Alternating Current

The root mean square (r.m.s) value of a time-varying current $i(t)$ over a period $T$ is defined by the formula :

$ \mathrm{i}_{\mathrm{rms}}=\sqrt{\frac{1}{\mathrm{~T}} \int\limits_0^{\mathrm{T}} \mathrm{i}^2 \mathrm{dt}} $

We are given $\mathrm{i}=\mathrm{i}_{\mathrm{o}}\left(\frac{\mathrm{t}}{\mathrm{T}}\right)$.

So, using the rms formula,

$ \mathrm{i}_{\mathrm{rms}}^2=\frac{1}{\mathrm{~T}} \int\limits_0^{\mathrm{T}}\left(\mathrm{i}_{\mathrm{o}}^2 \frac{\mathrm{t}^2}{\mathrm{~T}^2}\right) \mathrm{dt} $

Since $i_o$ and $T$ are constants, we can move them outside the integral :

$\mathrm{i}_{\mathrm{rms}}^2=\frac{\mathrm{i}_{\mathrm{o}}^2}{\mathrm{~T}^3} \int_0^{\mathrm{T}} \mathrm{t}^2 \mathrm{dt}=\frac{\mathrm{i}_{\mathrm{o}}^2}{\mathrm{~T}^3}\left[\frac{\mathrm{t}^3}{3}\right]_0^{\mathrm{T}}=\frac{\mathrm{i}_{\mathrm{o}}^2}{\mathrm{~T}^3} \times \frac{\mathrm{T}^3}{3}$

$\Rightarrow \mathrm{i}_{\mathrm{rms}}=\sqrt{\frac{\mathrm{i}_{\mathrm{o}}^2}{3}}$

$\Rightarrow $ $ \mathrm{i}_{\mathrm{rms}}=\frac{\mathrm{i}_{\mathrm{o}}}{\sqrt{3}} $

Hence, the correct option is (B).

The power factor ( $\cos \phi$ ) of an AC circuit is defined as the ratio of the true resistance (R) to the total impedance (Z) of the circuit.

Power Factor $=\cos \phi=\frac{\mathrm{R}}{\mathrm{z}}$

In a series $L C R$ circuit, the inductive reactance $\left(X_L\right)$ and capacitive reactance $\left(X_C\right)$ oppose each other. The net reactance is the difference between them.

$ \mathrm{X}=\left|\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right| $

Substituting the given values :

$ \mathrm{X}=|70-150| \Omega=80 \Omega $

The total impedance of the circuit is,

$ Z=\sqrt{R^2+X^2} $

$ \mathrm{Z}=\sqrt{(60)^2+(80)^2}=\sqrt{10000}=10 \Omega $

So, the power factor is $\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{60}{100}=\frac{6}{10}$

$ \Rightarrow \frac{\mathrm{a}}{10}=\frac{6}{10} $

$ \Rightarrow a=6 $

Therefore, the value of a is 6 . Hence, the correct option is (C).

When a charged capacitor is connected to an inductor, the energy oscillates between the electric field of the capacitor and the magnetic field of the inductor.

The charge q on the capacitor at any time t is given by :

$ \mathrm{q}(\mathrm{t})=\mathrm{Q}_0 \cos (\omega \mathrm{t}) $

Where :

• $\mathrm{Q}_0$ is the initial maximum charge.

• $\omega$ is the angular frequency, defined as $\omega=\frac{1}{\sqrt{\mathrm{LC}}}$.

Initially all energy is in the capacitor.

$ \mathrm{U}_0=\frac{\mathrm{Q}_0^2}{2 \mathrm{C}} $

Energy in capacitor at time t is :

$ U_C(t)=\frac{q(t)^2}{2 C}=\frac{\left(Q_0 \cos (\omega t)\right)^2}{2 C}=U_0 \cos ^2(\omega t) $

By conservation of energy the energy stored in inductor as magnetic field,

$\mathrm{U}_{\mathrm{L}}=\mathrm{U}_0-\mathrm{U}_{\mathrm{C}} $

$\Rightarrow $ $ \mathrm{U}_{\mathrm{L}}(\mathrm{t})=\mathrm{U}_0\left(1-\cos ^2(\omega \mathrm{t})\right)=\mathrm{U}_0 \sin ^2(\omega \mathrm{t})$

It is that $25 \%$ of the initial energy is transferred to the inductor :

$ \mathrm{U}_{\mathrm{L}}=\frac{1}{4} \mathrm{U}_0 $

$\Rightarrow $ $U_0 \sin ^2(\omega t)=\frac{1}{4} U_0$

$\Rightarrow $ $\sin ^2(\omega \mathrm{t})=\frac{1}{4} $

$\Rightarrow $$ \sin (\omega \mathrm{t})=\frac{1}{2} $

$\Rightarrow $$ \omega \mathrm{t}=\frac{\pi}{6} $

$\Rightarrow $$ \frac{\mathrm{t}}{\sqrt{\mathrm{LC}}}=\frac{\pi}{6} $

$\Rightarrow $ $ \mathrm{t}=\frac{\pi \sqrt{\mathrm{LC}}}{6}$

In an AC circuit, the power factor $(\cos \phi)$ is defined as the ratio of the resistance $(\mathrm{R})$ to the total impedance $(\mathrm{Z})$ of the circuit :

$ \text { Power Factor }=\cos \phi=\frac{\mathrm{R}}{\mathrm{Z}} $

For a series LCR circuit, the impedance is

$ \mathrm{Z}=\sqrt{\mathrm{R}^2+\left(\mathrm{X}_{\mathrm{C}}-\mathrm{X}_{\mathrm{L}}\right)^2} $

The value of resistance is $\mathrm{R}=80 \Omega$, the inductive reactance is $\mathrm{X}_{\mathrm{L}}=20.0 \Omega$, and the capacitive reactance is $\mathrm{X}_{\mathrm{C}}=80.0 \Omega$.

