Compute the total capacitance with parallel connection:
Three capacitors C_{1}, C_{2} and C_{3} contain capacitance 20 μf, 15 μf, 30 μf, respectively. compute:
1. Charge on each of the capacitor when connected in parallel with 220 V supply.
2. Total capacitance with parallel connection.
3. Voltage across each of the capacitor when connected in series and 220 V supply is given to this series connection.
Solution
(a) Capacitors are in parallel:
Figure
Q_{1} = C_{1} V = 20 × 10^{-6 } × 220 = 4400 × 10
Q_{2} = C_{2} V = 15 × 10^{- 6 }× 220 = 3300 × 10^{- 6} C
Q_{3} = C_{3} V = 30 × 10^{- 6 }× 220 = 6600 × 10^{- 6} C
(b) Total capacitance :
C = C_{1} + C_{2} + C_{3}
= (20 + 15 + 30) × 10^{- 6}
= 65 μF
(c) As capacitors are in series so charge Q remains same for all capacitors.
Figure
1 /C_{eq} = (1 / C_{1} )+ (1 / C_{2} ) +(1/C_{3})
1 / C_{eq} = 1 /20 + 1/15 + 1/30 = 0.05 + 0.0666 + 0.0333
C_{eq} = 6.667 μF
Total charge Q = C_{eq} × V
= 1466.74 μC
Now, Voltage across C_{1} is V_{1} = Q/ C_{1 }= 73.337 volt
Voltage across C is V = Q/ C_{2 }== 97.78 volt
Voltage across C is V = Q/C_{3 } == 48.89 volt