Alternating Current

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.

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.

Statement I: An AC circuit undergoes electrical resonance if it contains either a capacitor or an inductor.

This statement is incorrect. Electrical resonance occurs in an AC circuit when the capacitive reactance and inductive reactance are equal, causing the impedance of the circuit to be minimum. This typically happens in a series RLC circuit or a parallel RLC circuit. If the circuit contains only a capacitor or an inductor, it cannot undergo electrical resonance as there is no counterpart reactance to balance the impedance.

Statement II: An AC circuit containing a pure capacitor or a pure inductor consumes high power due to its non-zero power factor.

This statement is also incorrect. An AC circuit containing a pure capacitor or a pure inductor will have a power factor of 0, not a non-zero power factor. The power factor of a capacitor is -1, and the power factor of an inductor is +1, but when only considering the reactive components, the power factor is 0. In such a circuit, no real power is consumed, and the circuit only has reactive power. The energy is alternately stored and released by the capacitor and inductor, but no energy is dissipated as heat or used to perform work.

As both statements are incorrect, the correct answer would be an option that states both statements are false.

Statement I is true because in a series LCR circuit, the current first increases as the frequency increases, reaching a maximum value when the circuit is at resonance. At resonance, the inductive reactance (XL) and capacitive reactance (XC) cancel each other out, resulting in the lowest impedance (Z) and the highest current. As the frequency continues to increase beyond resonance, the current in the circuit decreases.

Statement II is also true because, at resonance in a series LCR circuit, the inductive reactance (XL) and capacitive reactance (XC) are equal and cancel each other out. This results in the impedance (Z) being purely resistive. The power factor at resonance is given by the cosine of the phase angle (θ), and since the phase angle is 0° at resonance, the power factor is 1.

Thus, both statements are true, and the correct answer is Both Statement I and Statement II are true.

In a series LCR circuit connected to an AC source, resonance occurs at a particular frequency at which the inductive reactance is equal to the capacitive reactance, resulting in the minimum impedance of the circuit. At this frequency, the circuit draws maximum current from the source, and thus, the maximum power is dissipated in the circuit. Therefore, Statement I is true.

In a circuit containing only a resistor, the power dissipated is given by P = VI = I$^2$R, where V is the voltage across the resistor, I is the current flowing through the resistor, and R is the resistance of the resistor. The voltage and current are in phase in a purely resistive circuit, which means that the power is maximized. Therefore, Statement II is also true.

For a capacitor connected to an AC source, the maximum current $I_\text{max}$ can be calculated using the formula:

$I_\text{max} = E_\text{max} \cdot \omega C$

where $E_\text{max}$ is the maximum voltage, $\omega$ is the angular frequency, and $C$ is the capacitance.

Given the emf equation: $E = 36 \sin(120\pi t) \, \text{V}$, we can determine that $E_\text{max} = 36\, \text{V}$ and $\omega = 120\pi \, \text{rad/s}$.

The capacitance is given as $150.0\, \mu\text{F} = 150.0 \times 10^{-6}\, \text{F}$.

Now, we can calculate the maximum current:

$I_\text{max} = 36 \cdot (120\pi) \cdot (150.0 \times 10^{-6})$

$I_\text{max} \approx 2\, \text{A}$

Thus, the correct answer is $2\, \text{A}$.

${I_{rms}} = {{{V_{rms}}} \over z} = {{200\sqrt 2 } \over {\sqrt {{{100}^2} + {{(200 - 100)}^2}} }}$

$ = {{200\sqrt 2 } \over {100\sqrt 2 }}$

= 2 A

${X_L} = R$

$ \Rightarrow {P_1} = {R \over {\sqrt {X_L^2 + {R^2}} }} = {1 \over {\sqrt 2 }}$

Now, ${X_L} = {X_C} = R$

$ \Rightarrow {P_2} = {R \over {\sqrt {{R^2} + {{({X_L} - {X_C})}^2}} }} = 1$

$ \Rightarrow {{{P_1}} \over {{P_2}}} = {1 \over {\sqrt 2 }}$

For (a), $i = {V \over R} = {{220} \over {40}} = 5.5\,A$

for (b), ${X_C} = {1 \over {2\pi fC}} = {1 \over {2\pi \times 50 \times 0.5 \times {{10}^{ - 6}}}} = {{{{10}^6}} \over {50\pi }}\Omega $

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

${X_C} > {X_L}$, hence impedance is greater than 40 $\Omega$.

${i_{rms}} = {{220} \over Z}$

$\therefore$ ${\left. {{i_{rms}}} \right|_b} < {\left. {{i_{rms}}} \right|_a}$