Question

A box contains L,C and R. When $250V$ dc is applied to the terminals of the box, a current of $1.0A$ flows in the circuit. When ac source of $250V$ rms at $2250rad/s$ connected, a current of $1.25A$ rms flows. It is observed that the current rises with frequency and becomes maximum at $4500$ rad/s. Find the values of L, C and R. Draw the circuit diagram.

Answer 1

In the box containing L, C and R a constant current of $1$ Amp can only flow when the circuit is as shown in the adjacent circuit.
Therefore, the effective circuit is $V=IR$
$\Rightarrow R=\frac{250}{1}=250\Omega ...\left(1\right)$
Now for ac as current is maximum at $4500$ rad/s
${\omega }_0=\frac{1}{\sqrt{LC}}\Rightarrow LC=\frac{1}{{\omega }_0^2}=\frac{1}{{4500}^2}...\left(2\right)$
and as for ac
$I_R=\frac{V}{R}\sin \omega t=\frac{250}{250}\sin \omega t=\sin \omega t;I_X=\frac{V}{X}\sin \omega t$
So, $I=I_R+I_X=\sin \omega t+\frac{V}{X}\sin \omega tI=I_0\sin (\omega t+\varphi )$
With $I_0\cos \varphi =1;I_0\sin \varphi =V/X$
ie., $I_0^2=1+{\left(\frac{V}{X}\right)}^{2}or{\left(1.25\right)}^2=1+{\left(\frac{250}{X}\right)}^2\Rightarrow X=\frac{250}{\sqrt{0.5625}}=\frac{1000}{3}\Omega$
ie., $\omega L\sim \frac{1}{\omega L}=\frac{1000}{3}(X=\omega L-\frac{1}{\omega C})$
or $2250L\sim \frac{1}{2250C}=\frac{1000}{3}$
Substituting the value of L in terms of C from equation (2) and (3) we have
$2250\frac{1}{C{\left(4500\right)}^2}\sim \frac{1}{2250C}=\frac{1000}{3}$
$\Rightarrow \frac{1}{C}\times \frac{3}{9000}=\frac{1000}{3}$
$C={10}^{-6}F=1\mu F$
and so from equation (2)
$L=\frac{1}{C{\omega }_0^2}=\frac{1}{{10}^{-6}\times {\left(4500\right)}^2}=\frac{4}{81}=0.049H$