A capacitor with capacitance C = 5.00×10-3 F is initially uncharged. It is in se
ID: 1592799 • Letter: A
Question
A capacitor with capacitance C = 5.00×10-3 F is initially uncharged. It is in series with a source of Emf of 7.00 volts, a resistor R, and a switch as shown in the figure below.
At t=0, the switch is closed. The graph below shows the potential difference across the capacitor as function of t, the time elapsed since the switch closed.
A) Calculate the value of R. Note that the curve passes through a grid intersection point.
(in ohm)
B) What can you tell about the maximum power dissipated in R?
The maximum power dissipated in R occurs at ...
Explanation / Answer
C = 5 x 10^-3 F = 0.005 F
for charging,
V = V0 [ 1 - e^(-t/T) ]
where T = time constant = RC
V0 = 7 volt
for 2sec, V = 3 V
3 = 7[ 1 - e^-(2/T)]
0.571 = e^-(2/T)
- 2 / T = ln(0.571)
T = 3.57
T = RC
3.57 = R x 0.005
R = 714.78 ohm
b) maximum current I = e/R = 7 / 714.78 = 0.0098 A
Power = 0.0098^2 x 714.78 = 0.0686 W
it will be at t = 0
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