Problem 19.62 In the circuit shown in (Figure 1) , the capacitors are all initia

Question

Problem 19.62

In the circuit shown in (Figure 1) , the capacitors are all initially uncharged and the battery has no appreciable internal resistance. Assume that E = 50.0 V and R = 18.5 .

Part B

After the switch is closed, find the maximum charge on the 20 pF capacitor.

Express your answer in picocoulumbs to three significant figures.

Part C

After the switch is closed, find the maximum charge on the 30 pF capacitor.

Express your answer in picocoulumbs to three significant figures.

Part D

After the switch is closed, find the maximum charge on the 40 pF capacitor.

Express your answer in picocoulumbs to three significant figures.

Part E

After the switch is closed, find the maximum potential difference across the 10 pFcapacitor.

Express your answer in volts to three significant figures.

Part F

After the switch is closed, find the maximum potential difference across the 20 pFcapacitor.

Express your answer in volts to three significant figures.

Part G

After the switch is closed, find the maximum potential difference across the 30 pFcapacitor.

Express your answer in volts to three significant figures.

Part H

After the switch is closed, find the maximum potential difference across the 40 pFcapacitor.

Express your answer in volts to three significant figures.

Part I

After the switch is closed, find the maximum reading of the ammeter A.

Express your answer in amperes to three significant figures.

Part J

After the switch is closed, find the time constant for the circuit.

Express your answer in picoseconds to three significant figures.

Explanation / Answer

1/Ceqa =1/C1+1/C2+1/C3

=1/20+1/30+1/40

Ceqa=9.23pf

There is no voltage across R because there is no current on R

50V across the 10 of and 50V across the series combination of capacitor s.All three capacitor s have the same charge

Q=CeqaV

=9.23*50=461.5pc

V=Q/C

V20=461.5/20=23.0v

V30=461.5/30=15.38v

V40=461.5/40=11.53v

V10=50v

I=V/R

= 50/18.5

=2.7A

T=RC

=18.5*19.23

=0.355ns

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