$ \mathrm{X}_{\text {net }}=\left|\mathrm{X}_{\mathrm{L}}-\mathrm{X}_{\mathrm{C}}\right|=|20-80|=60 \Omega $

So, the impedance is,

$ \mathrm{Z}=\sqrt{80^2+60^2}=\sqrt{6400+3600}=\sqrt{10000}=100 \Omega $

Thus, the power factor is,

$ \cos \phi=\frac{\mathrm{R}}{\mathrm{Z}}=\frac{80}{100}=0.8 $

Therefore, the correct option is (A).

For a series $RC$ circuit, the current is

$ I=\frac{V}{Z} $

where the impedance is

$ Z=\sqrt{R^2+X_C^2} $

and capacitive reactance is

$ X_C=\frac{1}{\omega C} $

At angular frequency $\omega$, let the reactance be

$ X=\frac{1}{\omega C} $

So current is

$ I=\frac{V}{\sqrt{R^2+X^2}} $

Now frequency is changed to $\omega/4$.

Then new capacitive reactance becomes

$ X_C'=\frac{1}{(\omega/4)C}=\frac{4}{\omega C}=4X $

So new current is

$ I'=\frac{V}{\sqrt{R^2+(4X)^2}}=\frac{V}{\sqrt{R^2+16X^2}} $

Given that

$ I'=\frac{I}{3} $

Substitute expressions for $I'$ and $I$:

$ \frac{V}{\sqrt{R^2+16X^2}}=\frac{1}{3}\cdot \frac{V}{\sqrt{R^2+X^2}} $

Cancel $V$:

$ \frac{1}{\sqrt{R^2+16X^2}}=\frac{1}{3\sqrt{R^2+X^2}} $

So,

$ 3\sqrt{R^2+X^2}=\sqrt{R^2+16X^2} $

Squaring both sides:

$ 9(R^2+X^2)=R^2+16X^2 $

$ 9R^2+9X^2=R^2+16X^2 $

$ 8R^2=7X^2 $

Therefore,

$ \frac{R^2}{X^2}=\frac{7}{8} $

Hence,

$ \frac{R}{X}=\sqrt{\frac{7}{8}} $

So the required ratio of resistance to reactance at frequency $\omega$ is

$ \boxed{\sqrt{\frac{7}{8}}} $

Therefore, the correct option is:

$ \boxed{\text{Option C}} $

To find the root mean square (RMS) value of the given alternating current, follow these steps:

The current is represented as:

$ i = 5\sqrt{2} + 10 \cos \left(650\pi t + \frac{\pi}{6}\right) \, \text{Amp} $

Here, the time-independent DC component is $5\sqrt{2}$ and the AC component is $10 \cos \left(650\pi t + \frac{\pi}{6}\right)$.

Calculate the square of the current, $i^2$:

$ i^2 = \left(5\sqrt{2}\right)^2 + \left(10 \cos \left(650\pi t + \frac{\pi}{6}\right)\right)^2 + 2 \times 5\sqrt{2} \times 10 \cos \left(650\pi t + \frac{\pi}{6}\right) $

Simplifying, we have:

$ i^2 = 50 + 100 \cos^2 \left(650\pi t + \frac{\pi}{6}\right) + 100\sqrt{2} \cos \left(650\pi t + \frac{\pi}{6}\right) $

Find the average value $\langle i^2 \rangle$:

The average value of $\cos$ terms over a period is zero, simplifying our equation to:

$ \langle i^2 \rangle = 50 + \frac{100}{2} + 0 $

This simplifies to:

$ \langle i^2 \rangle = 50 + 50 = 100 $

Calculate the RMS current:

The RMS value is the square root of the mean of the squares of the current:

$ \langle i \rangle = \sqrt{100} = 10 \, \text{Amp} $

Thus, the RMS value of the current is 10 Amps.

To solve the problem, we need to determine both the RMS value of the current and the frequency from the given equation:

$i(t)=100\sqrt{2}\sin(100\pi t).$

Here's how we do it step by step:

RMS Current Calculation:

For a sinusoidal current of the form $i(t)=I_0 \sin(\omega t),$ the RMS (root-mean-square) current is given by:

$I_{\text{RMS}} = \frac{I_0}{\sqrt{2}}.$

In our case, the amplitude $I_0 = 100\sqrt{2}\, \text{A}.$ Plugging into the formula:

$I_{\text{RMS}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100\, \text{A}.$

Frequency Calculation:

The argument of the sine function is $100\pi t.$ In the general form $\sin(\omega t),$ $\omega$ is the angular frequency which relates to the frequency $f$ by the equation:

$\omega = 2\pi f.$

Given $\omega = 100\pi,$ we can solve for $f$:

$f = \frac{\omega}{2\pi} = \frac{100\pi}{2\pi} = 50\, \text{Hz}.$

With these calculations, we find that the RMS current is $100\, \text{A}$ and the frequency is $50\, \text{Hz}.$

Thus, the correct option is:

Option D: $100\,\text{A}, \, 50\,\text{Hz}.$

To find the peak current through the bulb, follow these steps:

Step 1: Calculate the bulb's rms (root mean square) current:

The formula to find the rms current is:

$I_{\text{rms}}=\frac{P}{V}$

Here, $P = 100 \text{ W}$ and $V = 220 \text{ V}$. Substitute these values into the formula:

$I_{\text{rms}} = \frac{100\text{ W}}{220\text{ V}} \approx 0.455\text{ A}$

Step 2: Convert the rms current to peak current:

Use the following formula to find the peak current:

$I_{\text{peak}} = I_{\text{rms}} \times \sqrt{2}$

Substitute the value of $I_{\text{rms}}$ into the formula:

$I_{\text{peak}} = 0.455 \times 1.414 \approx 0.64\text{ A}$

The correct choice is 0.64 A (Option B).

