Question

A capacitor and a coil in series are connected to a $6\text{volt}$ ac source. By varying the frequency of the source, maximum current of $600\text{mA}$ is observed. If the same coil is now connected to a cell of emf $6\text{volt}$ and internal resistance of $2\Omega$, the current through it will be :

• A. $0.5\text{A}$
• B. $0.6\text{A}$
• C. $1\text{A}$
• D. $2\text{A}$

Answer 1

The correct option is A $0.5\text{A}$


Maximum current $\left[I_{max}\right])$ will flow in the condition of resonance
Impedence $\left[Z\right]=\sqrt{R^2+(X_L-X_C{)}^2}$ at resonance Z is minimum hence I is maximum $X_L=X_C$
$I_{max}=\frac{V}{R}$
$600\times {10}^{-3}A=\frac{6}{R}$
$R=10\Omega$
Now connected by D.C. source $[\omega =0]$

$I=\frac{v}{R_{eq}}=\frac{6}{10+2}=\frac{6}{12}=\frac{1}{2}=0.5A$
Hence option (A) is correct.