Q165 / 180 · COMEDK 2025 (25 May Shift) · Physics
A series LCR circuit is connected to an AC source of 120 V and 60 Hz . The value of resistance R in the circuit is $100 \Omega$.A voltmeter reads the voltage across R as 50 V , across the inductor as 132 V and across the capacitor as $\dfrac{70}{6.6} \mathrm{~V}$. The value of inductance and capacitance in the circuit is:
- Correct answerA0.7 H and $125 \mu \mathrm{~F}$
- B7 H and $0.0125 \mu \mathrm{~F}$
- C0.07 H and $0.0125 \mu \mathrm{~F}$
- D0.007 and $12.5 \mu \mathrm{~F}$
Solution
Given $V_{R} = 50 \text{ V}$, $V_{L} = 132 \text{ V}$, $V_{C} = \dfrac{70}{6.6} \text{ V} \approx 10.6 \text{ V}$, and $R = 100 \Omega$.
The current in the series LCR circuit is $I = \dfrac{V_{R}}{R} = \dfrac{50}{100} = 0.5 \text{ A}$.
The inductive reactance $X_{L}$ is given by $X_{L} = \dfrac{V_{L}}{I} = \dfrac{132}{0.5} = 264 \Omega$.
Since $X_{L} = 2\pi f L$, where $f = 60 \text{ Hz}$, we have $L = \dfrac{264}{2 \times \pi \times 60} = \dfrac{264}{377} \approx 0.7 \text{ H}$.
The capacitive reactance $X_{C}$ is given by $X_{C} = \dfrac{V_{C}}{I} = \dfrac{70 / 6.6}{0.5} = \dfrac{140}{6.6} \approx 21.21 \Omega$.
Since $X_{C} = \dfrac{1}{2\pi f C}$, we have $C = \dfrac{1}{2 \times \pi \times 60 \times (140 / 6.6)} = \dfrac{6.6}{2 \times \pi \times 60 \times 140} = \dfrac{6.6}{52778} \approx 1.25 \times 10^{-4} \text{ F} = 125 \mu \text{F}$.
Comparing these values with the options, the inductance is $0.7 \text{ H}$ and the capacitance is $125 \mu \text{F}$.
Answer: 0.7 H and $125 \mu \mathrm{~F}$
The current in the series LCR circuit is $I = \dfrac{V_{R}}{R} = \dfrac{50}{100} = 0.5 \text{ A}$.
The inductive reactance $X_{L}$ is given by $X_{L} = \dfrac{V_{L}}{I} = \dfrac{132}{0.5} = 264 \Omega$.
Since $X_{L} = 2\pi f L$, where $f = 60 \text{ Hz}$, we have $L = \dfrac{264}{2 \times \pi \times 60} = \dfrac{264}{377} \approx 0.7 \text{ H}$.
The capacitive reactance $X_{C}$ is given by $X_{C} = \dfrac{V_{C}}{I} = \dfrac{70 / 6.6}{0.5} = \dfrac{140}{6.6} \approx 21.21 \Omega$.
Since $X_{C} = \dfrac{1}{2\pi f C}$, we have $C = \dfrac{1}{2 \times \pi \times 60 \times (140 / 6.6)} = \dfrac{6.6}{2 \times \pi \times 60 \times 140} = \dfrac{6.6}{52778} \approx 1.25 \times 10^{-4} \text{ F} = 125 \mu \text{F}$.
Comparing these values with the options, the inductance is $0.7 \text{ H}$ and the capacitance is $125 \mu \text{F}$.
Answer: 0.7 H and $125 \mu \mathrm{~F}$
Hence, the correct answer is option: 'A'.