Question

Consider the situation of the previous problem. Find the average electric field energy stored in the capacitor and the average magnetic field energy stored in the coil.

Answer 1

Average Energy Stored is 5 mJ

Explanation:

• Given, R = 300 $\Omega$

C = $20 \mu F = 20 \times 10^-^6 F$  

L = 1H, Z = 500  

• Electric Energy stored in Capacitor = $\frac {1}{2} CV^2$

= $\frac {1}{2} \times 20 \times 10^-^6 \times 50 \times 50$  

= $25 \times 10^-^3$ J  

= 25 mJ

• Magnetic field energy stored in the coil = $(\frac {1}{2})LI_0^2$

= $(\frac {1}{2}) \times 1 \times (0.1)^2$

= $5 \times 10^-^3$ J  

= 5 mJ