In the circuit shown in (Figure 1) both capacitors are initially charged to 40.0 V Part...

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In the circuit shown in (Figure 1) both capacitors are initially charged to 40.0 V Part...

In the circuit shown in (Figure 1) both capacitors are initially charged to 40.0 V

For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Discharging a capacitor.

Part A

How long after closing the switch S will the potential across each capacitor be reduced to 15.0 V ?

Part B

What will be the current at that time?

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Answer 1

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Answer #1

A)

Resistance, R = 50 + 30 = 80 ohm

Capacitance, C = 20 + 15 = 35 uF

Time, t = - RC ln(V/Vo)

t = - (80 x 35 x 10^-6) ln(15/40)

t = 2.74 ms

B)

Current, I = 40/80 = 0.5 A

Current, I = 0.5 x e^(-0.00274/(80 x 35 x 10^-6))

I = 0.18 A

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Answer 2

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In the circuit shown in (Figure 1) both capacitors are initially charged to 40.0 V Part...

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In the circuit shown in (Figure 1) both capacitors are initially charged to 40.0 V Part...