Answer: Charge, Q= 22.131 columbs
Power, P = 196.65 Watts
Explanation:
I(t) = 10( 1 - e(-0.5t)) A
I = dQ/dt
dQ = 10( 1 - e(-0.5t))dt
Integrating we have
Q = 10t - (20/(-0.5))e(-0.5t)
taking the value of t at 1s
we have,
Q = 10 + 20e(-0.5)
Q = 10 + 20/1.6487
Q = 10 + 12.131
Q = 22.131 Columbs
(b) Power
Power = voltage x current
Power = 5cos(2t) x 10(1 - e(-0.5t))
Power at t= 1, we substitute t=1
Power = 5cos(2) x 10(1 - e(-0.5))
Power = 5(0.9995) x 10(1 - 0.60654)
Power = 39.35 x 4.9975 = 196.65 Watts
Power = 196.65 Watts
