CBSE 2025, Set 2

Question

(a) (i) Write the principle of working of an ac generator. Draw its labelled diagram and explain its working.
(ii) A resistor of $400\ \Omega$, an inductor of $(\frac{5}{\pi})	ext{ H}$ and a capacitor of $(\frac{50}{\pi})\ \mu	ext{F}$ are joined in series across an ac source $v = 140 \sin(100\pi)t	ext{ V}$. Find the rms voltages across these three circuit elements. The algebraic sum of these voltages is more than the rms voltage of source. Explain.

Edvaya · Accepted Answer

The correct answer is Subjective. **(i) AC Generator:**
* **Principle:** It works on the principle of **electromagnetic induction**. When a coil is rotated in a uniform magnetic field, the magnetic flux linked with it changes continuously, inducing an alternating emf in the coil.

![AC Generator](https://www.edvaya.com/images/cbse/physics/2025/set2/q33a-ac-generator-answer.webp)

* **Working:** An armature coil is mechanically rotated in a uniform magnetic field provided by strong permanent magnets. As the coil rotates, the angle $	heta$ between the area vector and the magnetic field changes ($	heta = \omega t$). This changes the magnetic flux ($\Phi = NBA \cos\omega t$), inducing an alternating emf ($\epsilon = -d\Phi/dt = NBA\omega \sin\omega t$). The induced current is drawn out through slip rings and carbon brushes to the external circuit.

**(ii) Series LCR Circuit Calculations:**
Given: $R = 400\ \Omega$, $L = \frac{5}{\pi}	ext{ H}$, $C = \frac{50}{\pi} 	imes 10^{-6}	ext{ F}$, $v = 140 \sin(100\pi t)	ext{ V}$.
From voltage equation: Peak voltage $V_0 = 140	ext{ V}$, Angular frequency $\omega = 100\pi	ext{ rad/s}$.

* Inductive Reactance: $X_L = \omega L = 100\pi 	imes (\frac{5}{\pi}) = 500\ \Omega$
* Capacitive Reactance: $X_C = \frac{1}{\omega C} = \frac{1}{100\pi 	imes (50	imes 10^{-6}/\pi)} = \frac{10^6}{5000} = 200\ \Omega$
* Impedance: $Z = \sqrt{R^2 + (X_L - X_C)^2} = \sqrt{400^2 + (500 - 200)^2} = \sqrt{400^2 + 300^2} = 500\ \Omega$

* RMS Source Voltage: $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{140}{\sqrt{2}}	ext{ V}$
* RMS Current: $I_{rms} = \frac{V_{rms}}{Z} = \frac{140/\sqrt{2}}{500} = \frac{0.28}{\sqrt{2}}	ext{ A}$

* **Voltage across Resistor:** $(V_{rms})_R = I_{rms} R = \left(\frac{0.28}{\sqrt{2}}ight) 	imes 400 = \boxed{56\sqrt{2}	ext{ V}}$
* **Voltage across Inductor:** $(V_{rms})_L = I_{rms} X_L = \left(\frac{0.28}{\sqrt{2}}ight) 	imes 500 = \boxed{70\sqrt{2}	ext{ V}}$
* **Voltage across Capacitor:** $(V_{rms})_C = I_{rms} X_C = \left(\frac{0.28}{\sqrt{2}}ight) 	imes 200 = \boxed{28\sqrt{2}	ext{ V}}$

**Explanation:** The algebraic sum of these voltages ($154\sqrt{2}	ext{ V}$) is greater than the source voltage ($70\sqrt{2}	ext{ V}$) because the voltages across $R, L$, and $C$ are **not in phase**. $V_L$ leads the current by $90^\circ$ and $V_C$ lags by $90^\circ$, so they must be added as phasors ($V = \sqrt{V_R^2 + (V_L - V_C)^2}$), not algebraically.

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