Question

A student in a lab took a coil and connected it to a $12V$ DC source. He measures the steady state current in the circuit to be $4A$. He then replaced the $12V$ DC source by a $12V$ $\left(\omega =50rad/s\right)$ AC source and observes that the reading in the ammeter is $2.4A$. He then decides to connect a $2500\mu F$ capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current in the circuit (with the capacitor and the battery in series).
Based on the readings taken by student, the power developed in the circuit when the capacitor of $2500\mu F$ is connected in series with the coil is $W$. (Upto two decimal digit).

Answer 1

$R=\frac{V_{dc}}{i}=\frac{12}{4}=3\Omega$
$z=\frac{V_{rms}}{i_{rms}}=\frac{12}{2.4}=5\Omega$
$$\begin{array}{c}z=\sqrt{R^2+x_L^2}=\frac{120}{24}=5\Omega \\ x_L^2=16{\Omega }^2\Rightarrow x_L=4\Omega \\ \text{Now}\cos \varphi =\frac{R}{\sqrt{R^2+{(x_C-x_L)}^2}}=\frac{3}{\sqrt{9+{(\frac{1}{\omega C}-\omega L)}^2}}\\ =\frac{3}{5}\end{array}$$ $$\begin{array}{cc}P& =V_{rms}{I}_{rms}\cos \varphi =12\times 2.4\times \frac{3}{5}\\ & =17.28W\end{array}$$