(a) An ac source v = vm sinω t is connected across an ideal capacitor. Derive the expression for (i) the curre

Question

v = v_m \sin\omega t

(i) the current flowing in the circuit, and

(ii) the reactance of the capacitor. Plot a graph of current

i

versus

\omega t

(OR)

(b) A series combination of an inductor

L

, a capacitor

C

and a resistor

R

is connected across an ac source of voltage in a circuit. Obtain an expression for the average power consumed by the circuit. Find the power factor for

(i) a purely inductive circuit, and

(ii) a purely resistive circuit.

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Part (a)Concept understanding — AC Through a Capacitor

AC Through a Capacitor — From Intuition to the Exact Statement

Imagine a capacitor as a tiny, two-plate storage tank for charge. When you connect it to a DC battery, it charges up quickly and then blocks any further current — that's why a capacitor is an open circuit for steady DC. But AC is different: the voltage keeps reversing, so the capacitor never gets a chance to settle. It is constantly being charged, discharged, charged the other way, discharged again — and that motion of charge is an alternating current.

The key intuition: current flows because the voltage is changing. If the voltage were steady, no current would flow. The faster the voltage changes, the larger the current. This is the opposite of a resistor, where current depends on the voltage itself, not its rate of change.

The Mathematical Link

For a capacitor, the charge stored is $Q = C V$ . Current is the rate of flow of charge: $I = dQ/dt$ . So:

I = C \frac{dV}{dt}

This single equation is the whole story. If the applied voltage is sinusoidal, say $V = V_0 \sin(\omega t)$ , then:

I = C \frac{d}{dt}[V_0 \sin(\omega t)] = C V_0 \omega \cos(\omega t)

Now compare the two waveforms:

Voltage: $V_0 \sin(\omega t)$ — starts at zero, rises to peak.
Current: $C V_0 \omega \cos(\omega t)$ — starts at its maximum value, then falls.

A cosine is a sine shifted forward by $90^\circ$ (or $\pi/2$ radians). So the current reaches its peak a quarter-cycle before the voltage does. That is the famous result: in a purely capacitive circuit, current leads voltage by $90^\circ$ .

Important

The phase relation: $I$ leads $V$ by $90^\circ$ in a pure capacitor. Equivalently, $V$ lags $I$ by $90^\circ$ .

Why "Leads" and Not "Lags"?

Think physically. At the instant you first apply the AC voltage, the voltage is zero but rising fastest (the slope of $\sin$ is maximum at zero). A fast-changing voltage means a large current. So the current is already at its peak while the voltage is still near zero. That is the meaning of "leading" — the current's peak comes first.

Later, when the voltage reaches its peak, it is momentarily not changing (slope = 0), so the current drops to zero. The current is always ahead of the voltage by exactly one quarter-cycle.

The Limiting Factor: Capacitive Reactance

From the current expression above, the peak current is:

I_0 = \omega C V_0

This looks like Ohm's law if we define an effective resistance-like quantity:

X_C = \frac{V_0}{I_0} = \frac{1}{\omega C}

This $X_C$ is called capacitive reactance. It has units of ohms, but it is not a resistance — it does not dissipate energy. It merely limits the current by the capacitor's opposition to changes in voltage.

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

Key points about $X_C$ :

It is inversely proportional to frequency. At high $f$ , the voltage changes rapidly, so the current is large — low reactance. At low $f$ , the voltage changes slowly, so the current is small — high reactance. At DC ( $f = 0$ ), $X_C \to \infty$ , which is the open-circuit behaviour you already know.
It is also inversely proportional to capacitance $C$ . A larger capacitor stores more charge per volt, so for the same voltage change it pushes more current — lower reactance.

The Complete Picture in One Table

Property	Resistor	Capacitor
Relation	$V = IR$	$I = C\,dV/dt$

Part (b)Concept understanding — Average Power Absorption

Average Power Absorption – From Intuition to Precision

Think of pushing a child on a swing. You don't push constantly — you push only when the swing is moving away from you, and you time your push to add energy each time. Some pushes land perfectly, others might be slightly off. Over several minutes, what matters is not the force at any single instant, but the net energy you transferred averaged over time.

That's the core idea behind average power absorption: how much energy, on average, is being delivered per unit time to a device or system, even when the instantaneous power fluctuates wildly.

The Intuitive Picture

Consider a light bulb connected to household AC supply. The voltage oscillates 50 times per second (in India). At the peak of the voltage cycle, the bulb glows brightest; when voltage crosses zero, the bulb goes dark for an instant. But you don't see flickering — your eyes average out the rapid changes. What you perceive as "brightness" corresponds to the average power the bulb absorbs.

