Question

The power developed in the circuit when the capacitor of $2500\mu F$ is connected in series with the coil is

• A. $28.8W$
• B. $23.04W$
• C. $17.28W$
• D. $9.6W$

Answer 1

The correct option is C $17.28W$

$V_{DC}=I_{DC}R$
$\therefore R=\frac{V_{DC}}{I_{DC}}=\frac{12}{4}=3\Omega$
$I_{AC}=\frac{V_{AC}}{Z}=\frac{V_{AC}}{\sqrt{R^2+X_L^2}}$
$2.4=\frac{12}{\sqrt{(3{)}^2+X_L^2}}$
Solving this equaion we get,
$X_L=4\Omega$
$X_C=\frac{1}{\omega C}=\frac{1}{50\times 2500\times {10}^{-6}}$
$=8\Omega$
$Z=\sqrt{R^2+(X_C-X_L{)}^2}=5\Omega$
$\therefore I=\frac{V_{DC}}{Z}=\frac{12}{5}=2.4A=I_{rms}$
$P=I_{rms}^{2}R=\left(2.4{)}^2\right(3)$
$=17.28W$
At
given frequency $X_C>X_L$. If $\omega$ is further decreased,
$X_C$ will increase $(as{X}_C\propto \frac{1}{\omega })$
and $X_L$ will increase $(as{X}_L\propto \omega )$.
Therefore $X_C-X_L$ and hence $Z$ will increase. So, current will decrease.