An inductor coil is connected to a capacitor and an AC source of rms voltage $8 \mathrm{~V}$ in series. The rms current in the circuit is $16 \mathrm{~A}$ and is in phase with emf. If this inductor coil is connected to $6 \mathrm{~V}$ DC battery, the magnitude of steady current is

• A

  $8 \mathrm{~A}$
• B

  $10 \mathrm{~A}$
• Correct answer
  C

  $12 \mathrm{~A}$
• D

  $16 \mathrm{~A}$

Solution

Given, rms voltage of AC source $V_{\text {rms }}=8 \mathrm{~V}$ and the rms current in the circuit, $I_{\mathrm{rms}}=16 \mathrm{~A}$ According to the question,
$\therefore$ Impedance of the circuit, $$ \begin{aligned} & Z=\frac{V_{\mathrm{rms}}}{I_{\mathrm{rms}}}=\frac{8}{16} \\ & Z=0.5 \Omega \end{aligned} $$ When the inductor coil is comected to a $6 \mathrm{~V} D C$ battery then $V_{\text {rms }}^{\prime}=6 \mathrm{~V}=V_{D C}$ $\therefore$ Magnitude of steady current, $$ I^{\prime}=\frac{V_{\text {rms }}^{\prime}}{Z}=\frac{6}{0.5}=12 \mathrm{~A} $$

Hence, the correct answer is option: 'C'.