Question

In a series $LCR$ circuit, voltage across $R$ is 100V and $R = 1k\Omega$ with $C = 2\mu F$. The resonant frequency $\omega$ is 200 rad/s. At resonance the voltage across $L$ is

• A. $4 \times 10^{-3}V$
• B. $2.5 \times 10^{-2}V$
• C. 40 V
• D. 250 V

Answer 1

Answer: (D)

250 V

Answer 2

In a series $LCR$ circuit at resonance, the inductive reactance ($X_L$) is equal to the capacitive reactance ($X_C$).
$X_L = X_C = \frac{1}{\omega C}$

The current flowing through the circuit at resonance can be found using the voltage across the resistor ($V_R$) and the resistance ($R$). Since it's a series circuit, the same current flows through all components.
Given $V_R = 100$ V and $R = 1$ k$\Omega = 1000$ $\Omega$.
The current $I$ is given by Ohm's law:
$I = \frac{V_R}{R} = \frac{100 \text{ V}}{1000 \text{ }\Omega} = 0.1 \text{ A}$

The voltage across the inductor ($V_L$) is given by the product of the current ($I$) and the inductive reactance ($X_L$).
$V_L = I \times X_L$

At resonance, $X_L = X_C = \frac{1}{\omega C}$.
Given $\omega = 200$ rad/s and $C = 2$ $\mu$F $= 2 \times 10^{-6}$ F.
Calculate the capacitive reactance $X_C$:
$X_C = \frac{1}{\omega C} = \frac{1}{(200 \text{ rad/s}) \times (2 \times 10^{-6} \text{ F})} = \frac{1}{400 \times 10^{-6}} = \frac{1}{4 \times 10^{-4}}$
$X_C = \frac{10^4}{4} = 2500 \text{ }\Omega$

Since $X_L = X_C$ at resonance, $X_L = 2500$ $\Omega$.

Now calculate the voltage across the inductor $V_L$:
$V_L = I \times X_L = (0.1 \text{ A}) \times (2500 \text{ }\Omega) = 250 \text{ V}$

The voltage across the inductor at resonance is 250 V.