The capacitor in the circuit shown above is initially charged. After closing the switch, how much time elapses until one-half of the capacitor’s initial stored energy is dissipated?
A. RC
B. RC/2
C. RC/4
D. RC ln 2
E. ½ RC ln 2
Ut = ½U0 , t = ?
Energy of capacitor: U = ½CV2
Ut = ½U0 → ½CVt 2 = ½ × ½CV02
Vt 2 = ½ V02
Potential difference of capacitor: Vt = V0e−t/RC
Vt 2 = V02e−2t/RC
½ V02 = V02e−2t/RC → ½ = e−2t/RC
ln ½ = − 2t/RC → ln 2 = 2t/RC
t = ½ RC ln 2
Answer: E
A. RC
B. RC/2
C. RC/4
D. RC ln 2
E. ½ RC ln 2
(GR9277 #11)
Solution:
Ut = ½U0 , t = ?
Energy of capacitor: U = ½CV2
Ut = ½U0 → ½CVt 2 = ½ × ½CV02
Vt 2 = ½ V02
Potential difference of capacitor: Vt = V0e−t/RC
Vt 2 = V02e−2t/RC
½ V02 = V02e−2t/RC → ½ = e−2t/RC
ln ½ = − 2t/RC → ln 2 = 2t/RC
t = ½ RC ln 2
Answer: E
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