Alternating Current

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} $

The rms value is found from

writing $i(t)=A\sin\omega t + B\cos\omega t$ with $A=3$, $B=4$,

using

$i_{\rm rms}=\sqrt{\langle i^2\rangle} =\sqrt{\frac{A^2+B^2}{2}} =\frac{\sqrt{3^2+4^2}}{\sqrt2} =\frac{5}{\sqrt2}\,\rm A\,. $

So the correct choice is Option C: $\displaystyle \frac{5}{\sqrt2}$ A.

Given:

Frequency, $ f = 50 \, \text{Hz} $

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

Resistance, $ R = 2 \, \Omega $

First, calculate the inductive reactance ($ X_L $) using the formula:

$ X_L = 2 \pi f L $

Substituting the given values:

$ X_L = 2 \pi \times 50 \times 10 \times 10^{-3} = \pi $

For the circuit to have a unit power factor, the inductive reactance ($ X_L $) must equal the capacitive reactance ($ X_C $). Therefore,

$ X_C = X_L $

The formula for capacitive reactance is:

$ X_C = \frac{1}{2 \pi f C} $

Setting $ X_C = X_L $, we have:

$ \pi = \frac{1}{2 \pi \times 50 \times C} $

Solving for $ C $:

$ C = \frac{1}{2 (\pi^2) \times 50} $

$ C = \frac{1}{2 \times 9.8696 \times 50} $

$ C = \frac{1}{973.036} $

$ C = 1.014 \times 10^{-3} \, \text{F} = 1.014 \, \text{mF} $

Thus, the required capacitance is $ 1.014 \times 10^{-3} \, \text{F} $.

Given an $L-C-R$ circuit in resonance, let’s analyze it:

Provided Values:

Inductive Reactance ($X_L$): $2R$

Resistance ($R$): $R$

Resonance Condition:

In a resonant circuit, the inductive reactance is equal to the capacitive reactance. Therefore:

$ X_L = X_C $

$ X_C = 2R $

Power Factor:

The power factor of a circuit is calculated as:

$ P = \frac{\text{Resistance}}{\text{Impedance (Z)}} $

The impedance in an $L-C-R$ circuit, when in resonance and when $X_L = X_C$, is:

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

Since $X_L = X_C$ in resonance:

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

Thus, the power factor is:

$ P = \frac{R}{R} = 1 $

Therefore, when the circuit is in resonance, the capacitive reactance $X_C = 2R$, and the power factor is 1.

Given:

The voltage expression: $ V = 144 \sin \left(100 \pi t + \frac{\pi}{2}\right) \, \text{V} $

The current expression: $ I = 6 \sin \left(100 \pi t + \frac{\pi}{6}\right) \, \text{A} $

Analysis

Standard Form Comparison:

The standard forms of voltage and current are:

$ V = V_0 \sin(\omega t + \phi_v) $

$ I = I_0 \sin(\omega t + \phi_i) $

Phase Difference:

Comparing the given equations, we identify the phases:

Voltage phase: $ \phi_v = \frac{\pi}{2} $

Current phase: $ \phi_i = \frac{\pi}{6} $

The phase difference, $\phi$, between voltage and current is:

$ \phi = \phi_v - \phi_i = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3} = 60^\circ $

Impedance Calculation:

The impedance $ Z $ of the circuit is given by:

$ Z = \frac{V_0}{I_0} = \frac{144}{6} = 24 \, \Omega $

Resistance Calculation:

The relationship between impedance $ Z $, resistance $ R $, and the phase angle $ \phi $ is given by:

$ \frac{R}{Z} = \cos \phi $

Solving for $ R $, we have:

$ R = \cos(60^\circ) \times Z = \frac{1}{2} \times 24 = 12 \, \Omega $

Therefore, the resistance of the resistor is $ 12 \, \Omega $.