$ I_{\text{rms}} = \sqrt{\frac{I_A^2 + I_B^2}{2}} $

To determine the root mean square (r.m.s) value of the alternating current given by

$ I(t) = I_A \sin (\omega t) + I_B \cos (\omega t), $

we start by squaring the current:

$ I(t)^2 = I_A^2 \sin^2 (\omega t) + I_B^2 \cos^2 (\omega t) + 2 I_A I_B \sin (\omega t) \cos (\omega t). $

The r.m.s value is defined as the square root of the average of this squared current over one complete period. When averaging over a full cycle, we utilize the known averages:

$ \langle \sin^2 (\omega t) \rangle = \frac{1}{2}, \quad \langle \cos^2 (\omega t) \rangle = \frac{1}{2}, \quad \langle \sin (\omega t) \cos (\omega t) \rangle = 0. $

Thus, the time-averaged square of the current becomes:

$ \langle I(t)^2 \rangle = I_A^2 \frac{1}{2} + I_B^2 \frac{1}{2} = \frac{I_A^2 + I_B^2}{2}. $

Taking the square root gives the r.m.s current:

$ I_{\text{rms}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}. $

This corresponds to Option B.

At resonance in a series LCR circuit, the reactive components (inductive and capacitive) cancel each other out, leaving only the resistance. Therefore, the amplitude of the current is given by:

$ I_0 = \frac{E}{R} $

Now, if the resistance is doubled, the new resistance becomes $2R$. The current amplitude at resonance then becomes:

$ I = \frac{E}{2R} = \frac{1}{2}\left(\frac{E}{R}\right) = \frac{I_0}{2} $

Thus, the amplitude of current at resonance when the resistance is doubled is $ \frac{I_0}{2} $.

The correct option is A.

$ \begin{aligned} & \Rightarrow i_0=\frac{v_0}{R}=\frac{300}{30}=10 \mathrm{~A} \\ & \text { and } \phi=0 \\ & \begin{array}{l} So (P) \rightarrow 3 \end{array} \end{aligned} $

$ \begin{aligned} z & =\sqrt{R^2+(\omega L)^2} \quad\left(\text { as } x_L=\omega L\right) \\ & =\sqrt{(30)^2+\left(400 \times 100 \times 10^{-3}\right)^2} \\ & =\sqrt{900+1600}=\sqrt{2500} \\ z=50 \Omega & \end{aligned} $

$ \begin{aligned} \text { So } & i_0=\frac{V_0}{2}=\frac{300}{50}=6 \mathrm{~A} \\ \phi & =\tan ^{-1}\left(\frac{x_L}{R}\right)=\tan ^{-1}\left(\frac{\omega L}{R}\right) \\ & =\tan ^{-1}\left(\frac{400 \times 106 \times 10^{-3}}{36}\right)=\tan ^{-1}\left(\frac{4}{3}\right) \\ \Rightarrow \quad \phi & =53^{\circ}(l \mathrm{ag}) \\ & \therefore Q \rightarrow 5 \end{aligned} $

$ \begin{aligned} z & =\sqrt{R^2+\left(x_C-x_L\right)^2} \\ & =\sqrt{R^2+\left(\frac{1}{\omega c}-\omega L\right)^2} \\ & =\sqrt{(30)^2+\left(\frac{1}{400 \times 50 \times 10^{-6}}-400 \times 25 \times 10^{-3}\right)^2} \\ z & =\sqrt{900+\left(\frac{1}{20 \times 10^{-3}}-10\right)^2} \end{aligned} $

$ \begin{aligned} &\begin{array}{r} z=\sqrt{900+(50-10)^2}=\sqrt{900+1600} \\ z=\sqrt{2500}=50 \Omega \\ i_0=\frac{v_0}{2}=\frac{300 }{50}=6 \mathrm{~A} \\ \phi=\tan ^{-1}\left[\frac{x_L-x_C}{R}\right]=\tan ^{-1}\left[\frac{10-50}{30}\right] \\ =\tan ^{-1}\left|\frac{40 }{30}\right|=53^{\circ} \text { (Lead) } \end{array}\\ &(R) \rightarrow 2 \end{aligned} $

$ \begin{aligned} z & =\sqrt{(60)^2+\left(\frac{1}{400 \times 50 \times 10^{-6}}-400 \times 125 \times 10^{-3}\right)^2} \\ & =\sqrt{(60)^2+(50-50)^2} \\ z & =60 \Omega \\ & i_0=\frac{300}{60}=5 \mathrm{~A} \\ \phi & =\tan ^{-1}\left|\frac{x_L-x_C}{R}\right|=\tan ^{-1}\left|\frac{0}{60}\right|=0^{\circ} \end{aligned} $

So $(s) \rightarrow 1$

Hence, option (A) is correct.

For a series $L$-$C$-$R$ circuit:

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

Step 1: What does “purely resistive” mean?

If the circuit is purely resistive, it means only the resistor affects the circuit. So, the total impedance $Z$ should equal the resistance $R$.

Step 2: Set impedance equal to resistance:

$ Z = R $

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

Step 3: Solve for $X_L$ and $X_C$:

$ R^2 = R^2 + (X_L - X_C)^2 $

Subtract $R^2$ from both sides: $ 0 = (X_L - X_C)^2 $

This means: $ X_L = X_C $

Conclusion:

If the circuit is purely resistive, the inductive reactance $X_L$ must be equal to the capacitive reactance $X_C$.

