Question

An $\text{LCR}$ series circuit containing a resistance of $120\Omega$ has angular resonance frequency $4\times {10}^5\text{rad/s}$. At resonance, the voltage across resistance and inductance are $60\text{V}$ and $40\text{V}$ respectively. Then the values of $\text{L}$ and $\text{C}$ are respectively.

• A. $0.4\text{mH};\frac{1}{32}\mu \text{F}$
• B. $0.2\text{mH};\frac{1}{16}\mu \text{F}$
• C. $0.4\text{mH};\frac{1}{16}\mu \text{F}$
• D. $0.2\text{mH};\frac{1}{32}\mu \text{F}$

Answer 1

The correct option is D $0.2\text{mH};\frac{1}{32}\mu \text{F}$

Given , $R=120\Omega ,{\omega }_0=4\times {10}^5{\text{rads}}^{-1},V_R=60\text{V};V_L=40\text{V}$

At resonance , $V_{rms}=V_R=60\text{V}$

$\therefore I_{rms}=\frac{V_R}{R}=\frac{60}{120}=0.5\text{A}$

Thus, $V_L=V_C$

$V_L=I_{rms}\times L\omega =0.5\times L\times 4\times {10}^5$

$40=2\times {10}^5\times L$

$\Rightarrow L=2\times {10}^{-4}\text{H}=0.2\text{mH}$

$V_C=I_{rms}\times \frac{1}{C\omega }$

$40=0.5\times \frac{1}{4\times {10}^{-5}C}$

$\therefore C=\frac{0.5}{16}\times {10}^{-6}=\frac{1}{32}\mu \text{F}$

Hence, option (d) is the correct answer.