A potential difference V = 4.400×10 2 V is applied to two capacitors connected i
ID: 1421566 • Letter: A
Question
A potential difference V = 4.400×102V is applied to two capacitors connected in series. One capacitor, C1, is 4.40 F and the other, C2, is 7.50 F. What is the charge on the positive plate of C1?
What is the potential difference across C1?
What is the charge on the positive plate of C2?
What is the potential difference across C2?
The charged capacitors are disconnected carefully from each other and from the battery. They are then reconnected, positive plate to positive plate and negative plate to negative plate, with no external voltage being applied. What is the charge on the positive plate of C1?
What is the potential difference across C1?
What is the charge on the positive plate of C2?
What is the potential difference across C2?
Suppose the charged capacitors had been reconnected with plates of opposite sign together. What then would be the steady-state potential difference for C1 ? (Do not try this at home!)
Tries 0/10Explanation / Answer
When capacitors are connected in series.
their equivalent Ceq will be
1/Ceq= 1/C1 + 1/C2 = 1/4.40 + 1/7.50
Ceq = 2.77 uF
When capacitor are connected in series then charge on both will be same.
Q = Ceq V = 2.77 x 440 = 1218.8 uC ........ANs (On both capacitor)
PD across C1 = Q/C1 = 1218.8 / 4.40 =277 volt ....Ans
PD across C2 = Q/C2 = 1218.8 / 7.50 = 163 volt .........Ans
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now capacitors are connected in parallel connection.
so PD across both capacitors will be same.
Ceq = C1 + C2 = 4.40 + 7.50 = 11.9 uF
stored energy = C V^2 / 2 = 11.9V^2 / 2
initial energy = Q^2 / 2C = (1218.8uC) / 2(2.77 uF) = 268136 J
11.9V^2 / 2 = 268136
V = 212.28 Volt .....Ans (for both )
Q1 = C1V = 4.40 x 212.28 = 934.04 uC
Q2 = C2V = 7.50 x 212.28 = 1592.14 uC
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