A resistor $R=40\ \Omega$ and a capacitor $C=100\ \mu\text{F}$ are connected in series to a $200\ \text{V}$ (rms), $50\ \text{Hz}$ AC source. Calculate (a) the capacitive reactance, (b) the impedance of the circuit, (c) the rms current, and (d) the phase angle by which the current leads the voltage.

**Given:** $R=40\ \Omega$, $C=100\ \mu\text{F}$, $V_{rms}=200\ \text{V}$, $f=50\ \text{Hz}$. **(a) Capacitive reactance.** $$X_C = \frac{1}{2\pi(50)(100\times10^{-6})} = \frac{1}{0.0314} \approx 31.83\ \Omega$$ **(b) Impedance.** $$Z = \sqrt{R^2+X_C^2} = \sqrt{40^2+31.83^2} = \sqrt{1600+1013.2} = \sqrt{2613.2} \approx 51.12\ \Omega$$ **(c) RMS current.** $$I_{rms} = \frac{200}{51.12} \approx 3.91\ \text{A}$$ **(d) Phase angle.** …

A resistor R=40Ω and a capacitor C=100muF are connected in series to a 200V (rms), 50Hz AC source. Calculate (

Numerical · Q27

R=40\ \Omega

(a) the capacitive reactance,

(b) the impedance of the circuit,

(d) the phase angle by which the current leads the voltage.

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Given: $R=40\ \Omega$ , $C=100\ \mu\text{F}$ , $V_{rms}=200\ \text{V}$ , $f=50\ \text{Hz}$ .

Capacitive reactance. $X_C = \frac{1}{2\pi(50)(100\times10^{-6})} = \frac{1}{0.0314} \approx 31.83\ \Omega$
Impedance. $Z = \sqrt{R^2+X_C^2} = \sqrt{40^2+31.83^2} = \sqrt{1600+1013.2} = \sqrt{2613.2} \approx 51.12\ \Omega$
RMS current. $I_{rms} = \frac{200}{51.12} \approx 3.91\ \text{A}$
Phase angle. …

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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. …
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Q.A resistor $R=40\ \Omega$ and a capacitor $C=100\ \mu\text{F}$ are connected in series to a $200\ \text{V}$ (rms), $50\ \text{Hz}$ AC source. Calculate

Concept understanding — Impedance of a CR Circuit

Q.A resistor R=40 ΩR=40\ \OmegaR=40 Ω and a capacitor C=100 μFC=100\ \mu\text{F}C=100 μF are connected in series to a 200 V200\ \text{V}200 V (rms), 50 Hz50\ \text{Hz}50 Hz AC source. Calculate

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Q.A resistor $R=40\ \Omega$ and a capacitor $C=100\ \mu\text{F}$ are connected in series to a $200\ \text{V}$ (rms), $50\ \text{Hz}$ AC source. Calculate

Unlock everything free for 14 days