A resistor and an ideal inductor are connected in series to a $100\sqrt{2}$ V, 50 Hz ac source. When a voltmeter is connected across the resistor or the inductor, it shows the same reading. The reading of the voltmeter is : (A) $100\sqrt{2}$ V (B) $100$ V (C) $50\sqrt{2}$ V (D) $50$ V

A resistor and an ideal inductor are connected in series to a 100√(2) V, 50 Hz ac source. When a voltmeter is

Question

100\sqrt{2}

CBSECBSE Class XII Board 2024MCQ· 1mImportance★★★★★

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Concept understanding — AC Circuit Analysis

AC Circuit Analysis: A First Look

The Core Intuition

Imagine you're pushing a child on a swing. You don't push once and stop — you push rhythmically, back and forth, matching the swing's natural motion. That rhythmic push is like Alternating Current (AC).

In a DC circuit (like a battery), current flows in one direction — steady, like a river. In an AC circuit, the current reverses direction many times per second — like the swing going forward, then backward, then forward again.

Why does this matter? Because when current keeps changing direction, components like coils and capacitors behave very differently than they do with steady DC. They "resist" change, store energy, and create phase shifts — the current and voltage don't peak at the same time.

The Precise Statement

AC Circuit Analysis is the study of how voltage, current, and power behave in circuits where the source voltage varies sinusoidally with time:

v(t) = V_m \sin(\omega t + \phi)

Where:

$V_m$ = peak voltage (maximum value)
$\omega$ = angular frequency ( $2\pi f$ , where $f$ is frequency in Hz)
$\phi$ = phase angle (where the wave starts)
$t$ = time

The current in such a circuit is also sinusoidal, but may be shifted in time relative to the voltage:

i(t) = I_m \sin(\omega t + \phi - \theta)

Here $\theta$ is the phase difference between voltage and current.

Three Key Behaviors (The "Why")

1. Resistor (R) — No phase shift

Voltage and current rise and fall together
$v(t) = R \cdot i(t)$ holds at every instant
Phase difference $\theta = 0^\circ$

2. Inductor (L) — Current lags voltage

An inductor resists changes in current
When voltage peaks, current is still catching up
Current lags voltage by $90^\circ$ ( $\theta = +90^\circ$ )
Opposition to AC is called inductive reactance: $X_L = \omega L = 2\pi f L$

3. Capacitor (C) — Current leads voltage

A capacitor resists changes in voltage
When voltage is zero, current is maximum (charging)
Current leads voltage by $90^\circ$ ( $\theta = -90^\circ$ )
Opposition to AC is called capacitive reactance: $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$

The Big Idea: Impedance

Instead of just "resistance," AC circuits use impedance ( $Z$ ), which combines:

Resistance ( $R$ ) — energy loss (real part)
Reactance ( $X$ ) — energy storage (imaginary part)

Z = R + jX

Where $j = \sqrt{-1}$ (engineers use $j$ instead of $i$ to avoid confusion with current).

Ohm's Law for AC:

V = I \cdot Z

But here $V$ and $I$ are phasors — complex numbers that represent magnitude and phase.

Why This Matters for Exams

Concept	What to remember
RMS value	$V_{rms} = \frac{V_m}{\sqrt{2}}$ — the "effective" DC equivalent
Power	$P = V_{rms} I_{rms} \cos\theta$ — only the in-phase part does real work
Resonance	When $X_L = X_C$ , impedance is minimum, current is maximum

Start your 14-day free trial to unlock the full solution →

In a series RL circuit, equal voltmeter readings across R and L imply equal voltage drops, which forces the impedance triangle to be isosceles — the common reading is $100$ V.

The key idea here is that the voltmeter reads the RMS voltage across each component. For a series RL circuit driven by an AC source, the source voltage $V_s$ is the phasor sum of $V_R$ and $V_L$ , not the arithmetic sum — because the resistor voltage and inductor voltage are $90^\circ$ out of phase.

When the voltmeter shows the same reading across R and L, we have $V_R = V_L = V$ (say). The source RMS voltage is given as $100\sqrt{2}$ V. The phasor relationship is:

V_s^2 = V_R^2 + V_L^2 = V^2 + V^2 = 2V^2

So $V = \frac{V_s}{\sqrt{2}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100$ V.

That is the reading on the voltmeter — $100$ V.

Understand the circuit and the source.

The source is $100\sqrt{2}$ V, 50 Hz AC. The voltmeter reads RMS values (standard for AC voltmeters). The resistor and ideal inductor are in series.
Phase relationship.

