Welcome to Physics Problems Q&A, where you can ask questions and receive answers from other members of the community.

Heat generated in LCR circuit

1 vote
66 views

I couldnot get how to approach this problem.

asked May 15, 2017 in Physics Problems by Green (564 points)
edited May 27 by sammy gerbil

1 Answer

0 votes

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 $$\frac12 CV_1^2=\frac12\times 1mF \times \frac{1}{25}V^2=20 \mu J$$

answered May 28 by sammy gerbil (26,096 points)
edited May 28 by sammy gerbil
...