PRACTICE IT! An uncharged capacitor and a resistor are connected in series to a
ID: 1539235 • Letter: P
Question
PRACTICE IT!
An uncharged capacitor and a resistor are connected in series to a battery, as shown in the figure above. If e m f = 15.0 V, C = 5.50 µF, and R = 7.90 105 , find the following.
(a) the time constant of the circuit
4.375s
(b) the maximum charge on the capacitor
82.5µC
(c) the charge on the capacitor after 5.50 s
59.03µC
(d) the potential difference across the resistor after 5.50 s
-4.267 V
(e) the current in the resistor at that time
5.4e-6.
Use the values from PRACTICE IT to help you work this exercise.
(a) Find the charge on the capacitor after 2.10 s have elapsed.
Q = C
(b) Find the magnitude of the potential difference across the capacitor after 2.10 s.
VC = V
(c) Find the magnitude of the potential difference across the resistor at that same time.
VR = V
Explanation / Answer
a)The magnitude if charge after time t is given by:
Q = CVb (1 - e^-t/RC)
Vb is the battery voltage
RC = 7.9 x 10^5 x 5.5 x 10^-6 = 4.345 s
Q = 5.5 x 10^-6 x 15 (1 - e^-2.1/4.345)
Q = 31.62 x 10^-6 = 31.62 micro C
Hence, Q = 31.62 micro C
b)V = V0 (1 - e^-t/RC)
V = 15 (1 - e^-2.1/4.345) = 5.75 Volts
Hence, V = 5.75 Volts
c)The drop across the resistor would be:
Vr = emf - V = 15 - 5.75 = 9.25 V
Hence, V = 9.25 Volts
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