$ \begin{aligned} &\\ &\text { } \begin{aligned} \tan \phi & =\frac{X_C}{R}=\frac{1}{\omega C R} \\ C & =\frac{1}{\omega R \tan \phi} \\ & =\frac{1}{200 \times 100 \sqrt{3} \tan 30^{\circ}} \\ & =\frac{1}{2 \times 10^4}=50 \times 10^{-6} \mathrm{~F}=50 \mu \mathrm{~F} \end{aligned} \end{aligned} $

When iron rod is inserted into interior of the inductor, then its self inductance increases.

Since, $X_L=\omega L$

i.e., $X_L \propto L$

Thus, due to increase in the value of $L$, $X_L$ also increases.

Hence, current decreases.

Thus, the glow of the bulb decreases.

• In an L-R series circuit,

• Voltage across resistor: $ V_R = 240 \, \text{V} $

• Voltage across inductor: $ V_L = 180 \, \text{V} $

We are asked to find the supply voltage $ V $.

Step 1: Relation between voltages in series L-R circuit

In a series L-R circuit, the resistor voltage $ V_R $ and the inductor voltage $ V_L $ are 90° out of phase (since $ V_R $ is in phase with current, and $ V_L $ leads the current by 90°).

Hence, the total voltage is obtained by the vector sum:

$ V = \sqrt{V_R^2 + V_L^2} $

Step 2: Substitute given values

$ V = \sqrt{(240)^2 + (180)^2} $

$ V = \sqrt{57600 + 32400} = \sqrt{90000} = 300 \, \text{V} $

✅ Final Answer:

300 V

Option A

$ \begin{aligned} &\text { } V=V_0 \sin \omega t\\ &\begin{aligned} & \Rightarrow \frac{V_0}{2}=V_0 \sin \omega t \\ & \Rightarrow \frac{1}{2}=\sin \omega t \Rightarrow \omega t=\frac{\pi}{6} \\ & \Rightarrow 2 \pi f t=\frac{\pi}{6} \Rightarrow 2 \pi \times 50 \times t=\frac{\pi}{6} \\ & \Rightarrow \quad t=\frac{1}{600} \mathrm{~s} \end{aligned} \end{aligned} $

First, find the angle by which the current lags the voltage. This angle is called $\phi$.

Find $\tan \phi$:

The formula is $\tan\phi = \frac{X_L}{R}$, where $X_L$ is the inductive reactance and $R$ is resistance.

Here, $X_L = \frac{1}{\sqrt{3}}~\Omega$ and $R = 1~\Omega$.

So, $\tan\phi = \frac{\frac{1}{\sqrt{3}}}{1} = \frac{1}{\sqrt{3}}$.

We know that $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$. So, $\phi = \frac{\pi}{6}$.

Find the time period $T$ of the AC supply:

The frequency $f = 50$ Hz, so $T = \frac{1}{f} = \frac{1}{50}$ seconds.

Find the time lag $\Delta t$:

The time lag is given by $\Delta t = \frac{\phi}{2\pi} \times T$.

Put the values in: $\Delta t = \frac{\frac{\pi}{6}}{2\pi} \times \frac{1}{50}$.

Simplify it: $\Delta t = \frac{1}{12} \times \frac{1}{50} = \frac{1}{600}$ seconds.

$ \begin{aligned} &\text { For proper matching, } R_p=R_{\text {source }}\\ &\begin{aligned} & R_{\text {source }}=\left(\frac{N_P}{N_S}\right)^2 R_{\text {load }} \\ & \Rightarrow 1000=\left(\frac{N_P}{N_S}\right)^2 \times 10 \\ & \quad \frac{N_P}{N_S}=10 \Rightarrow 10: 1 \end{aligned} \end{aligned} $

The voltages across the capacitor ($V_C$), resistor ($V_R$), and inductor ($V_L$) have the ratio 2:3:6.

This means $V_C = 2k$, $V_R = 3k$, and $V_L = 6k$ for some number $k$.

The total voltage across the circuit is given by: $V = \sqrt{V_R^2 + (V_L - V_C)^2}$ This formula is used because in an LCR circuit, resistor voltage is in phase with supply, while inductor and capacitor voltages are 180° out of phase with each other.

We are told the source voltage is 240 V: $240 = \sqrt{(3k)^2 + (6k - 2k)^2}$

Simplifying inside the square root: $(3k)^2 = 9k^2 \\ (6k - 2k)^2 = (4k)^2 = 16k^2$ So, $240 = \sqrt{9k^2 + 16k^2} = \sqrt{25k^2} = 5k$

Now solve for $k$: $5k = 240 \implies k = \frac{240}{5} = 48$

Now, calculate the voltage across the inductor: $V_L = 6k = 6 \times 48 = 288~\mathrm{V}$

The formula to find the average (real) power in an AC circuit is:

$ P = V_{\mathrm{rms}} \times I_{\mathrm{rms}} \times \cos \phi $ where $\phi$ is the phase difference between voltage and current.

The voltage is $50 \sin(50t)$, so its maximum (peak) value is $50~\mathrm{V}$.

The current is $50 \sin\left(50t + \frac{\pi}{4}\right)$, so its peak value is $50~\mathrm{mA} = 50 \times 10^{-3}~\mathrm{A}$.

The phase difference $\phi$ between voltage and current is $\frac{\pi}{4}$.

To get the RMS (root mean square) values from the peak values, divide by $\sqrt{2}$:

$V_{\mathrm{rms}} = \frac{50}{\sqrt{2}}$

$I_{\mathrm{rms}} = \frac{50 \times 10^{-3}}{\sqrt{2}}$

Now, put these values into the power formula:

$ P = \frac{50}{\sqrt{2}} \times \frac{50 \times 10^{-3}}{\sqrt{2}} \times \cos\left(\frac{\pi}{4}\right) $

$\cos\left(\frac{\pi}{4}\right)$ is $\frac{1}{\sqrt{2}}$.

