A capacitor of capacitance C= 7.5 F is initially uncharged. It is connected in s

Question

A capacitor of capacitance C= 7.5 F is initially uncharged. It is connected in series with a switch of negligible resistance, a resistor of resistance R = 14 k, and a battery which provides a potential difference of Vg = Mi V. Calculate the time constant r for the circuit in seconds. After a very long time after the switch has been closed, what is the voltage drop V_Q across the capacitor in terms of V_B? Calculate the charge Q on the capacitor a very long time after the switch has been closed in C. Calculate the current I a very long time after the switch has been closed in A. Calculate the time t after which the current through the resistor is one-third of its maximum value in s. Calculate the charge Q on the capacitor when the current in the resistor equals one third its maximum value in C.

Explanation / Answer

a)time constant =T = RC = (7.5 x 10-6 )(14 x 103) = 0.105sec

b) Vc = VB

so Vc= 145 Volts

c)Q = CV = (7.5 x 10-6 ) 145

Q=1.0875x10-3 C

d) Q = Qo (1-e-t/T)

taking derivative both side

dQ/dt = Qo (1/T) e-t/T

i = (Qo/T) e-t/T

at t= infinity (after a long time)

i= Qo/T

i=1.0875x10-3/0.105

i=0.0104 A

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