# Heat generated in LCR circuit

1 vote
241 views I couldnot get how to approach this problem.

asked May 15, 2017
edited May 27, 2018

First calculate the impedance of the circuit and then divide the applied voltage between the RC and LC combinations to find the voltage across C when $t=\pi$.
Here $\omega=100 rad/s, R_1=10\Omega, C=1mF, R_2=20\Omega, L=0.2H$. The applied voltage is $$V_0=10\sin(\omega t)=10\cos(\frac12\pi-\omega t)=Re[ 10e^{j(\frac12 \pi - \omega t)}]$$. The impedances of the RC and RL parallel combinations are $$Z_1=\frac{R_1/j\omega C}{R_1+1/j\omega C}=\frac{R_1(1-j\omega CR_1)}{1+(\omega CR_1)^2}=\frac{10(1-j)}{1+1}=5(1-j)$$ $$Z_2=\frac{j\omega LR_2}{R_2+j\omega L}=\frac{200j}{10+20j}=\frac{20}{1+2j}=4(1-2j)$$ The voltage across the RC combination, and therefore across the capacitor, is $V_1$ where $$\frac{V_1}{V_0}=\frac{Z_1}{Z_1+Z_2}=\frac{5(1-j)}{9-13j}=\frac{5(1-j)}{81+169}=\frac{1}{50}(1-j)=\frac{\sqrt2}{50}(\frac{1}{\sqrt2}-j\frac{1}{\sqrt2})=\frac{\sqrt2}{50}e^{-\frac14 \pi j}$$ $$V_1=\frac{\sqrt2}{50} e^{-\frac14 \pi j} \times 10 e^{j(\frac12 \pi -\omega t)}=\frac{\sqrt2}{5}e^{j(\frac14 \pi -\omega t)}$$ When $t=\pi$ the voltage across C is $$Re[V_1(\pi)]=\frac{\sqrt2}{5}\cos(\frac14\pi - 100\pi)=\frac{\sqrt2}{5}\cos(\frac14\pi)=\frac15 V$$
The last step is to find the energy stored in the capacitor, which equals the heat generated. The energy stored is $$\frac12 CV_1^2=\frac12\times 1mF \times \frac{1}{25}V^2=20 \mu J$$