Three capacitors C₁, C₂ and C₃ are connected to voltage source V as shown in figure. The voltage across C₃ will be (a) (C₃V)/((C₁+C₂+C₃)) (b) ((C₁+C₂)V)/(C₃) (c) ((C₂+C₃)V)/(C₁+C₂) (d) ((C₁+C₂)V)/((C₁+C₂+C₃))

Answer: (d). C₁ and C₂ are in parallel: C₁₂ = C₁ + C₂, and this is in series with C₃. In series the charge is the same: q = ((C₁+C₂)C₃)/(C₁+C₂+C₃)V. V₃ = q/C₃ = ((C₁+C₂)V)/(C₁+C₂+C₃). So the answer is (d).

Three capacitors C₁, C₂ and C₃ are connected to voltage source V as… | MHT CET 2026 (17 Apr, Shift 1)

Q6MHT CET 2026 · 17 APR · SHIFT 1Easy

Combination of Capacitors

Three capacitors $C_1, C_2$ and $C_3$ are connected to voltage source V as shown in figure. The voltage across $C_3$ will be

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Answer: (d)

$C_1$ and $C_2$ are in parallel: $C_{12} = C_1 + C_2$ , and this is in series with $C_3$ .
In series the charge is the same: $q = \frac{(C_1+C_2)C_3}{C_1+C_2+C_3}V$ .
$V_3 = \frac{q}{C_3} = \frac{(C_1+C_2)V}{C_1+C_2+C_3}$ .
So the answer is (d).

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Combination of Capacitors question from MHT CET 2026 (17 Apr, Shift 1)

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