STRATEGY Finding the time constant in part (a) requires substitution into the ti
ID: 1430598 • Letter: S
Question
STRATEGY Finding the time constant in part (a) requires substitution into the time constant of an RC circuit. For part (b), the maximum charge occurs after a long time, when the current has dropped to zero. By Ohm's law, ?V = IR, the potential difference across the resistor is also zero at that time, and Kirchhoff's loop rule then gives the maximum charge. Finding the charge at some particular time, as in part (c), is a matter of substituting into the proper equation. Kirchhoff's loop rule and the capacitance equation can be used to indirectly find the potential drop across the resistor in part (d), and then Ohm's law yields the current.
(a) the time constant of the circuit
__________ s
(b) the maximum charge on the capacitor
____________ µC
(c) the charge on the capacitor after 5.90 s
____________µC
(d) the potential difference across the resistor after 5.90 s
____________ V
(e) the current in the resistor at that time
____________ A
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)
time constant = R*C
= 8*10^5 * 5*10^-6
= 4 s
b)
Q = C*E
= 5*10^-6 * 12
= 6*10^-5 C
= 60 µC
C)
USE:
Q = QO* (1 - e^(-t/tau))
=60*(1-e^(-5.9/4))
= 46.3 µC
d)
V across capacitor = Q/C
= 46.3 µC / 4.5 µF
=10.3 V
V across R = E - V across capacitor
= 12 - 10.3
= 1.7 V
i am allowed to answer only 4 parts at a time
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