Navjyot Kalra
Last Activity: 10 Years ago
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