So,

$ P = \frac{50}{\sqrt{2}} \times \frac{50}{\sqrt{2}} \times 10^{-3} \times \frac{1}{\sqrt{2}} $

Multiply the numbers:

$ = \frac{50 \times 50 \times 10^{-3}}{\sqrt{2} \times \sqrt{2} \times \sqrt{2}}$

$\sqrt{2} \times \sqrt{2} \times \sqrt{2} = 2\sqrt{2}$

$= \frac{2500 \times 10^{-3}}{2\sqrt{2}}$

$= \frac{2.5}{2\sqrt{2}}$

This is approximately equal to $0.884~\mathrm{W}$.

Given:

In a series LCR circuit at resonance, the power dissipated is maximum.

We need to find:

For the power dissipated to become half of its maximum value, what is the current amplitude?

Step 1: Relation between Power and Current

The average power dissipated in an AC circuit is given by:

$ P = I_{\text{rms}}^{2} R $

At resonance, the impedance is purely resistive ($Z = R$), and the current is maximum ($I = I_{\max}$).

So maximum power is:

$ P_{\max} = I_{\text{rms,max}}^{2} R = \left(\frac{I_{\text{max}}}{\sqrt{2}}\right)^2 R = \frac{I_{\text{max}}^2 R}{2} $

But we only need relative power, so we can use $P \propto I^2$.

Step 2: Condition for half power

We are told $P = \frac{1}{2} P_{\max}$

So:

$ I^2 R = \frac{1}{2} I_{\max}^2 R $

Simplify:

$ I = \frac{I_{\max}}{\sqrt{2}} $

✅ Final Answer:

Option A: $\displaystyle \frac{1}{\sqrt{2}}$ times its maximum value.

So the current amplitude becomes $1/\sqrt{2}$ times its maximum value when the power is half of the maximum.

$ \begin{aligned} &\text { Impedance, }\\ &\begin{aligned} Z & =\sqrt{R^2+\left(X_L-X_C\right)^2} \\ & =\sqrt{4^2+(9-6)^2}=\sqrt{16+9}=5 \Omega \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Power factor }\\ &\begin{aligned} \cos \phi & =\frac{R}{Z} \\ 0.6 & =\frac{R}{\sqrt{R^2+X_L^2}} \\ \Rightarrow \quad 0.36 & =\frac{R^2}{R^2+(2 \pi f L)^2} \end{aligned} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad 0.36=\frac{(450)^2}{(450)^2+\left(2 \pi \times \frac{75}{\pi} L\right)^2} \\ & \Rightarrow \quad 0.36=\frac{(450)^2}{450^2+(150 L)^2} \end{aligned}\\ &\text { Solving, we get }\\ &L=4 \mathrm{H} \end{aligned} $

For better tunning of a series $L C R$ circuit in a communication system. Quality factor ' $Q$ ' should be high.

i.e.

$ Q=\frac{1}{R} \sqrt{\frac{L}{C}} $

So, for high value of $Q$, value of $R$ and $C$ should be minimum and value $L$ should be maximum.

$ \begin{aligned} &\text { AC voltage applied to the circuit, }\\ &\begin{aligned} V & =\sqrt{V_R^2+\left(V_L-V_C\right)^2} \\ & =\sqrt{40^2+(60-30)^2} \\ & =\sqrt{1600+900}=\sqrt{2500} \\ & =50 \mathrm{~V} \end{aligned} \end{aligned} $

Resonance frequency

$ f_0=\frac{1}{2 \pi \sqrt{L C}} $

When dielectric material is introduced, then new capacitance,

$ C^{\prime}=C_0 K=16 C_0=16 C $

$\therefore$ New resonance frequency

$ \begin{aligned} f_0^{\prime} & =\frac{1}{2 \pi \sqrt{L C^{\prime}}}=\frac{1}{2 \pi \sqrt{L \times 16 C}} \\ & =\frac{1}{4} \times \frac{1}{2 \pi \sqrt{L C}}=\frac{f_0}{4} \end{aligned} $

Clearly first element is resistance and second element is inductor

$ \begin{aligned} R & =X_L=\frac{200}{5}=40 \Omega \\ \therefore Z & =\sqrt{R_f^2+X_L^2}=40 \sqrt{2} \Omega \\ I_{\mathrm{rms}} & =\frac{V_{\mathrm{rms}}}{Z}=\frac{200 / \sqrt{2}}{40 \sqrt{2}}=\frac{5}{2} \mathrm{~A} \end{aligned} $

To understand the impact of placing a dielectric between the plates of the capacitor on the glow of the bulb, we need to consider the properties and behavior of capacitors in an AC circuit.

When a capacitor is connected in series with a bulb in an AC circuit, the impedance of the capacitor plays a significant role in determining the current through the circuit. The impedance $ Z_C $ of a capacitor in an AC circuit is given by:

$ Z_C = \frac{1}{\omega C} $

where:

• $ Z_C $ is the capacitive reactance (impedance of the capacitor).
• $ \omega $ is the angular frequency of the AC supply ( $ \omega = 2 \pi f $ , where $ f $ is the frequency).
• $ C $ is the capacitance of the capacitor.

When we place a dielectric between the plates of the capacitor, the capacitance $ C $ increases. The capacitance with a dielectric can be described as:

$ C' = \kappa C $

where:

• $ C' $ is the new capacitance with the dielectric present.
• $ \kappa $ is the dielectric constant ( $ \kappa > 1 $ ).

