Question

A coil and a capacitor and an ac source of $rms$ voltage $24V$ are connected in series. By varying the frequency of the source, a maximum $rms$ current of $6A$ is observed. If this coil is connected to a battery of emf $12V$ and internal resistance $4\Omega$, the current through it will be

• A. $1.5A$
• B. $2.5A$
• C. $3.5A$
• D. $4.5A$

Answer 1

Answer: (A)

Let $R$ and $L$ be resistance and inductance of a coil.

As this circuit is a series $LCR$ circuit, the current will be maximum at resonance.
Impedance of the circuit, $Z=\sqrt{R^2+{\left(X_L-X_C\right)}^2}$
At resonance, $X_L=X_C$
$\therefore Z=R$
$\therefore {\left(I_{\text{rms}}\right)}_{\text{max}}=\frac{V_{\text{rms}}}{Z}=\frac{V_{\text{rms}}}{R}$
Here, $V_{\text{rms}}=24V,{\left(I_{\text{rms}}\right)}_{\text{max}}=6A$
$\therefore 6A=\frac{24V}{6A}=4\Omega$
When this coil is connected to battery of emf $12V$ and internal resistance $4\Omega$, the current through it
$I=\frac{12V}{\left(4\Omega +4\Omega \right)}=\frac{12V}{8\Omega }=1.5A$