Similarly, when you charge a phone battery, the power drawn isn't constant — it's high when the battery is low, then tapers off. The "charging speed" you care about is the average power over the charging session.

The Precise Definition

P_{\text{avg}} = \frac{1}{T} \int_{0}^{T} p(t) \, dt

Where:

$p(t)$ is the instantaneous power at time $t$ (in watts)
$T$ is the time period over which we average (in seconds)

For a resistor with a sinusoidal voltage $v(t) = V_m \sin(\omega t)$ and current $i(t) = I_m \sin(\omega t)$ (since they're in phase), the instantaneous power is:

p(t) = v(t) \cdot i(t) = V_m I_m \sin^2(\omega t)

This is always positive (since $\sin^2$ is never negative) but it oscillates between $0$ and $V_m I_m$ . The average over one complete cycle gives:

P_{\text{avg}} = \frac{V_m I_m}{2}

Why This Matters for Exams

The most common mistake students make is confusing peak power with average power. A 100 W bulb doesn't draw 100 W at every instant — it draws about 200 W at the voltage peak and 0 W at the zero crossing. The 100 W rating is the average power it's designed to dissipate safely.

Watch out

Never use $P = VI$ directly with AC peak values unless you divide by 2 (for sinusoidal waveforms). The correct formula for average power in a resistor is $P_{\text{avg}} = \frac{V_m I_m}{2} = V_{\text{rms}} I_{\text{rms}}$ , where $V_{\text{rms}} = V_m / \sqrt{2}$ .

The General Case (Phase Differences)

When voltage and current are not in phase — as in circuits with inductors or capacitors — the instantaneous power can become negative during parts of the cycle (energy flows back to the source). The average power then becomes:

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

Why this formula?

Average Power Absorption: Why the Formula Holds

Let's build this from first principles — understanding why average power is what it is, not just memorising the formula.

1. Instantaneous Power — The Starting Point

For any circuit element, instantaneous power is always:

p(t) = v(t) \cdot i(t)

This is the fundamental definition: power at an instant is voltage times current at that same instant.

2. Why We Need an Average

In AC circuits, both $v(t)$ and $i(t)$ vary sinusoidally with time. So $p(t)$ also varies — often at twice the frequency of the original signals.

Instantaneous power oscillates between zero and a peak value.
What matters for real energy consumption is the average over a complete cycle.

Hence, we define:

P_{\text{avg}} = \frac{1}{T} \int_0^T p(t) \, dt

where $T$ is the time period of the AC waveform.

3. The Key Derivation (Step-by-Step)

Step 1: Write the sinusoidal forms

Let:

$v(t) = V_m \cos(\omega t + \theta_v)$
$i(t) = I_m \cos(\omega t + \theta_i)$

Here $\theta_v$ and $\theta_i$ are phase angles. The phase difference is:

\phi = \theta_v - \theta_i

Step 2: Instantaneous power

p(t) = V_m I_m \cos(\omega t + \theta_v) \cos(\omega t + \theta_i)

Use the trigonometric identity:

\cos A \cos B = \frac{1}{2}[\cos(A-B) + \cos(A+B)]

So:

p(t) = \frac{V_m I_m}{2} \left[ \cos(\theta_v - \theta_i) + \cos(2\omega t + \theta_v + \theta_i) \right]

Step 3: Average over one cycle

The average of $\cos(2\omega t + \text{constant})$ over a full cycle is zero — because it's a sinusoid symmetric about zero.

Only the constant term survives:

P_{\text{avg}} = \frac{V_m I_m}{2} \cos(\phi)

4. The Standard Form Using RMS Values

Recall:

$V_{\text{rms}} = \frac{V_m}{\sqrt{2}}$
$I_{\text{rms}} = \frac{I_m}{\sqrt{2}}$

Therefore:

\frac{V_m I_m}{2} = V_{\text{rms}} I_{\text{rms}}

So the final formula is:

\boxed{P_{\text{avg}} = V_{\text{rms}} \, I_{\text{rms}} \, \cos \phi}

5. What $\cos \phi$ Really Means

$\phi$ is the phase difference between voltage and current.
$\cos \phi$ is called the power factor.
Why it appears: Only the component of current in phase with voltage contributes to average power. The quadrature (90° out-of-phase) component averages to zero.

| $\phi$ | $\cos \phi$ | Interpretation | …

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Part (a): For an ideal capacitor $i=\omega C v_m\cos\omega t$ leads $v$ by $90^\circ$ , and $X_C=1/(\omega C)$ .

Part (b): For a series LCR circuit $P_{av}=V_{rms}I_{rms}\cos\phi$ ; the power factor is $0$ for a purely inductive circuit and $1$ for a purely resistive one.

