(50%) Problem 1: A capacitor of capacitance C= 8.5 F is initially uncharged. It
ID: 1781123 • Letter: #
Question
(50%) Problem 1: A capacitor of capacitance C= 8.5 F is initially uncharged. It is connected in series with a switch of negligible resistance, a resistor of resistance R = 5.5 kQ, and a battery which provides a potential difference of V, = 55 V. Otheex 17 % Part (a) Calculate the time constant for the circuit in seconds. 17% Part (b) After a very long time after the switch has been closed, what is voltage drop Vc across the capacitor in terms of VB? #17% Part (c) Calculate the charge Q on the capacitor a very long time after t has been closed in C. > 17% Part (d) Calculate the current I a very long time after the switch has be n A 0 Hint I give up! Hints: 0% deduction per hint. Hints remaning: 3 Feedback: 0% deduction per feedback. 17% Part (e) Calculate the time t after which the current through the resistor third of its maximum value in s 1 7% Part (f) Calculate the charge Q on the capacitor when the current in the equals one third its maximum value in C.Explanation / Answer
d] after a very long time, i = 0 A
e] i = i0 * e^(-t/RC)
i/i0 = e^(-t/RC) = 1/3
1/3 = e^(-t/(8.5e-6*5.5e3))
ln (1/3) = -t/(8.5e-6*5.5e3)
t = (8.5e-6*5.5e3) * ln 3
= 0.05136 s answer
f] Q = Q0*(1- e^(-t/RC))
= CVB (1 - 1/3)
= 8.5E-6*55*2/3
= 0.0003117 C answer
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