Question

A circuit is set up by connecting inductance L = 100 mH, resistor R = 100 Ω and a capacitor of reactance 200 Ω in series. An alternating emf of \[150\sqrt{2}\]  V, 500/π Hz is applies across this series combination. Calculate the power dissipated in the resistor.

Answer 1

Given the r.m.s value of emf in the circuit E_v = 150√2 V
The impedance of the LCR- circuit is given by

\[Z = \sqrt{R^2 + \left( X_L - X_C \right)^2}\]

\[ = \sqrt{R^2 + \left( 2\pi fL - 200 \right)^2}\]

\[ = \sqrt{{100}^2 + \left( 2\pi \times \frac{500}{\pi} \times 0 . 1 - 200 \right)^2}\]

\[ = 100\sqrt{2} \Omega\]

The r.m.s value of current I_v in the circuit is given by

\[I_v = \frac{E_v}{Z} = \frac{150\sqrt{2}}{100\sqrt{2}} = 1 . 5 A\]

Power dissipated in the resistor = E_vI_v = 318.2 W