Part (a): AC source across an ideal capacitor

Unlike a resistor, a capacitor's current depends on the rate of change of voltage, so voltage and current are not in phase.

1. Charge and current. The capacitor voltage equals the source: $v=v_m\sin\omega t$ , and $q=Cv$ . Current is

i=\frac{dq}{dt}=C\frac{dv}{dt}=C\frac{d}{dt}(v_m\sin\omega t)=\omega C v_m\cos\omega t.

Writing $\cos\omega t=\sin(\omega t+\pi/2)$ ,

i=i_m\sin\!\left(\omega t+\frac{\pi}{2}\right),\qquad i_m=\omega C v_m.

Important

In a purely capacitive circuit the current leads the voltage by $\pi/2$ (mnemonic CIVIL: in a C, I leads V).

2. Capacitive reactance. By analogy with Ohm's law,

X_C=\frac{v_m}{i_m}=\frac{1}{\omega C}=\frac{1}{2\pi f C}.

3. Graph of $i$ vs $\omega t$ .

$\omega t$	$v=v_m\sin\omega t$	$i=i_m\sin(\omega t+\pi/2)$
$0$	$0$	$+i_m$
$\pi/2$	$+v_m$	$0$
$\pi$	$0$	$-i_m$
$3\pi/2$	$-v_m$	$0$
$2\pi$	$0$	$+i_m$

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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 31 on this concept.

CBSE 2026Set A1 markMCQ
$Q. In an ac circuit of capacitance the current from potential difference is (A) forward (B) backward (C) both are in same phase (D) none of these$
›Reveal solutionSolution

In a pure capacitor the current leads the applied voltage by π/2, i.e. current is ahead (forward).

For a capacitor driven by $V=V_0\sin\omega t$ , the charge is $q=CV$ and the current is

$I=\frac{dq}{dt}=\omega C V_0\cos\omega t=\omega C V_0\sin\!\left(\omega t+\tfrac{\pi}{2}\right).$

…

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CBSE 2025Set D1 markMCQ
$Q. In an a.c. circuit containing only capacitor, the phase difference between current and voltage is (A) 0° (B) 90° (C) 180° (D) 45°$
›Reveal solutionSolution

In a pure capacitor the current leads the applied voltage by a quarter cycle, i.e. a phase difference of 90°.

For an a.c. circuit containing only a capacitor, the charging current is maximum when the voltage is zero and vice versa. The current leads the voltage by

…

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CBSE 2025Set ANNUAL1 mark
$Q. Draw the phasor diagram to represent current and supply voltage for an AC circuit containing capacitance only.$
›Reveal solutionSolution
Figure — Explicit 'Draw the phasor diagram ... AC circuit containing capacitance only' hard gate. Catalog fig-7-8 is ex

In a pure capacitor, $I$ leads $V$ by $\pi/2$ .

In an AC circuit containing only a capacitor, the current leads the applied voltage by a phase angle of $90°$ ( $\pi/2$ radians) — this follows from $I = C\dfrac{dV}{dt}$ , since the current is proportional to the rate of change of voltage, which is largest when V is crossing zero and zero when V is at its peak. In the phasor diagram, the voltage phasor $V_0$ is drawn along the reference (horizontal) axis, and the current phasor $I_0$ is drawn rot …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. The average power dissipated in a pure inductor is$
(i) VI^2
(ii) zero
(iii) (1/2)VI
(iv) VI^2/4
›Reveal solutionSolution

A pure inductor has phase angle 90 degrees, so average power = V I cos(90) = 0.

Average power in an AC circuit is $P_{avg} = V_{rms} I_{rms}\cos\phi$ . In a purely inductive circuit the current lags the applied voltage by exactly $90^\circ$ , so $\cos\phi = \cos 90^\circ = 0$ . Hence $P_{avg} = 0$ : over a full cycle e …

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CBSE 2024Set FS1 markMCQ
$Q. The power consumed in alternating current in circuit containing only capacitor will be:$
(i) $P = -1$
(ii) $P = 0$
(iii) $P = +1$
(iv) None of the above
›Reveal solutionSolution

In a purely capacitive AC circuit current leads voltage by $90^\circ$ , so $\cos\phi=0$ and the average power consumed is zero — option (ii).

Concept. Average AC power is $P=V_{rms}I_{rms}\cos\phi$ , where $\phi$ is the phase angle between voltage and current.

