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`        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.`
3 years ago

Navjyot Kalra
654 Points
```										Sol R = 300Ω
C = 20 μF = 20 * 10^-6 F
L = 1H, Z = 500 (from 14)
Ε base 0 = 50 V, I base 0 = E base 0/Z = 50/500 = 0.1 A
Electric Energy stored in Capacitor = (1/2) CV^2 = (1/2) × 20 × 10^–6 × 50 × 50 = 25 × 10^–3 J = 25 mJ
Magnetic field energy stored in the coil = (1/2) L I base 0^2 = (1/2) * 1 * (0.1)^2 = 5 * 10^-3 J = 5 mJ

```
3 years ago
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