Question

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

• A. $17.28W$
• B. $8.64W$
• C. $10W$
• D. $15W$

Answer 1

The correct option is A $17.28W$

impedance of coil=$R+jL\omega$.
when DC is applied,
$I_{dc}=\frac{V_{dc}}{R}\Rightarrow 4=\frac{12}{R}\Rightarrow R=3\Omega$
when AC (12 V , 50rad/s ) is applied,

$I_{ac}=\frac{V_{ac}}{\sqrt{R^2+{\left(L\omega \right)}^2}}\Rightarrow L=\sqrt{\frac{1}{{\omega }^2}(\frac{V_{ac}^2}{I_{ac}^2}-R^2)}=\frac{2}{25}=.08H$
$X_L=L\omega$ $X_C=\frac{1}{C\omega }$

$I_{rms}=\frac{V_{ac}}{\sqrt{R^2+{(X_L-X_C)}^2}}=2.4A$

Power developed in the circuit = $I_{rms}^{2}R=17.28W$