Capacitors of capacitance 7.00 microfarads and 13.0 microfarads are connected in
ID: 1431832 • Letter: C
Question
Capacitors of capacitance 7.00 microfarads and 13.0 microfarads are connected in series with a 12.0 V battery.
(a) What is the equivalent capacitance? (F)
(b) What is the charge on the 7.00 microfarad capacitor? (C)
(c) What is the potential difference across the 13.0 microfarad capacitor? (V)
Capacitors of capacitance 7.00 microfarads and 13.0 microfarads are connected in series with a 12.0 V battery.
(a) What is the equivalent capacitance? (F)
(b) What is the charge on the 7.00 microfarad capacitor? (C)
(c) What is the potential difference across the 13.0 microfarad capacitor? (V)
Explanation / Answer
As given in the question,
C1 = 7 µF = 7*10^-6 F, C2 = 13 µF = 13*10^-6 F and V = 12 V
(a) Since C1 and C2 are connected in Series combination, so the equivalent capacitance:
1 / C(eqv) = 1 / C1 + 1 / C2
=> C(eqv) = C1*C2 / (C1 + C2)
= 7*13 / (7 + 13) = 4.55 µF = 4.55*10^-6 F
(b) The net charge in the circuit,
Q(net) = C(eqv)*V = (4.55*10^-6)*(12) = 5.46*10^-5 C
Since C1 and C2 are connected in Series combination, so the charges on each capacitor will be same as the total charge in the circuit.
So, the charge on the 7 µF capacitor: Q1 = Q(net) = 5.46*10^-5 C
(c) The potential difference across the 13 µF capacitor,
V2 = Q2 / C2 = Q(net) / C2 = (5.46*10^-5) / (13*10^-6) = 4.2 V