PHYS HELP! A capacitor of capacitance C = 5 muF is initially uncharged. It is co

Question

PHYS HELP!

A capacitor of capacitance C = 5 muF is initially uncharged. It is connected in series with a switch of negligible resistance, a resistor of resistance R = 13.5 kohm, and a battery which provides a potential difference of VB = 55 V. Immediately after the switch is closed, what is the voltage drop VC across the capacitor? Immediately after the switch is closed, what is the voltage drop VR across the resistor in terms of VB? Immediately after the switch is closed, what is the current 1 through the resistor in terms of VB and R? Find an expression for the time t1/2 after the switch is closed when the current in the resistor equals half its maximum value. Find an expression for the charge Q on the capacitor when the current in the resistor equals one half its maximum value.

Explanation / Answer

a)

Vc = 0 capacitor behaves as short circuit just after switch is closed.

b)

VR = VB

c)

I = VB/R

d)

I = VB/R * e(-t/RC)

at t1/2 I = VB/2R

so 1/2 = e(-t/RC)

t1/2 = RC*ln(2) = 13.5*5*ln(2) millisec

t1/2 = 46.78 msec = .04678 sec.

e)

Q = CV

V = VB(1-e(-ln(2)RC/RC))

V = 55(1-.5)

V = 27.5

so Q = 5*27.5 uC

= 137.5 uC

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