1. Three capacitors, C1 = 2 μF, C2 = 4 μF, C3 = 4 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination.
Known :
Capacitor C1 = 2 μF
Capacitor C2 = 4 μF
Capacitor C3 = 4 μF
Wanted : The equivalent capacitance (C)
Solution :
Capacitor C2 and C3 connected in parallel. The equivalent capacitance :
CP = C2 + C3 = 4 + 4 = 8 μF
Capacitor C1 and Cp connected in series. The equivalent capacitance :
1/C = 1/C1 + 1/CP = 1/2 + 1/8 = 4/8 + 1/8 = 5/8
C = 8/5 μF
2. Five capacitors, C1 = 2 μF, C2 = 4 μF, C3 = 6 μF, C4 = 5 μF, C5 = 10 μF, are connected in series and parallel. Determine the capacitance of a single capacitor that will have the same effect as the combination.
Known :
Capacitor C1 = 2 μF
Capacitor C2 = 4 μF
Capacitor C3 = 6 μF
Capacitor C4 = 5 μF
Capacitor C5 = 10 μF
Wanted : The equivalent capacitance (C)
Solution :
Capacitor C2 and C3 are connected in parallel. The equivalent capacitance :
CP = C2 + C3
CP = 4 + 6
CP = 10 μF
Capacitor C1, CP, C4 and C5 are connected in series. The equivalent capacitance :
1/C = 1/C1 + 1/CP + 1/C4 + 1/C5
1/C = 1/2 + 1/10 + 1/5 + 1/10
1/C = 5/10 + 1/10 + 2/10 + 1/10
1/C = 9/10
C = 10/9 μF
3. C1 = 3 μF, C2 = 4 μF and C3 = 3 μF, are connected in series and parallel. Determine the electric energy on the circuits.
Known :
Capacitor C1 = 3 μF
Capacitor C2 = 4 μF
Capacitor C3 = 3 μF
Wanted : The equivalent capacitance (C)
Solution :
Capacitor C2 and C3 are connected in parallel. The equivalent capacitance :
CP = C2 + C3
CP = 4 + 3
CP = 7 μF
Capacitor C1 and CP are connected in series. The equivalent capacitance :
1/C = 1/C1 + 1/CP
1/C = 1/3 + 1/7
1/C = 7/21 + 3/21
1/C = 10/21
C = 21/10
C = 2.1 μF
C = 2.1 x 10-6 F
The electric energy on the circuits :
E = ½ C V2
E = ½ (2.1 x 10-6)(122)
E = ½ (2.1 x 10-6)(144)
E = (2.1 x 10-6)(72)
E = 151.2 x 10-6 Joule
E = 1.5 x 10-4 Joule
