With the switch in position A, the capacitor becomes charged. After a long time,

Question

With the switch in position A, the capacitor becomes charged. After a long time, the capacitor charge approaches q = C ?.
Now, suppose the capacitor is fully charged, when, at time t = 0, the switch is moved from A to B.
(A) Determine the charge q(t) and current i(t) for t > 0.
(B) Determine the energy of the capacitor U(t) as a function of time t. The initial energy, at t = 0, is U0. Calculate the energy at t = RC and at t = 0.26 RC. Express the answers as fractions of the initial energy. (No units because the answers are a ratio of energy/energy.)
(C) Determine the power P(t) dissipated in the resistor. Calculate the power at t = RC and t = 0.26 RC. Express the answers as fractions of U0/?.

Explanation / Answer

(A) Determine the charge q(t) and current i(t) for t > 0.

q(t) = q(0) e^(-t/RC)

i(t) = (/R) (1-e^(-t/RC))

(B) Determine the energy of the capacitor U(t) as a function of time t. The initial energy, at t = 0, is U0. Calculate the energy at t = RC and at t = 0.26 RC. Express the answers as fractions of the initial energy. (No units because the answers are a ratio of energy/energy.)

U(t) = q^2/C = ((q(0))2/C) e^(-2t/RC)

U(RC) = ((q(0))2/C) e^(-2) = 0.135 ((q(0))2/C) = 0.135 C 2

U(0.26 RC) = ((q(0))2/C) e^(-2*0.26) = 0.595 C 2

(C) Determine the power P(t) dissipated in the resistor. Calculate the power at t = RC and t = 0.26 RC. Express the answers as fractions of U0/?.

P = R i2 = R (/R)^2 (1-e^(-t/RC))^2 = (2/R) (1-e^(-t/RC))^2

P(RC) = (2/R) (1-e^(-1))^2 = 0.40 (2/R)

P(0.26 RC) = (2/R) (1-e^(-0.26))^2 = 0.0524 (2/R)

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