Question

A series LCR circuit containing a resistance of 120 ohm has angular resonance frequency 4 x ${10}^{3}rad{s}^{-1}$. At resonance, the voltage across resistance and inductance are 60V and 40V respectively. The values of L and C are respectively :

• A. 20 mH, 25/8 $\mu F$
• B. 2 mH, 1/35 $\mu F$
• C. 20 mH, 1/40 $\mu F$
• D. 2 mH, 25/8 nF

Answer 1

The correct option is A 20 mH, 25/8 $\mu F$

Resistance$,R=120\Omega$

angular resonance frequency $w=4\times {10}^{3}rad/s$

voltage across resistance $V_R=60V$ and the voltage across inductance i$,V_L=40V$

Now$,$ potential difference across $R$ is

$V_R=IR$

$\Rightarrow 60=I\times 120$

$\Rightarrow I=\frac{1}{2}A$

Now$,$ potential difference across $L$ is$,$

$V_L=IwL$

$\Rightarrow L=\frac{V_L}{Iw}=\frac{40}{\frac{1}{2}\times 4\times {10}^3}=20mH$

Now at resonance$,$ $,wL=\frac{1}{wC}$

$\Rightarrow C=\frac{1}{w^{2}L}=\frac{1000}{\left(4\times {10}^3\right)\times 20}=\frac{25}{8}\mu F$

Hence,

option $\left(A\right)$ is correct answer.