T=RC = 250K * 8.5uF = 2125ms = 2.125s.
t/T = 0.32S / 2.125S = 0.1506.
Q = Qo / e^(t/T) = 8.
Qo / e^(0.1506) = 8,
Cross multiply:
Qo = 8*e^(0.1506),
Qo = 8 * 1.1625 = 9.300C. @ t = 0.
t/T = 1S / 2.125s = 0.4706.
Q = Qo / e^(t/T),
Q = 9.30 / e^(0.4706),
Q = 9.30 / 1.60 = 5.809C after 1sec.
Charge Q coulombs at time t seconds is given by the differential equation:
0
C
Q
dt
dQ
R where C is the capacitance in farads and R is the resistance in
ohms. Solve the equation for Q given that Q=Q0 when t=0.
A circuit possesses a resistance of 250kΩ and a capacitance of 8.5μF, and
after 0.32 seconds the charge falls to 8.0C. Determine the initial charge and the
charge after 1 second, each correct to 3 significant figures.
2 answers
t/T=1S /2.125s=0.4706.