Question

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

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

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, $V_{\text{rms}} = 24\, V, \left(I_{\text{rms}}\right)_{\text{max}} = 6\, A$
$ \therefore 6\, A =\frac{ 24\, V}{6\, A} = 4 \,\Omega$
When this coil is connected to battery of emf $12 \,V$ and internal resistance $4 \Omega$, the current through it
$I =\frac{12 V}{\left(4\Omega+4\Omega\right)} =\frac{ 12\, V}{8\Omega} = 1.5A$