Question

An $LCR$ series circuit containing a resistance of $120\Omega$ has angular resonance frequency $4\times {10}^5$ rad $S^{-1}$. At resonance the voltage across resistance and inductance are $60V$ and $40V$ respectively. Then, the values of $L$ and $C$ are respectively.

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

Answer 1

Answer: (A)

$0.2mH$, $\frac{1}{32}\mu F$

$i_R=\frac{V_R}{R}=i_C=i_L$

$=\frac{60}{120}=0.5A$

$i_L=0.5A=\frac{V_L}{X_L}$

$X_L=\frac{V_L}{0.5}$

$=\frac{40}{0.5}=80$

$X_L=wL$

$80=4\times {10}^{5}L$

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

$X_C=\frac{V_L}{0.5}$

$=\frac{40}{0.5}$ $(\because$ At resonance $V_C=V_L)$

$X_C=\frac{1}{\omega C}$

$C=\frac{1}{\omega X_C}$

$=\frac{1}{4\times {10}^5\times 80}$

$=\frac{{10}^{-6}}{32}=\frac{1}{32}\mu F$