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
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Correct Answer: Option D
Explanation:
C = Q/V => Q = CV : ∴V = Q/C
∴V1 = Q/C[sub]1[/sub]; V2 = Q/C;
V3 = Q/C[sub]3[/sub]
for series 1/C = 1/C[sub]1[/sub] + 1/C[sub]2[/sub] + 1/C[sub]3[/sub]
= 1/2 + 1/6 + 1/3
= 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/C[sub]1[/sub]
= 6V
V2 = Q/C[sub]2[/sub]
= 2V
V3 = Q/C[sub]3[/sub]
∴ 6V, 2V and 4V
C = Q/V => Q = CV : ∴V = Q/C
∴V1 = Q/C[sub]1[/sub]; V2 = Q/C;
V3 = Q/C[sub]3[/sub]
for series 1/C = 1/C[sub]1[/sub] + 1/C[sub]2[/sub] + 1/C[sub]3[/sub]
= 1/2 + 1/6 + 1/3
| _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/C[sub]1[/sub]
| 2 x 10-6 |
= 6V
V2 = Q/C[sub]2[/sub]
| 6 x 10-6 |
= 2V
V3 = Q/C[sub]3[/sub]
| 3 x 10-6 |
∴ 6V, 2V and 4V