In the circuit shown in the figure (Figure 1) , the capacitors are all initially

Question

In the circuit shown in the figure (Figure 1) , the capacitors are all initially uncharged and the battery has no appreciable internal resistance. Assume that E = 45.0 V and R = 17.5 ? .

Part A

After the switch S is closed, find the maximum charge on each capacitor.

q10=____pC

q20=_____pC

q30=____pC

q40=____pC

Part B

After the switch S is closed, find the maximum potential difference across each capacitor.

V10=_____V

V20=_____V

V30=______V

V40=______V

Part C

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

?=_____ps

Explanation / Answer

part A) solution
The equivalent capacitance for the series combination of the 20pF, 30pF,40pF is Ceq
1/Ceq = [1/20*10-12 + 1/30*10-12 + 1/40*10-12] => Ceq = 9.23pF

There is no voltage across R because there is no current in R.
Thus we know there is 45.0 V across the 10pF so Q = 45*10 pF = 450 pC

45.0 V across the series combination of capacitors. All three capacitors have the same charge on them

Q = Ceq*V = 9.23pF*45.0 V = 415.35 pC

q10= 450 pC

q20= 415.35 pC

q30= 415.35 pC

q40= 415.35 pC

Part B

We can calculate the voltage on each using Q = CV => V = Q/C
V20= 415.35*10-12 C / 20*10-12 = 20.76 V
V30 = 415.35*10-12 C / 30*10-12 = 13.84 V
V40 = 415.35*10-12 C / 40*10-12 = 10.38 V
Notice 20.76 V +13.84 V + 10.38 V = 45.0 V
V10 = 45.0 V because it has the same voltage as the voltage supply since the resistor has no voltage across it

part C
The capacitance the voltage supply sees is 10pf+9.23pF = 19.23pF

because capacitors in parallel add. The equivalent capacitance for the series combination is 9.23pf as calculated above

Therefore the time constant for the circuit is RC = 17.5*19.23pF = 336.525 ps

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