Question

A resistance of $100\ \Omega$, inductor of self-inductance $\left(\dfrac{4}{\pi^2}\right)\ \text{H}$ and a capacitor of unknown capacitance are connected in series to an a.c. source of $200\ \text{V}$ and $50\ \text{Hz}$. When the current and voltage are in phase, the capacitance and power dissipated respectively are

• A. $2.5 \times 10^{-5}\ \text{F},\ 400\ \text{W}$
• B. $1.5 \times 10^{-5}\ \text{F},\ 200\ \text{W}$
• C. $2.0 \times 10^{-5}\ \text{F},\ 100\ \text{W}$
• D. $3.0 \times 10^{-5}\ \text{F},\ 50\ \text{W}$

Hint:

At resonance in LCR circuit, impedance is minimum and equals resistance.

Answer 1

The Correct Option is A

Step 1: Condition for current and voltage in phase.
For series LCR circuit, current and voltage are in phase at resonance: \[ X_L = X_C \] Step 2: Write expressions for reactances.
\[ X_L = \omega L,\quad X_C = \dfrac{1}{\omega C} \] Step 3: Equate reactances.
\[ \omega L = \dfrac{1}{\omega C} \] \[ C = \dfrac{1}{\omega^2 L} \] Step 4: Substitute given values.
\[ \omega = 2\pi f = 100\pi \] \[ L = \dfrac{4}{\pi^2} \] \[ C = \dfrac{1}{(100\pi)^2 \cdot \dfrac{4}{\pi^2}} = 2.5 \times 10^{-5}\ \text{F} \] Step 5: Calculate power dissipated.
At resonance, impedance $Z = R = 100\ \Omega$
\[ P = \dfrac{V^2}{R} = \dfrac{200^2}{100} = 400\ \text{W} \]