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Amit Saxena Grade: upto college level
`        In a series RC circuit with an AC source, R = 300 Ω, C = 25 μF, ℰ^0 = 50 V and v = 50/π Hz. Find the peak current and the average power dissipated in the circuit.`
3 years ago

Jitender Pal
365 Points
```										Sol. R = 300Ω, C = 25 μF = 25 * 10^-6 F, ε base 0 = 50 V, f = 50 Hz
X base c = 1/ωc = 1/50/π  *2π *25 *10^-6 = 10^4/25
Z = √R^2 + X base c^2 = √(300)^2 + (10^4/25)^2 = √(300)^2 + (400)^2 = 500
(a) Peak current = E base 0/Z = 50/500 = 0.1 A
(b) Average Power dissipitated, = E base rms I base rms Cos ∅
= E base 0/√2 * E base 0/√2Z * R/Z = E base 0^2/2Z^2 = 50 * 50*300/2* 500 *500 = 3/2 = 1.5 ω.

```
3 years ago
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