Physics · Electrostatics
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Two capacitors C1 = 2µ and C2 = 4µ F are connected in series across in a 100V supply. Find the effective capacitance.
- A
½ µF
- B
3/2 µF
- C
5/2 µF
- D
4/3 µF
- E
2 µF
Recall that the effective capacitance of capacitors in series is the reciprocal of the sum of reciprocals of individual capacitors (just like resistors in parallel)
Therefore,
1/Ceffective = 1/C1 + 1/C2
1/Ceffective = ½ +¼
1/Ceffective = ¾
C effective= 4/3 µF
Hence option D is the correct answer.
This option is incorrect.Recall that the effective capacitance of capacitors in series is the reciprocal of the sum of reciprocals of individual capacitors (just like resistors in parallel)
Therefore,
1/Ceffective = 1/C1 + 1/C2
1/Ceffective = ½ +¼
1/Ceffective = ¾
C effective= 4/3 µF
This option is incorrect.
Recall that the effective capacitance of capacitors in series is the reciprocal of the sum of reciprocals of individual capacitors (just like resistors in parallel)
Therefore,
1/Ceffective = 1/C1 + 1/C2
1/Ceffective = ½ +¼
1/Ceffective = ¾
C effective= 4/3 µF
This option is incorrect.
Recall that the effective capacitance of capacitors in series is the reciprocal of the sum of reciprocals of individual capacitors (just like resistors in parallel)
Therefore,
1/Ceffective = 1/C1 + 1/C2
1/Ceffective = ½ +¼
1/Ceffective = ¾
C effective= 4/3 µF
Recall that the effective capacitance of capacitors in series is the reciprocal of the sum of reciprocals of individual capacitors (just like resistors in parallel)
Therefore,
1/Ceffective = 1/C1 + 1/C2
1/Ceffective = ½ +¼
1/Ceffective = ¾
C effective= 4/3 µF
This option is incorrect.
Recall that the effective capacitance of capacitors in series is the reciprocal of the sum of reciprocals of individual capacitors (just like resistors in parallel)
Therefore,
1/Ceffective = 1/C1 + 1/C2
1/Ceffective = ½ +¼
1/Ceffective = ¾
C effective= 4/3 µF
Tagged under Physics · Electrostatics · 2009