100 \\ & =70.72 \% \end{aligned}$