(21%) Problem 5: A capacitor of capacitance C 9.5 LuF is initially uncharged. It
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(21%) Problem 5: A capacitor of capacitance C 9.5 LuF is initially uncharged. It is connected in series with a switch of negligible resistance, a resistor of resistance R 1.5 k2, and a battery which provides a potential difference of VB 75 Otheexpertta.com A 17% Part (a) Calculate the time constant t for the circuit in seconds Grade Summary Deductions 0% Potential 100% sino coso I tano 7 8 9 HOME Submissions Attempts remaining: 5 4 5 6 cotan per attempt 1 2 3 atano acotano si detailed view cosh00 tanho cotanh0 IVO BACKSPACE CLEAR Degrees O Radians Submit Hints 0% deduction per hint. Hints remaining 1 Feedback: 2% deduction per feedback. A 17% Part (b) After a very long time after the switch has been closed, what is the voltage drop Vc across the capacitor in terms of VB? A 17% Part (c) Calculate the charge Q on the capacitor a very long time after the switch has been closed inExplanation / Answer
a) time constant = RC = 7.5*103*9.5*10-6
T = 0.07125 sec
b) VC= VB
c)Q= C*V= ( 9.5*10-6)* 75V= 7.125*10-4 coulomb
d) I= 0 beacause circuit is opened
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