 The diagram above shows the capacitors C 1, C 2 and C 3 2μF, 6μF and 3μF respectively. The potential difference across C 1, C 2 and C 3 respectively are
A. 4V, 6V and 2V B. 2V, 6V and 4V C. 6V, 4V and 2V D. 6V, 2V and 4V Correct Answer: DExplanationC = Q/ V => Q = CV : ∴V = Q/ C∴V 1 = Q/ C[sub]1[/sub]; V 2 = Q/ C; V 3 = 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= 3+1+2 = 6/ 6 = 1 ∴ C = 1μF = 1.0 x 10 -6F ∴ Q = CV = 1.0 x 10 -6 x 12 = 12.0 x 10 -6C ∴V 1 = Q/ C[sub]1[/sub]= 12.0x10 -6= 6V V 2 = Q/ C[sub]2[/sub]= 12.0x10 -6= 2V V 3 = Q/ C[sub]3[/sub]= 12.0x10 -6∴ 6V, 2V and 4V |