Answer:
- 0.3sin6000t A
Explanation:
Voltage, v = 10 cos 6000t V
Capacitance = 5-uF
Current flowing through, i(t)
i(t) = c * d/dt (V)
c = 5-uF = 5 * 10^-6 F
i(t) = (5 * 10^-6) * d/dt(10 cos 6000t)
d/dt(10 cos 6000t) = (10 * 6000) * (-sin 6000t)
Hence,
i(t) = (5*10^-6) * (10*6000) * (-sin 6000t)
i(t) = 5*10^-6 * 6*10^4 * - sin6000t
i(t) = 30 * 10^-2 * - sin6000t
i(t) = 0.3*-sin6000t
i(t) = - 0.3sin6000t Ampere
