1. Four capacitors, C1 = 2 μF, C2 = 1 μF, C3 = 3 μF, C4 = 4 μF, are connected in parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination.
Known :
Capacitor C1 = 2 μF
Capacitor C2 = 1 μF
Capacitor C3 = 3 μF
Capacitor C3 = 4 μF
Wanted : The equivalent capacitance
Solution :
The equivalent capacitance :
C = C1 + C2 + C3
C = 4 μF + 2 μF + 3 μF = 9 μF
The equivalent capacitance of the entire combination is 9 μF.
2. Determine the charge on capacitor C2 if the potential difference between point A and B is 9 Volt…
Known :
Capacitor C1 = 20 μF = 20 x 10-6 F
Capacitor C2 = 30 μF = 30 x 10-6 F
Potential difference between point A and B (VAB) = 9 Volt
Wanted : the charge on capacitor C2 (Q2)
Solution :
Potential difference :
Capacitors are connected in parallel so that the potential difference between A and B (VAB) = the potential difference on capacitor C1 (V1) = the potential difference on capacitor C2 (V2) = 9 Volt.
Electric charge on capacitor C2 :
Q2 = C2 V2 = (30 x 10-6)(9) = 270 x 10-6 C
Q2 = 270 μC
The electric charge on capacitor C2 is 270 μC.
3. Three capacitors, C1 = 4 μF, C2 = 2 μF, C3 = 3 μF, are connected in parallel. The capacitor are charged. The potential difference on capacitor C2 is 4 Volt. Determine
(a) Electric charge on capacitor C1, C2 and C3
(b) Electric charge on the equivalent capacitor of the entire combination
Known :
Capacitor C1 = 4 μF = 4 x 10-6 F
Capacitor C2 = 2 μF = 2 x 10-6 F
Capacitor C3 = 3 μF = 3 x 10-6 F
Potential difference on capacitor C2 (V2) = 4 Volt
Wanted : Electric charge on capacitor C3 (Q3)
Solution :
(a) Electric charge on capacitor C3
The potential difference on capacitor C3 :
Capacitors are connected in parallel so that the potential difference on capacitor C3 (V3) = the potential difference on capacitor C2 (V2) = the potential difference on capacitor C1 (V1) = the potential difference on equivalent capacitor (V) = 4 Volt
Electric charge on capacitor C1 :
Q1 = C1 V1 = (4 x 10-6)(4) = 16 x 10-6 C
Q1 = 16 μC
Electric charge on capacitor C2 :
Q2 = C2 V2 = (2 x 10-6)(4) = 8 x 10-6 C
Q2 = 8 μC
Electric charge on capacitor C3 :
Q3 = C3 V3 = (3 x 10-6)(4) = 12 x 10-6 C
Q3 = 12 μC
(b) Electric charge on equivalent capacitor
Q = Q1 + Q2 + Q3
Q = 16 μC + 8 μC + 12 μC = 36 μC
Alternative solution :
The equivalent capacitance :
C = C1 + C2 + C3
C = 4 μF + 2 μF + 3 μF = 9 μF
C = 9 x 10-6 F
The potential difference on the equivalent capacitor :
V1 = V2 = V3 = V = 4 Volt
The electric charge on the equivalent capacitor :
Q = C V = (9 x 10-6)(4) = 36 x 10-6 C
Q = 36 μC