Given:

Resistance: $ R = 20 \, \Omega $

Alternating potential: $ V = 200 \sin (10 \pi t) $

The expression for the alternating current $ I $ can be derived as follows:

$ I = \frac{V}{R} = \frac{200 \sin (10 \pi t)}{20} = 10 \sin (10 \pi t) $

We know that the root mean square (RMS) current is related to the peak current by the equation:

$ I_{\text{RMS}} = \frac{I_{\text{peak}}}{\sqrt{2}} $

Thus, the RMS current is also expressed as:

$ I_{\text{RMS}} = I_{\text{peak}} \sin (10 \pi t) $

Setting the RMS and the peak relation,

$ \frac{I_{\text{peak}}}{\sqrt{2}} = I_{\text{peak}} \sin (10 \pi t) $

This simplifies to:

$ \frac{1}{\sqrt{2}} = \sin (10 \pi t) $

The sine of $ \frac{\pi}{4} $ is known to be $ \frac{1}{\sqrt{2}} $, so:

$ \sin\left(\frac{\pi}{4}\right) = \sin (10 \pi t) \Rightarrow 10 \pi t = \frac{\pi}{4} $

Solving for $ t $:

$ t = \frac{1}{40} \Rightarrow t = 2.5 \times 10^{-2} \, \text{s} $

In a full-wave rectifier, both the positive and negative halves of the AC input are used, effectively doubling the frequency of the output ripple compared to the mains frequency. Here’s a step-by-step explanation:

The mains frequency is $50\, \text{Hz}.$

The full-wave rectifier inverts the negative half cycle, so the ripple frequency becomes:

$f_{\text{ripple}} = 2 \times 50\, \text{Hz} = 100\, \text{Hz}.$

Therefore, the fundamental frequency in the ripple output is $100\, \text{Hz},$ which is Option C.

In the given LCR circuit, the components have the following values:

Inductance, $ L = 10 \, \text{H} $

Capacitance, $ C = 40 \, \mu\text{F} = 4 \times 10^{-5} \, \text{F} $

Resistance, $ R = 60 \, \Omega $

The circuit is connected to a variable frequency source of 240 V.

To find the resonant frequency ($\omega_r$), we use the formula:

$ \omega_r = \frac{1}{\sqrt{LC}} $

Substituting the given values:

$ \omega_r = \frac{1}{\sqrt{10 \times 4 \times 10^{-5}}} = \frac{1}{2 \times 10^{-2}} = 50 \, \text{rad/s} $

At resonance, the impedance $ Z $ of the circuit is minimized to the resistance $ R $ only. Thus, the current $ I $ at resonant frequency can be calculated by:

$ I = \frac{V}{Z} = \frac{V}{R} = \frac{240}{60} = 4 \, \text{A} $

We know that, $X_C=1 / \omega C$ and $X_L=\omega L$

According to figure,

$ \begin{aligned} & C=10 \mathrm{pF}=10 \times 10^{-12} \mathrm{~F} \\ & L=100 \mathrm{mH}=0.1 \mathrm{H} \end{aligned} $

Here, $C \ll L$

$ \text { So, } \quad X_C \gg X_L $

$ \therefore \quad i_A \gg i_B $

Hence, bulb $A$ will glow brighter.

To determine the phase angle $\phi$ between the current and the applied voltage in an $R-L-C$ circuit, we need to use the following relationship:

$\tan \phi = \frac{X_L - X_C}{R}$

where:

• $X_L$ is the inductive reactance


• $X_C$ is the capacitive reactance


• $R$ is the resistance

Given:

• Resistance, $R = 150 \Omega$
• Capacitance, $C = 20 \mu \mathrm{F} = 20 \times 10^{-6} \mathrm{F}$
• Inductance, $L = 500 \mathrm{mH} = 500 \times 10^{-3} \mathrm{H}$
• Angular frequency, $\omega = 400 \mathrm{rad} \, \mathrm{s}^{-1}$

The inductive reactance $X_L$ is calculated as:

$X_L = \omega L$

Substituting the values:

$X_L = 400 \times 500 \times 10^{-3}$

$X_L = 200 \, \Omega$

The capacitive reactance $X_C$ is calculated as:

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

Substituting the values:

$X_C = \frac{1}{400 \times 20 \times 10^{-6}}$

$X_C = 125 \, \Omega$

Now, we can find the phase angle using the equation:

$\tan \phi = \frac{X_L - X_C}{R}$

Substituting the values:

$\tan \phi = \frac{200 - 125}{150}$

$\tan \phi = \frac{75}{150}$

$\tan \phi = 0.5$

Therefore, the phase angle $\phi$ is:

$\phi = \tan^{-1}(0.5)$

So, the correct answer is:

Option D $\tan^{-1}(0.5)$

Given capacitive reactance,

$X_{C_1}=3 \mathrm{k} \Omega$

we know that,

$X_C=\frac{1}{\omega C}=\frac{1}{2 \pi f C}$

$ \begin{aligned} &\begin{aligned} \Rightarrow X_C & \propto \frac{1}{f} \Rightarrow \frac{X_{C_2}}{X_{C_1}}=\frac{f_1}{f_2} \\ & =\frac{f_1}{2 f_1}=\frac{1}{2} \quad \left(\because f_2=2 f_1\right)\\ \Rightarrow X_{C_2} & =\frac{X_{C_1}}{2}\\ & =\frac{3 \mathrm{k} \Omega}{2}=1.5 \mathrm{k} \Omega \end{aligned}\\ \end{aligned}$

Given, Inductance, $L=0.1 \mathrm{~H}$

Resistance, $R=110 \Omega$;

Potential difference, $V=110 \mathrm{~V}$

Frequency, $f=350 \mathrm{~Hz}$

Phase difference between maximum voltage and maximum current is given as

$ \begin{aligned} \tan \phi & =\frac{X_L}{R} \\ & =\frac{2 \pi f L}{R}=\frac{2 \times \frac{22}{7} \times 350 \times 01}{110}=2 \\ \Rightarrow \quad \tan \phi & =2 \\ \Rightarrow \quad \phi & =\tan ^{-1}(2) \end{aligned}$

Given, $V_{\mathrm{rms}}=20 \mathrm{~V}$

Voltage across resistance $=12 \mathrm{~V}$

Let voltage across inductor be $V_L$

As we know that,

$\begin{aligned} V^2 & =V_R^2+V_L^2 \\ \Rightarrow \quad V_L^2 & =V^2-V_R^2=(20)^2-(12)^2 \\ & =400-144=256 \mathrm{~V} \\ V_L & =16 \mathrm{~V} \end{aligned}$

Given, resistance of bulb, $R=280 \Omega$

Supply voltage $V_{\text {rms }}=200 \mathrm{~V}$

As, $\quad V_{\mathrm{rms}}=\frac{V_0}{\sqrt{2}}=$ root mean square speed

$\Rightarrow \quad V_0=\sqrt{2} V_{\mathrm{rms}} $

where, $V_0$ is peak voltage,

As, $\quad V=I R$

$\therefore \quad I_0=\frac{V_0}{R}=\frac{\sqrt{2} \times 200}{280}=\frac{5 \sqrt{2}}{7}=1.010 \mathrm{~A}$

Which is nearly 1A.

Given,

Initial resonance frequency, $f_i=f$

Initial capacitance and inductance be $C_1=C$ and $L_1=L$

Final capacitance and inductance one $C_2=4 C$ and $L_2=L$

Let final frequency be $f^{\prime}$.

Since, $\omega=\frac{1}{\sqrt{L C}}=2 \pi f$

where, $\omega$ and $f$ and angular frequency and frequency.

$\begin{aligned} & \therefore \quad f \propto \frac{1}{\sqrt{L C}} \Rightarrow \frac{f}{f^{\prime}}=\sqrt{\frac{L^{\prime} C^{\prime}}{L C}}=\frac{\sqrt{L 4 C}}{\sqrt{L C}}=\sqrt{4}=2 \\ & \therefore \quad f^{\prime}=\frac{f}{2} \end{aligned}$