The diagram above shows the capacitors C1, C2 and C3 2μF, 6μF and 3μF respectively. The potential difference across C1, C2 and C3 respectively are
- 4V, 6V and 2V
- 2V, 6V and 4V
- 6V, 4V and 2V
-
6V, 2V and 4V
The correct answer is: D
Explanation
C = Q/V => Q = CV : ∴V = Q/C∴V1 = Q/C1; V2 = Q/C;
V3 = Q/C3
for series
1/C = 1/C1 + 1/C2 + 1/C3
= 1/2 + 1/6 + 1/3
| = | 3+1+2 |
| _6 |
= 6/6 = 1
∴ C = 1μF = 1.0 x 10-6F
∴ Q = CV = 1.0 x 10-6 x 12
= 12.0 x 10-6C
∴V1 = Q/C1
| = | 12.0x10-6 |
| 2 x 10-6 |
= 6V
V2 = Q/C2
| = | 12.0x10-6 |
| 6 x 10-6 |
= 2V
V3 = Q/C3
| = | 12.0x10-6 |
| 3 x 10-6 |
∴ 6V, 2V and 4V
There is an explanation video available .