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

Answer: (A)

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

Answer 2

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, $\omega_{0}=\frac{1}{\sqrt{L C}}$
$4 \times 10^{5}=\frac{1}{\sqrt{0.2 \times 10^{-3} C}}$
$\Rightarrow C=\frac{1}{32} \mu F$