In the circuit shown in (Figure 1) , the capacitors are all initially uncharged
ID: 1439611 • Letter: I
Question
In the circuit shown in (Figure 1) , the capacitors are all initially uncharged and the battery has no appreciable internal resistance. Assume that E = 48.5 V and R = 23.0 ? .
Problem 19.62 In the circuit shown in (Figure 1), the capacitors are a initially uncharged and the battery has no appreciable internal resistance. Assume that E 48.5 V and R 23.0 S2 Figure 1 of 1 20.0 pF 20.00 30.0 pF 0.0 pF 40.0 pF Part A After the switch is closed, find the maximum charge on the 10 pF capacitor. Express your answer in picocoulumbs to three significant figures pC 910 My Answers Give U Submit 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 pC q20 My Answers Give U Submit 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 pC 930 Submit My Answers Give U 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 pC 940 Submit My Answers Give U Part E After the switch is closed, find the maximum potential difference across the 10 pF capacitor. Express your answer in volts to three significant figures. 10Explanation / Answer
A) The equivalent capacitance for the series combination of the 20pF, 30pF,40pF is Ceq
1/Ceq = [1/20e-12 + 1/30e-12 + 1/40e-12] => Ceq = 9.23pF
After some time the caps are completely charged and current from the voltage supply ceases.
There is no voltage across R because there is no current in R.
Thus we know there is 48.5V across the 10pF and 48.5 across the series combination of capacitors. All three caps have the same charge on them Q = Ceq*V = 9.23pF*48.5 = 447.66 pC
We can calculate the voltage on each using Q = CV => V = Q/C
F) V20= 447.66e-12C/20e-12 = 22.4V
G) V30 = 447.66e-12/30e-12 = 14.9V
H) V40 = 447.66e-12/40e-12 = 11.2V
Notice 22.4V +14.9V + 11.2V = 48.5V
V10 = 48.5V because it has the same voltage as the voltage supply since the resistor has no voltage across it.
I)Uncharged caps are modeled as short circuits. Voltage will appear across them after some time but at t = 0, they are modeled as short circuits. Therefore all the supply voltage appears across the 22 resistor which means the current in the ammeter A is 48.5/23 = 2.106A
J)The capacitance the voltage supply sees is 10pf+9.23pF = 19.23pF because caps in parallel add. We already calculated the equivalent capacitance of the series combination = 9.23pF
Therefore the time constant for the circuit is RC = 23*19.23pF =442.290 ps