Why zero. In a circuit with only a capacitor, the current leads the voltage by exactly $\phi=90^\circ$ . Then …

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CBSE 2024Set A1 markMCQ
$Q. If the phase difference between alternating current and e.m.f. is φ, then the value of power factor is (A) cos φ (B) cos^2 φ (C) sin φ (D) tan φ$
›Reveal solutionSolution

Power factor = cos φ.

The average power dissipated in an AC circuit is

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

where φ is the phase difference between the applied emf and the current. The factor $\cos\phi$ that determines what fraction of the apparent power $V_{rms}I_{rms}$ is actually consumed is called the power factor.

…

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CBSE 2024Set A1 markMCQ
$Q. In AC circuit, power is lost in only (A) resistance (B) inductance (C) capacitance (D) all of these$
›Reveal solutionSolution
Only resistance dissipates (real) power in AC; ideal L and C do not.

In an AC circuit the average power is $P = V_{rms}I_{rms}\cos\phi$ .

In a resistor, current and voltage are in phase (φ = 0), so cos φ = 1 and power $I_{rms}^2R$ is dissipated as heat. …
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CBSE 2024Set A1 mark
$Q. Write the value of power factor for a pure resistive circuit.$
›Reveal solutionSolution

In a pure resistor, V and I are in phase, so power factor cos φ = cos 0° = 1.

The average power dissipated in an AC circuit is $P_{avg} = V_{rms}I_{rms}\cos\phi$ , where $\cos\phi$ is called the power factor and φ is the phase difference between the current and the voltage.

In a purely resistive circuit, the current is always in phase with the applied voltage (φ = 0°), because a resistor offers no reactance. Therefore:

$\cos\phi = \cos 0^\circ = 1$ …

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CBSE 2024Set ANNUAL1 markMCQ
$Q. For an alternating current i = I0 sinωt passing through a resistor R, how much is the average power loss due to Joule heating ?$
(a) I0²R
(b) (1/2) I0²R
(c) 4 I0²R
(d) 2 I0²R
›Reveal solutionSolution

Average power in a resistor over one AC cycle is Irms²R, and Irms = I0/√2 for a sinusoidal current.

For i = I0 sin ωt through a resistor R, the instantaneous power is p = i²R = I0²R sin²ωt.

Averaging sin²ωt over a full cycle gives 1/2, so:

Pavg = I0²R × (1/2) = (1/2) I0²R …

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CBSE 2024Set ANNUAL1 mark
$Q. What is wattless current?$
›Reveal solutionSolution

The quadrature (reactive) current component, $I_{rms}\sin\phi$ , that transfers no net energy.

In an AC circuit with a phase difference $\phi$ between voltage and current, the average power is $P = V_{rms}I_{rms}\cos\phi$ . The current can be resolved into two components: one in phase with the voltage, $I_{rms}\cos\phi$ (which does work / consumes power), and one $90°$ out of phase with the voltage, $I_{rms}\sin\phi$ (the quadrature component). This second component does no net work over a complete cycle — it is called the wattless current (or idle current). It occurs purely in ideal (resistance-free) inductive or capacitive circuits, w …

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CBSE 2023Set MODEL1 markMCQ
$Q. In an LCR circuit, the power factor becomes 0 (zero) when:$
(a) $R = 0$
(b) $\omega L = \omega C$
(c) $\omega L = \dfrac{1}{\omega C}$
(d) $\left(\omega L - \dfrac{1}{\omega C}\right) = R$
›Reveal solutionSolution

Power factor is zero only for a purely reactive (resistance-less) AC circuit.

In an LCR circuit, the power factor is $\cos\phi = \dfrac{R}{Z}$ , where $Z=\sqrt{R^2+(\omega L-\tfrac{1}{\omega C})^2}$ . The power factor becomes zero (i.e. $\phi = 90^\circ$ , purely reactive circuit, average power consumed is zero) only when $R=0$ . …

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CBSE 2023Set F1 markMCQ
$Q. The phase-difference between current and voltage in only capacitive alternating current circuit is (A) 0° (B) 90° (C) 180° (D) 45°$
›Reveal solutionSolution

In a purely capacitive AC circuit, current leads voltage by a phase angle of 90°.

For a pure capacitor, if $V = V_0\sin\omega t$ then the charge $q = CV_0\sin\omega t$ and the current

$I = \dfrac{dq}{dt} = \omega C V_0\cos\omega t = I_0\sin\!\left(\omega t + \dfrac{\pi}{2}\right)$

…

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Q.(a) An ac source v=vmsin⁡ωtv = v_m \sin\omega tv=vm​sinωt is connected across an ideal capacitor. Derive the expression for

Part (a): AC source across an ideal capacitor

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Q.(a) An ac source $v = v_m \sin\omega t$ is connected across an ideal capacitor. Derive the expression for

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