For a series CR circuit connected to an AC source, derive the expression for the impedance Z and the phase ang

Exercise · Q11

Z

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✓ Free question

Setting up the phasor diagram. With the shared current $I$ as reference, $V_R=IR$ is in phase with $I$ . The capacitor's voltage $V_C=IX_C$ lags the current by $90^\circ$ (since the current itself leads the voltage across a pure capacitor), so its phasor is drawn perpendicular to $V_R$ , but on the OPPOSITE side from where the inductor's $V_L$ would sit.

Adding the phasors.

V = \sqrt{V_R^2+V_C^2} = I\sqrt{R^2+X_C^2} \quad\Rightarrow\quad Z=\sqrt{R^2+X_C^2}

Phase angle.

\tan\phi = \frac{V_C}{V_R} = \frac{X_C}{R}

with the CURRENT LEADING the applied voltage by this angle -- the exact geometric mirror image of the LR circuit's result, with $X_C$ in place of $X_L$ and the sense of the phase difference reversed.

✓Final answer

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

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

CBSE 2020Set 55/1/13 marks
$R$
(a) The peak value of the voltage across the
(i) resistance and
(ii) capacitor.
(b) The phase difference between the applied voltage and current. Which of them is ahead ?
›Reveal solutionSolution
In an RC series circuit driven by an AC source, the current is the same through both components, but the voltages across R and C are out of phase with each other. The peak voltage across R is $V_R = I_0 R$ , across C is $V_C = I_0 / (\omega C)$ , and the current leads the applied voltage by a phase angle $\phi = \tan^{-1}(1/(\omega RC))$ .

We start with the physics: In a series RC circuit, the same current flows through the resistor and the capacitor. But the voltage across a resistor is in phase with the current, while the voltage across a capacitor lags the current by $90^\circ$ (or equivalently, the current leads the capacitor voltage by $90^\circ$ ). This phase difference is the key to everything that follows.

The applied voltage $V = V_0 \sin \omega t$ must equal the sum of the instantaneous voltages across R and C. Because these voltages are not in phase, we cannot simply add their peak values — we must use phasor addition.
$Z = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}$
Let’s work through the problem step by step.

Find the peak current in the circuit. The impedance $Z$ of the series RC combination is the phasor sum of resistance $R$ and capacitive reactance $X_C = 1/(\omega C)$ .

$Z = \sqrt{R^2 + X_C^2} = \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}$
The peak value of the current is therefore:
$I_0 = \frac{V_0}{Z} = \frac{V_0}{\sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}}$

Peak voltage across the resistor. Since the resistor obeys Ohm’s law instantaneously, the peak voltage across it is simply:

$V_R = I_0 R = \frac{V_0 R}{\sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}}$
This is the answer for part (a)(i).

Peak voltage across the capacitor. For a capacitor, $V_C = I X_C$ (using peak values). So:

$V_C = I_0 \cdot \frac{1}{\omega C} = \frac{V_0}{\omega C \sqrt{R^2 + \left(\frac{1}{\omega C}\right)^2}}$
This is the answer for part (a)(ii).

Tip

Notice that $V_R$ and $V_C$ do NOT add up to $V_0$ directly. Instead, $V_0^2 = V_R^2 + V_C^2$ , which is a direct consequence of the $90^\circ$ phase difference between the two voltages.

Phase difference between applied voltage and current. In an RC circuit, the current leads the voltage. Why? Because the capacitor opposes changes in voltage — current must flow first to build up charge. The phase angle $\phi$ by which the current leads the voltage is given by:

$\tan \phi = \frac{X_C}{R} = \frac{1}{\omega RC}$
So:
$\phi = \tan^{-1}\left(\frac{1}{\omega RC}\right)$
This is the phase angle between the applied voltage and the current, with the current ahead of the voltage.

Watch out

A common mistake is to think the voltage leads the current in an RC circuit. That’s true for an RL circuit (inductor), but for a capacitor, it’s the opposite: current leads voltage.

Which is ahead? As stated, the current leads the applied voltage by $\phi$ . So the current waveform reaches its peak earlier than the voltage waveform.

✓Final answer
$\frac{V_0 R}{\sqrt{R^2 + (1/\omega C)^2}}$

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Q.For a series CR circuit connected to an AC source, derive the expression for the impedance $Z$ and the phase angle $\phi$ by which the current leads the voltage.

Concept understanding — Impedance of a CR Circuit

Q.For a series CR circuit connected to an AC source, derive the expression for the impedance ZZZ and the phase angle ϕ\phiϕ by which the current leads the voltage.

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Q.For a series CR circuit connected to an AC source, derive the expression for the impedance $Z$ and the phase angle $\phi$ by which the current leads the voltage.

Unlock everything free for 14 days