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`        A capacitor of capacitance 10 μF is connected to an oscillator giving an output voltage ℰ = (10 V) sin ωt. Find the peak currents in the circuit for ω = 10 s^-1, 100 s^-1, 500 s^-1, 1000 s^-1.`
3 years ago

Navjyot Kalra
654 Points
```										Sol. C = 10 μF = 10 * 10^-6 F = 10^-5 F
E = (10 V) Sin ωt
a) I = E base 0/Xc = E base 0/(1/ωC) = 10/(1/10*10^-5) = 1 * 10^-3 A
b) ω = 100 s^-1
I = E base 0/(1/ωC) = 10/(1/100 * 10^-5) = 1 10^-2 A = 0.01
c) ω = 500 s^-1
I = E base 0/(1/ωC) = 10/(1/500 * 10^-5) = 5 * 10^-2 A = 0.05 A
d) ω = 1000 s^-1
I = E base 0/(1/ωC) = 10/(1/1000 * 10^-5) = 1 * 10^-1 A = 0.1 A

```
3 years ago
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