In a series RL circuit, current $I$ is common. Voltage across resistor $V_R = IR$ is in phase with current. Voltage across inductor $V_L = I X_L$ leads current by $90^\circ$ . So $V_R$ and $V_L$ are perpendicular phasors.
Given condition: equal readings.

The voltmeter reads the same value across R and L. So $V_R = V_L = V$ (say).
Apply phasor addition for the source voltage.

The source RMS voltage $V_s$ is the magnitude of the phasor sum:

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

Solve for $V$ . …

Start your 14-day free trial to unlock the full solution →

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 26 on this concept.

CBSE 2026Set DS1 mark
$V$
›Reveal solutionSolution
In-phase $V$ and $I$ ⇒ resistive circuit; current leading voltage by $90^\circ$ ⇒ capacitive circuit.

Concept. The phase relation between voltage and current identifies the circuit element:

Pure resistor: current is in phase with voltage — both reach their maxima and zeros at the same instant.

Pure capacitor: current leads the voltage by a quarter cycle ( $90^\circ$ , i.e. $\pi/2$ ) — the current curve peaks before the voltage curve.

Reading the graphs.

In the left graph the current curve $I$ and the voltage curve $V$ rise together and reach their peaks at the same time — they are in phase → this is the resistive circuit. …
Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 mark
$Q. Can a battery be charged by using alternating current? Write the reason.$
›Reveal solutionSolution

Charging a battery needs current flowing in one fixed direction; AC reverses direction every half-cycle, so used directly it undoes on the second half-cycle whatever it did on the first.

Charging a cell/battery means forcing conventional current into it against its own e.m.f. so chemical energy is stored, which requires a steady, one-directional (DC) current. Alternating current changes direction periodically (e.g. every 1/100 s for 50 Hz mains), so during one half-cycle it would tend to charge the battery, but during the very next half-cycle the reversed current would discharge it by an equal amount — the net effect over a full cycle is essentially zero charging, and repeatedly reversing cu …

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$Q. The time taken by an AC of frequency 50 Hz to reach from zero to the maximum value is$
(a) $50\times10^{-3}$ s
(b) $5\times10^{-3}$ s
(c) $1\times10^{-3}$ s
(d) $2\times10^{-3}$ s
›Reveal solutionSolution

An alternating current varies as $i=I_0\sin\omega t$ . It starts at zero and first reaches its maximum ( $I_0$ ) after one quarter of a full cycle, i.e. $t=T/4$ .

Step 1 — Find the time period

$T=\frac{1}{f}=\frac{1}{50\text{ Hz}}=0.02\text{ s}=20\times10^{-3}\text{ s}$

Step 2 — Express the current as a function of time

Since the current is zero at $t=0$ and increases from there, it is described by

$i(t)=I_0\sin(\omega t), \qquad \omega=\frac{2\pi}{T}$

$i(t)$ reaches its maximum value $I_0$ when $\sin\omega t=1$ , i.e. when

…

Start your 14-day free trial to see the full solution →
CBSE 2026Set ANNUAL1 markMCQ
$E$
(a) This phase relationship is for a resistor when AC is passed through it.
(b) This phase relationship is for a capacitor when AC is passed through it.
(c) This phase relationship is for an inductor when AC is passed through it.
(d) This phase relationship is for an L-C-R circuit.
›Reveal solutionSolution
The graph shows current lagging voltage by a quarter cycle ( $\pi/2$ ). Comparing this to the known phase behaviour of R, L, C elements in AC circuits identifies the element as a pure inductor.

Phase relationships of R, L, C in AC circuits (recap)

Pure resistor: $V=IR$ at every instant, so $I$ and $V$ (or $E$ ) are always in phase (zero phase difference).

Pure capacitor: $I=C\,dV/dt$ ; when $V=V_0\sin\omega t$ , $I=\omega C V_0\cos\omega t=\omega C V_0\sin(\omega t+\pi/2)$ — current leads voltage by $\pi/2$ .

Pure inductor: $V=L\,dI/dt$ ; when $I=I_0\sin\omega t$ , $V=\omega L I_0\cos\omega t = \omega L I_0 \sin(\omega t+\pi/2)$ — equivalently, current lags voltage by $\pi/2$ : $I=I_0\sin(\omega t-\pi/2)$ relative to $V=V_0\sin\omega t$ .

Series L-C-R circuit: the phase difference $\phi$ between $V$ and $I$ is given by $\tan\phi=(X_L-X_C)/R$ , which generally lies between $0$ and $\pi/2$ (not exactly $\pi/2$ ) unless $R=0$ .