Since $ C' > C $, the new capacitive reactance $ Z_C' $ can be given as:

$ Z_C' = \frac{1}{\omega C'} $

Because $ C' = \kappa C $, we have:

$ Z_C' = \frac{1}{\omega \kappa C} = \frac{1}{\kappa} \left( \frac{1}{\omega C} \right) = \frac{Z_C}{\kappa} $

Since $\kappa > 1$, $ Z_C' < Z_C $. This means the impedance of the capacitor decreases when a dielectric is placed between its plates. In a series circuit, the overall impedance decreases when the impedance of one component decreases, leading to an increase in the current through the circuit.

Thus, with an increase in current, the bulb will glow brighter. Therefore, the correct option is:

Option C: increases

To find the inductance of the coil, we need to analyze the given circuit and use the information provided. The circuit consists of a resistor and an inductor in series, connected to an AC supply. Here are the given values:

1. Resistance, $R = 90 \Omega$

2. Supply voltage, $V_{\text{total}} = 120 \mathrm{~V}$

3. Frequency, $f = 60 \mathrm{~Hz}$

4. Voltage across the resistor, $V_R = 36 \mathrm{~V}$

First, we calculate the current through the resistor (which is the same as the current through the inductor, since they are in series) using Ohm's law:

$I = \frac{V_R}{R} = \frac{36}{90} = 0.4 \mathrm{~A}$

Next, we find the total impedance $Z$ of the series combination from the total supply voltage:

$V_{\text{total}} = I \cdot Z$

$Z = \frac{V_{\text{total}}}{I} = \frac{120}{0.4} = 300 \Omega$

We know that the total impedance in a series circuit consisting of a resistor and an inductor is given by:

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

where $X_L$ is the inductive reactance. Rearrange this to solve for $X_L$:

$X_L = \sqrt{Z^2 - R^2} = \sqrt{(300)^2 - (90)^2} = \sqrt{90000 - 8100} = \sqrt{81900} \approx 286 \Omega$

Now, we use the inductive reactance formula to find the inductance $L$:

$X_L = 2\pi f L$

$L = \frac{X_L}{2\pi f} = \frac{286}{2 \pi \cdot 60} \approx \frac{286}{376.99} \approx 0.76 \mathrm{~H}$

Thus, the inductance of the coil is:

Option B: 0.76 H

To solve this problem, we need to understand the relationship between the current amplitude in a series LCR circuit at resonance and the resistance $ R $. At resonance, the impedance $ Z $ of the series LCR circuit is equal to the resistance $ R $, and thus:

$ Z = R $

The amplitude of the current $ I_0 $ at resonance is given by Ohm's law:

$ I_0 = \frac{V_0}{R} $

where $ V_0 $ is the amplitude of the voltage supplied.

Now, if the resistance $ R $ is halved while keeping the voltage amplitude $ V_0 $, capacitance $ C $, and inductance $ L $ the same, the new resistance becomes:

$ R_{new} = \frac{R}{2} $

The new current amplitude $ I_0' $ at resonance is given by the modified Ohm's law:

$ I_0' = \frac{V_0}{R_{new}} = \frac{V_0}{\frac{R}{2}} = \frac{2V_0}{R} = 2 \times I_0 $

Therefore, the current amplitude at resonance will be doubled. Hence, the correct answer is:

Option D: double

Statement I : True

Statement II : True

Current in purely resistive circuit is equal to current in LCR circuit with same resistance at resonance, otherwise more.

$\begin{aligned} &V_L=I(\omega L)=31.4\\ &\begin{aligned} \Rightarrow \quad I & =\frac{31.4}{2 \times 3.14 \times 50 \times 10 \times 10^{-3}} \\ & =10 \mathrm{~A} \end{aligned} \end{aligned}$

Let's calculate the maximum displacement current between the plates of the capacitor when an alternating voltage is applied directly across it.

The formula for the capacitive reactance $X_C$ of a capacitor is given by:

$X_C = \frac{1}{2\pi fC}$

where

• $f$ is the frequency of the alternating voltage, and
• $C$ is the capacitance of the capacitor.

Given:

• The amplitude of the voltage ($V_{max}$) = $40$ V
• Frequency ($f$) = $4000$ Hz (or $4 \mathrm{~kHz}$)
• Capacitance ($C$) = $12 \mu \mathrm{F} = 12 \times 10^{-6} \mathrm{~F}$

First, we calculate $X_C$:

$X_C = \frac{1}{2\pi(4000)(12 \times 10^{-6})} \approx \frac{1}{2\pi \cdot 4 \cdot 12 \cdot 10^{-3}} \approx \frac{1}{96\pi \cdot 10^{-3}}$

$X_C \approx \frac{1}{3.14 \cdot 96 \cdot 10^{-3}} \approx \frac{1}{0.30144} \approx 3.316 \Omega$

Now, the maximum current ($I_{max}$) in the circuit can be calculated using Ohm's law, considering the maximum voltage across the capacitor and its capacitive reactance:

$I_{max} = \frac{V_{max}}{X_C}$

$I_{max} = \frac{40}{3.316} \approx 12.06 \mathrm{~A}$

Hence, the maximum displacement current between the plates of the capacitor is nearly:

Option D: 12 A

For pure capacitive circuit, $I$ lead by $90^{\circ}$ to $\mathrm{V}$

For pure inductive circuit, $V$ lead by $90^{\circ}$ to $I$

At series LCR resonance, $I$ and $V$ are in same phase.

For LCR series circuit, $V$ and I may suffer some phase difference.

To understand which answer is correct, we must consider the relationship between voltage and current in various types of circuits, namely circuits with pure inductors, pure capacitors, pure resistors, and a combination of inductors and capacitors (LC circuits).

In a purely resistive circuit, the voltage and current are in phase, meaning when the voltage is maximum, the current is also maximum. Therefore, option C (pure resistor) cannot result in the instantaneous current being zero when the instantaneous voltage is maximum.

