Question

An inductor coil, a capacitor and an AC source of $24\text{V}\left(rms\right)$ are connected in series. When the frequency of the source is varied, a maximum rms current of $6\text{A}$ is observed. If the inductor coil is connected to a DC battery of emf $12\text{V}$ and internal resistance $4\Omega$, what will be steady state current through the battery.

• A. $1.5\text{A}$
• B. $3\text{A}$
• C. $4.5\text{A}$
• D. $6\text{A}$

Answer 1

The correct option is A $1.5\text{A}$

If $R$ is the resistance of the inductor coil.


At resonance, $Z_{min}=R$, so the maximum current in the circuit is,

$i_{rms}=\frac{V_{rms}}{Z_{min}}=\frac{24}{R}$

$\Rightarrow 6=\frac{24}{R}$

$\Rightarrow R=4\Omega$

Now , this inductor coil is connected with a battery of emf $12\text{V}$.


Here, the inductor coil itself having resistance $R=4\Omega .$

Across $DC$ source, the steady current in the circuit is given as,

$I=\frac{12}{4+R}=\frac{12}{8}=1.5\text{A}$

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