Question

A current of $4\text{A}$ flows in a coil when connected to a $12\text{V}\text{DC}$ source. If the same coil is connected to a $12\text{V},50\text{rad/s}$ $\text{AC}$ source, a $\text{RMS}$ current of $2.4\text{A}$ flows in the circuit. Find the power developed in the circuit if a $2500\mu \text{F}$ capacitor is connected in series with the coil.

• A. $10.25\text{W}$
• B. $17.28\text{W}$
• C. $22.22\text{W}$
• D. $30.34\text{W}$

Answer 1

The correct option is B $17.28\text{W}$

When the coil is connected to a $\text{DC}$ sources, its resistance $R$ is given as,

$R=\frac{V}{I}=\frac{12}{4}=3\Omega$

When it is connected to $\text{AC}$ sources, the impedance $Z$ of the coil is given as,

$Z=\frac{V_{rms}}{I_{rms}}=\frac{12}{2.4}=5\Omega$

For a coil its impedance is given as,

$Z=\sqrt{[R^2+(\omega L{)}^2]}$

$\Rightarrow 5=\sqrt{\left[\right(3{)}^2+\left(50L{)}^2\right]}$

$\Rightarrow 25=\left[\right(3{)}^2+\left(50L{)}^2\right]$

$\Rightarrow L=0.08\text{H}$

When the coil is connected with a capacitor in series, its impedance $Z^{\prime }$ is given by,

$Z^{\prime }=\sqrt{[R^2+{(\omega L-\frac{1}{\omega C})}^2]}$

$={\left[\right(3{)}^2+{(50\times 0.08-\frac{1}{50\times 2500\times {10}^{-6}})}^2]}^{\frac{1}{2}}$

$\therefore Z^{\prime }=5\Omega$

Average power developed in the circuit is given as,

$P=V_{rms}{I}_{rms}\cos \varphi$

Where power factor of the circuit is given as,

$\cos \varphi =\frac{R}{Z^{\prime }}=\frac{3}{5}=0.6$

$P=12\times 2.4\times 0.6=17.28\text{W}$

Hence, $\left(\text{B}\right)$ is the correct answer.