In a purely inductive circuit, the current lags the voltage by $90^\circ$, or in other words, when the voltage is at its maximum, the current is zero. This is because the inductor opposes changes in current, leading to this phase difference.

In a purely capacitive circuit, the current leads the voltage by $90^\circ$. This means when the voltage is at its maximum value, the current through the capacitor is zero because the current reaches its maximum or minimum before the voltage does.

In an LC circuit (inductor-capacitor), under certain conditions like at its resonant frequency, the circuit behaves as if it is purely resistive in nature where voltage and current are in phase. However, it's more nuanced because the question likely implies a condition not at resonance but rather at a general characteristic of LC circuits. In LC circuits, aside from the resonance condition, the presence of both inductive and capacitive components can lead to scenarios where the inductive and capacitive reactances cancel each other, particularly in the case where oscillations are involved, and indeed, there can be moments when the voltage is maximum and the current is minimum (zero in the ideal case) due to the energy being alternately stored in the magnetic field of the inductor and the electric field of the capacitor.

Thus, reflecting on the scenarios:

• Pure inductor - Fits the description: instantaneous current is zero when the instantaneous voltage is maximum.
• Pure capacitor - Also fits the description for the reason described, taking into account ideal conditions.
• Pure resistor - Does not fit as voltage and current are in phase.
• Combination of an inductor and capacitor (LC circuit) - Can fit the description under certain non-resonance conditions where the behavior of energy exchange between L and C at a specific instant can result in the instant current being zero when the voltage is maximum, but this is more complex and dependent on specific conditions unlike the straightforward lag or lead in purely inductive or capacitive circuits.

Therefore, the correct answer is Option C: A, B, and D only.

To solve this problem, we need to understand how to calculate the rms current in an AC circuit containing a capacitor, as well as the concept of displacement current.

First, the rms (root mean square) value of the current ($I_{\text{rms}}$) in a capacitor when connected to an AC supply is given by:

$I_{\text{rms}} = V_{\text{rms}} \cdot \omega C$

where $V_{\text{rms}}$ is the rms voltage, $\omega$ is the angular frequency, and $C$ is the capacitance.

For an AC supply, the rms voltage is related to the peak voltage ($V_0$) by:

$V_{\text{rms}} = \frac{V_0}{\sqrt{2}}$

However, we've been given the rms voltage directly, which is $230\, \text{V}$. Therefore, we can use this value directly in our calculations.

Given:

$C = 200\, \text{pF} = 200 \times 10^{-12}\, \text{F}$

$\omega = 300\, \text{rad/s}$

$V_{\text{rms}} = 230\, \text{V}$

Substitute these values into the equation for rms current:

$I_{\text{rms}} = V_{\text{rms}} \cdot \omega C = 230 \times 300 \times 200 \times 10^{-12} = 13.8 \times 10^{-6}\, \text{A}$

Therefore, the rms value of the conduction current is $13.8\, \mu \text{A}$.

Now, for the displacement current (which is essentially the current that would "flow" through the dielectric of the capacitor as a result of the changing electric field). In an AC circuit, the displacement current and the conduction current are the same. That's because the displacement current is necessary to sustain the changing electric field between the plates of the capacitor, and this changing field is what causes the conduction current in the leads and the rest of the circuit.

Hence, the rms value of the displacement current is the same as the conduction current, which is $13.8\, \mu \text{A}$.

So, the correct option is:

Option B

$13.8\, \mu \text{A}$ and $13.8\, \mu \text{A}$

The resonance frequency $ f_0 $ of an LCR circuit is given by:

where:

• $ L $ is the inductance.
• $ C $ is the capacitance.

To keep the resonance frequency unchanged when the capacitance is changed from $ C $ to $ 4C $, we must adjust the inductance $ L $ to a new value $ L' $ such that:

Now solving for $ L' $:

Squaring both sides, we have:

Dividing both sides by $ 4C $, we get:

Thus, the new inductance $ L' $ is one fourth of the original inductance $ L $. Therefore, to achieve the same resonance frequency with the capacitance increased to $ 4C $, the inductance should be reduced to a quarter of its initial value:

This means we have reduced the inductance by $ \frac{3}{4}L $, so the correct answer is:

Option C: reduced by $ \frac{3}{4}L $

$\begin{aligned} & <\mathrm{P}>=\mathrm{IV} \cos \phi \\ & =\frac{20}{\sqrt{2}} \times \frac{10}{\sqrt{2}} \times \cos 60^{\circ} \\ & =50 \mathrm{~W} \end{aligned}$

Rising half to peak

$\begin{aligned} & \mathrm{t}=\mathrm{T} / 6 \\ & \mathrm{t}=\frac{2 \pi}{6 \omega}=\frac{\pi}{3 \omega}=\frac{\pi}{300 \pi}=\frac{1}{300}=3.33 \mathrm{~ms} \end{aligned}$

$\begin{aligned} & \frac{\varepsilon_1}{\varepsilon_2}=\frac{N_1}{N_2}=\frac{100}{10} \Rightarrow \varepsilon_2=22 \mathrm{~V} \\ & I=\frac{22}{22 \times 10^3}=1 \mathrm{~mA}, V_0=7 \mathrm{~V} \end{aligned}$

$\begin{aligned} & E=25 \sin (1000 t) \\\\ & \cos \theta=\frac{1}{\sqrt{2}} \end{aligned}$

LR circuit

$\begin{aligned} & \text { Initially } \frac{R}{\omega_1 L}=\frac{1}{\tan \theta}=\frac{1}{\tan 45^{\circ}}=1 \\\\ & X_L=\omega_1 L \\\\ & \omega_2=2 \omega_1, \text { given } \\\\ & \tan \theta^{\prime}=\frac{\omega_2 L}{R}=\frac{2 \omega_1 L}{R} \\\\ & \tan \theta^{\prime}=2 \\\\ & \cos \theta^{\prime}=\frac{1}{\sqrt{5}} \end{aligned}$

