Question

An $L C R$ 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 $60\, V$ and $40 \,V$ respectively. The values of $L$ and $C$ are

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

Answer 1

The Correct Option is A

At resonance, voltage across resistance is $60 V$
$\Rightarrow 60=I_{0} R$
$\Rightarrow I_{0}=\frac{60}{120}=0.5 \,A$
Also, voltage across inductance is $40\, V$
$\Rightarrow 40=I_{0} X_{L} $
$\Rightarrow 80=L\left(4 \times 10^{5}\right) $
$\Rightarrow L=0.2\, m H $
Since, $\omega_{0}=\frac{1}{\sqrt{L C}} $
$4 \times 10^{5}=\frac{1}{\sqrt{0.2 \times 10^{-3} C}}$
$\Rightarrow C=\frac{1}{32} \mu F$