(13%) Problem 5: A bank of batteries, total emf = 4.5 V, is in a circuit with re
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(13%) Problem 5: A bank of batteries, total emf = 4.5 V, is in a circuit with resistor R = 120 kQ, capacitor C = 680 nF, and two-pole switch S as shown. The switch is initially set to point a so that the batteries, resistor, and capacitor are in series. The switched is left in this position for a sufficiently long time so that the capacitor is fully charged Randomized Variables =45V R=120k C = 680 nF ©theexpertta.com 17% Part (a) Calculate the maximum charge Q on the capacitor (in coulombs) 17% Part (b) The switch is moved from point a, but not connected to point b. What will happen to the voltage across the capacitor? 17% Part (c) The switch is now (at 0) moved to point b. Determine the current through resistor R (in amperes) the instant the switch is closed. You may assume the direction of this current is positive 17% Part (d) Determine an expression for the voltage across the capacitor as a function of time, with the switch at position b, in terms of the emf . Grade Summary Deductions Potential V(t) = |4.5 e-t/0.08 161 3% 97% 78 9 HOME Submissions Attempts remaining: 2 (1% per attempt) detailed view 1|2|3| 0 1% 1% 1% tVO BACKSPACE DE CLEAR Submit Feedback I give up! Hints: 2 for a 0% deduction. Hints remaining: 0 Feedback: 1 for a 0% deduction s that there may be a sign error associated with at least How is the potential across the capacitor related to the charge and capacitance? The charge is not constant here - how does the charge on a disc It appears one term in your expression Your answer is very close and contains something recognizable for all correct terms. Unfortunately there are additional terms in your equation that are not necessary harging capacitor vary with time? 17% Part (e) Calculate the time (in seconds), after contact is made by the switch at position b, that it will take the capacitor to discharge halfway 17% Part (f) Calculate the current through the resistor (in amperes) at time t = 105 ms after the switch is connected to point bExplanation / Answer
a)
Q = CV = (680 x 10-9)(4.5) = 3.06 x 10-6 C
d)
V(t) = E e -t/RC
e)
Q(t) = (3.06 x 10-6 ) e-t/0.0816
Given that : Q(t) = 3.06 x 10-6 /2
0.5 = e-t/0.0816
t = 0.057 sec
f)
V(t) = 4.5 e -t/0.0816
at t = 0.105 s
V(0.105) = 4.5 e -0.105/0.0816
V(0.105) = 1.24 Volts
i = V(0.105) /R = 1.24/120000 = 10.3 x 10-6 A
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