Question

A box $P$ and a coil $Q$ are connected in series with a $10\text{V}$ AC source whose frequency can be varied. Box $P$ contains a capacitance of $1\mu \text{F}$ in series with a resistance of $32\Omega$. Coil $Q$ has a self inductance $4.9\text{mH}$ and a resistance of $68\Omega$. The frequency is adjusted so that the maximum current flows in $P$ and $Q$. Find the value of angular frequency of the source and the maximum current in the circuit, respectively.

• A. $10\text{rad/s},2\text{A}$
• B. $10\text{rad/s},3\text{A}$
• C. $\frac{{10}^5}{7}\text{rad/s},2\text{A}$
• D. $\frac{{10}^5}{7}\text{rad/s},0.1\text{A}$

Answer 1

The correct option is D $\frac{{10}^5}{7}\text{rad/s},0.1\text{A}$

The current flowing in the $LCR$ series circuit is given as,

$i=\frac{V}{Z}=\frac{V}{\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}}$

The current is maximum at resonance, when,

$(\omega L-\frac{1}{\omega c})=0$

$\Rightarrow \omega L=\frac{1}{\omega C}$

$\Rightarrow {\omega }^2=\frac{1}{LC}$

$\Rightarrow \omega =\frac{1}{\sqrt{(4.9\times {10}^{-3})\left({10}^{-6}\right)}}$

$\Rightarrow \omega =\frac{{10}^5}{7}\text{rad/s}$

Also, $Z=R=R_1+R_2$

Maximum current,

$i_{max}=\frac{V}{R_1+R_2}=\frac{10}{(32+68)}$

$\Rightarrow i_{max}=\frac{1}{10}=0.1\text{A}$

Hence, option $\left(D\right)$ is the correct answer.