Reading the graph

…
Start your 14-day free trial to see the full solution →
CBSE 2025Set 55/4/11 markMCQ
$X_L = X_C$
›Reveal solutionSolution
At resonance, the inductive and capacitive reactances cancel exactly, making the circuit purely resistive. The reason correctly states the condition $X_L = X_C$ and the resonant frequency formula, so both are true and the reason is the correct explanation.

Concept and Intuition

In a series LCR circuit, the total impedance is $Z = R + j(X_L - X_C)$ , where $X_L = \omega L$ and $X_C = 1/(\omega C)$ . The phase angle $\phi$ between voltage and current is given by $\tan \phi = (X_L - X_C)/R$ . For the circuit to behave as a pure resistor, the imaginary part must vanish — that is, $X_L = X_C$ . This condition defines resonance. When it holds, the impedance is minimum ( $Z = R$ ), current is maximum, and voltage and current are in phase ( $\phi = 0$ ). The reason correctly identifies this condition and the resulting angular frequency $\omega = 1/\sqrt{LC}$ .

Watch out

A common mistake is to think that at resonance the circuit has zero impedance. It does not — the impedance equals $R$ , which is purely resistive but not zero (unless $R=0$ ). The key is that the reactive parts cancel, not that reactance disappears.

Step-by-step reasoning

Assertion (A) states that a series LCR circuit behaves as a pure resistive circuit at resonance. This is true because at resonance, $X_L = X_C$ , so the net reactance is zero. The impedance becomes $Z = R + j0 = R$ , a purely real quantity. Hence voltage and current are in phase, exactly as in a resistor.

Reason (R) states that at resonance, $X_L = X_C$ , which gives $\omega = 1/\sqrt{LC}$ . This is also true. Setting $X_L = X_C$ :

$\omega L = \frac{1}{\omega C} \quad\Rightarrow\quad \omega^2 = \frac{1}{LC} \quad\Rightarrow\quad \omega = \frac{1}{\sqrt{LC}}$
(taking the positive root, since frequency is positive). …
Start your 14-day free trial to see the full solution →
CBSE 2025Set X11 markMCQ
$Q. The variation of voltage and current through an a.c. circuit with time is as shown in the figure. Along with the a.c. source, the circuit :$
(a) has a series combination of resistance and capacitance
(b) has only inductance
(c) has only capacitance
(d) may have only resistance or may have a suitable series combination of inductance $(L)$ , capacitance $(C)$ and resistance $(R)$
›Reveal solutionSolution
The printed graph shows voltage $v$ and current $i$ crossing zero together and peaking at the same $\omega t$ — they are in phase (phase difference $\phi = 0$ ). A zero phase angle means the net reactance is zero, which is true for a pure resistor or for a series $LCR$ circuit at resonance. Hence the correct choice is (d).

What the graph tells us

Both sinusoids start at the origin, rise together, reach their peaks ( $v_0$ and $i_0$ ) at the same instant, and cross the $\omega t$ axis at the same points. So:
$v = v_0 \sin(\omega t), \qquad i = i_0 \sin(\omega t)$
The phase difference between voltage and current is $\phi = 0$ . (Only the amplitudes differ, $v_0 > i_0$ , which is irrelevant to the phase.)

Linking phase to the circuit

For a general a.c. circuit, $\tan\phi = \dfrac{X_L - X_C}{R}$ . A phase difference of zero requires $X_L - X_C = 0$ , i.e. the net reactance vanishes. This can happen in two ways:

Only resistance ( $R$ ): there is no reactance at all, so $v$ and $i$ are always in phase.

A series $LCR$ circuit at resonance: here $X_L = X_C$ , the reactances cancel, the impedance becomes purely resistive ( $Z = R$ ), and $v$ and $i$ come back into phase.

Why the other options are wrong …
Start your 14-day free trial to see the full solution →
CBSE 2025Set JS1 markMCQ
$14\sqrt{2}$
›Reveal solutionSolution

Impedance of a series R–L circuit is $Z=\sqrt{R^2+X_L^2}=\sqrt{64+36}=10\ \Omega$ — option (D).

Concept. In a series AC circuit the resistance $R$ and reactance $X_L$ add like the two perpendicular sides of a right triangle (because the voltage across $R$ is in phase with the current while the voltage across $L$ leads it by $90^\circ$ ). The magnitude of the total opposition, the impedance, is the hypotenuse: …

Start your 14-day free trial to see the full solution →
CBSE 2025Set JS1 mark
$V$
›Reveal solutionSolution
Figure — A single sine curve of emf/voltage vs time

Peak $40$ V, period $T=0.02$ s ( $20$ ms); plot a sine curve rising to $+40$ V at $t=5$ ms, back to $0$ at $10$ ms, down to $-40$ V at $15$ ms, and returning to $0$ at $20$ ms.