$\begin{aligned} & P_{\text {avg }}=V_{\text {rms }} I_{r m s} \cos (\Delta \phi) \\ & =\frac{100}{\sqrt{2}} \times \frac{100 \times 10^{-3}}{\sqrt{2}} \times \cos \left(\frac{\pi}{3}\right) \\ & =\frac{10^4}{2} \times \frac{1}{2} \times 10^{-3} \\ & =\frac{10}{4}=2.5 \mathrm{~W} \end{aligned}$

By energy conservation

$\begin{aligned} & \frac{1}{2} \mathrm{CV}^2=\frac{1}{2} \mathrm{LI}_{\text {max }}^2 \\ & \mathrm{I}_{\max }=\sqrt{\frac{\mathrm{C}}{\mathrm{L}}} \mathrm{V} \\ & =\sqrt{\frac{100 \times 10^{-6}}{6.4 \times 10^{-3}}} \times 12 \\ & =\frac{12}{8}=\frac{3}{2}=1.5 \mathrm{~A} \end{aligned}$

$\begin{aligned} & \frac{V_1}{V_2}=\frac{N_1}{N_2} \\ & \frac{230}{V_2}=\frac{10}{1} \\ & V_2=23 \mathrm{~V} \end{aligned}$

Power consumed $=\frac{\mathrm{V}_2^2}{\mathrm{R}}$

$=\frac{23 \times 23}{46}=11.5 \mathrm{~W}$

(P) When $K_1$ is closed current in $R_0$ is $I_1$

At $\mathrm{t}=0$; circuit will be

$\begin{aligned} & \mathrm{I}_1=0 \\ & \mathrm{P} \rightarrow(1) \end{aligned}$

(Q) After long time inductor behave as a wire so $I_2$

$\begin{aligned} & \mathrm{I}_2=\frac{20}{5}=4 \mathrm{~A} \\ & \mathrm{Q} \rightarrow(3) \end{aligned}$

$\text { (R) When } K_2 \text { is closed and } K_1 \text { open }$

$\begin{aligned} & \omega_0=\frac{1}{\sqrt{\mathrm{LC}}} \\ & \omega_0=\frac{1}{\sqrt{25 \times 10^{-3} \times 10 \times 10^{-6}}}=\frac{1}{5 \times 10^{-4}} \\ & \omega_0=2 \times 10^3 \mathrm{rad} / \mathrm{s} \\ & \omega_0=2 \mathrm{kilo}-\mathrm{radian} / \mathrm{s} \\ & \mathrm{R} \rightarrow(2) \end{aligned}$

(S) Now $K_2$ is closed and $K_1$ open

$\begin{aligned} & \frac{1}{2} \mathrm{LI}_2^2=\frac{1}{2} \mathrm{CV}_0^2 \\ & 25 \times 10^{-3} \times(4)^2=10 \times 10^{-6} \times \mathrm{V}_0^2 \\ & \mathrm{~V}_0^2=2500 \times 16 \\ & \mathrm{~V}_0=50 \times 4=200 \mathrm{~V} \\ & \mathrm{~S} \rightarrow(5) \end{aligned}$

Given,

The ratio of the capacitive reactance $ X_{C} $ to the resistance $ R $ is $ 4: 3 $.

Let, $ X_{C}=4 k $

$ R=3 k $

Then, the impedance $ Z $ of the series $ R-C $ circit is given by

$ \begin{array}{l} Z=\sqrt{R^{2}+X_{C}^{2}} \\\\ Z=\sqrt{(3 k)^{2}+(4 k)^{2}} \\\\ Z=5 k \end{array} $

We know that, the power factor is the cosine of the phase angle $ \phi $ between the total impedance and the resistive part of the impedance.

$ \begin{array}{l} \cos \phi=\frac{R}{Z} \\\\ \cos \phi=\frac{3 k}{5 k} \\\\ \cos (\phi)=0.6 \end{array} $

Given,

Inductive reactance, $ X_{L}=R $

Capacitive reactance, $ X_{C}=2 R $

Resistor $ =R $

Power factor, $ \cos \phi $ is given by

$ \begin{array}{l} \cos \phi=\frac{R}{Z} \\\\ \cos \phi=\frac{R}{Z}=\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}} \\\\ \cos \phi=\frac{R}{\sqrt{R^{2}+(R-2 R)^{2}}} \\\\ \cos \phi=\frac{R}{\sqrt{2 R^{2}}}=\frac{1}{\sqrt{2}} \end{array} $

Given, $ V_{L}=6 \mathrm{~V} $

$ V_{\text {RMS }}=10 \mathrm{~V} $

As the circuit consists only $ L $ and $ R $

$ V_{\mathrm{RMS}}=\sqrt{V_{R}^{2}+V_{L}^{2}} $

Squaring both sides

$ \begin{aligned} V_{\mathrm{RMS}}^{2} & =V_{R}^{2}+V_{L}^{2} \\\\ V_{R}^{2} & =V_{\mathrm{RMS}}^{2}-V_{L}^{2} \\\\ & =(10)^{2}-(6)^{2} \\\\ V_{R}^{2} & =100-36=64 \\\\ V_{R} & =8 \mathrm{~V} \end{aligned} $

In a series $L-C-R$ circuit where the current leads the voltage source, the reactance $(X)$ is predominantly capacitive, meaning the capacitive reactance ( $X_C$ ) is greater than the inductive reactane $\left(X_L\right)$.

Mathematically, it can be expressed as $X_C>X_L$ This imbalance the capacitive and inductive reactance values results in the leading current behaviour.