Concept. For $V=V_0\sin(\omega t)$ , the amplitude is $V_0$ and the angular frequency is $\omega$ , giving period $T=\dfrac{2\pi}{\omega}$ .

Read off the values.

$V_0=40\ \text{V},\qquad \omega=100\pi\ \text{rad s}^{-1},$

$T=\frac{2\pi}{\omega}=\frac{2\pi}{100\pi}=0.02\ \text{s}=20\ \text{ms},\qquad f=\frac1T=50\ \text{Hz}.$

Key points of one cycle (scale: $x$ -axis $1$ div $=5$ ms, $y$ -axis $1$ div $=10$ V):

| $t$ (ms) | 0 | 5 | 10 | 15 | 20 | …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. The value of alternating voltage in the following circuit is$
(a) 2 V
(b) 4 V
(c) 8 V
(d) 10 V
›Reveal solutionSolution

In a series R-L-C a.c. circuit, $V_L$ and $V_C$ are 180 degrees out of phase with each other and both are 90 degrees out of phase with $V_R$ , so the source voltage is the phasor sum $V=\sqrt{V_R^2+(V_L-V_C)^2}$ , not the plain arithmetic sum.

In a series a.c. circuit, the voltage across R is in phase with the current, the voltage across L leads the current by $90^\circ$ , and the voltage across C lags the current by $90^\circ$ ( $V_L$ and $V_C$ are therefore antiphase to each other). The applied (source) voltage is t …

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 markMCQ
$Q. The unit of reactance is$
(a) ohm
(b) mho
(c) farad
(d) ampere
›Reveal solutionSolution

Reactance X (whether inductive, XL = wL, or capacitive, XC = 1/wC) has the same dimensions as resistance, since it appears in V = IX exactly like Ohm's law.

…

Start your 14-day free trial to see the full solution →
CBSE 2025Set ANNUAL1 mark
$I = 50\sin100t$
›Reveal solutionSolution

Reading off $\omega$ from the given equation and using $f=\omega/2\pi$ gives about 15.9 Hz; for the alternative, $X_L=2\pi fL$ gives about 8.8 $\Omega$ .

Finding the frequency

The standard form of an alternating current is

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

Comparing with the given equation $I = 50\sin(100t)$ :

$\omega = 100\ \text{rad/s}$

Since $\omega = 2\pi f$ ,

$f = \frac{\omega}{2\pi} = \frac{100}{2\pi} \approx 15.9\ \text{Hz}$

…

Start your 14-day free trial to see the full solution →
CBSE 2024Set 55/2/11 markMCQ
$100\sqrt{2}$
›Reveal solutionSolution
In a series RL circuit, equal voltmeter readings across R and L imply equal voltage drops, which forces the impedance triangle to be isosceles — the common reading is $100$ V.

The key idea here is that the voltmeter reads the RMS voltage across each component. For a series RL circuit driven by an AC source, the source voltage $V_s$ is the phasor sum of $V_R$ and $V_L$ , not the arithmetic sum — because the resistor voltage and inductor voltage are $90^\circ$ out of phase.

When the voltmeter shows the same reading across R and L, we have $V_R = V_L = V$ (say). The source RMS voltage is given as $100\sqrt{2}$ V. The phasor relationship is:
$V_s^2 = V_R^2 + V_L^2 = V^2 + V^2 = 2V^2$
So $V = \frac{V_s}{\sqrt{2}} = \frac{100\sqrt{2}}{\sqrt{2}} = 100$ V.

That is the reading on the voltmeter — $100$ V.

Understand the circuit and the source.

The source is $100\sqrt{2}$ V, 50 Hz AC. The voltmeter reads RMS values (standard for AC voltmeters). The resistor and ideal inductor are in series.

Phase relationship.

In a series RL circuit, current $I$ is common. Voltage across resistor $V_R = IR$ is in phase with current. Voltage across inductor $V_L = I X_L$ leads current by $90^\circ$ . So $V_R$ and $V_L$ are perpendicular phasors.

Given condition: equal readings.

The voltmeter reads the same value across R and L. So $V_R = V_L = V$ (say).

Apply phasor addition for the source voltage.

The source RMS voltage $V_s$ is the magnitude of the phasor sum:

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

Solve for $V$ . …
Start your 14-day free trial to see the full solution →

🎓Unlock everything free for 14 days

Unlock everything free for 14 days