Question:medium
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
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
Show Hint
\(C_1\) and \(C_2\) are in parallel, and that pair is in series with \(C_3\) and the source.
Updated On: Oct 1, 2026
- \(\frac{C_3V}{(C_1+C_2+C_3)}\)
- \(\frac{(C_1+C_2)V}{C_3}\)
- \(\frac{(C_2+C_3)V}{C_1+C_2}\)
- \(\frac{(C_1+C_2)V}{(C_1+C_2+C_3)}\)
Show Solution
The Correct Option is D
Solution and Explanation
Step 1: Use the voltage divider rule for capacitors
For two capacitors in series, the voltage on one is proportional to the capacitance of the other: $V_3=V\cdot\dfrac{C_{12}}{C_{12}+C_3}$.
Step 2: Substitute
With $C_{12}=C_1+C_2$ this gives $V_3=\dfrac{(C_1+C_2)V}{C_1+C_2+C_3}$, option (D).
Final Answer:
Option (D) gives the voltage on $C_3$. \[ \boxed{\dfrac{(C_1+C_2)V}{C_1+C_2+C_3}} \]
For two capacitors in series, the voltage on one is proportional to the capacitance of the other: $V_3=V\cdot\dfrac{C_{12}}{C_{12}+C_3}$.
Step 2: Substitute
With $C_{12}=C_1+C_2$ this gives $V_3=\dfrac{(C_1+C_2)V}{C_1+C_2+C_3}$, option (D).
Final Answer:
Option (D) gives the voltage on $C_3$. \[ \boxed{\dfrac{(C_1+C_2)V}{C_1+C_2+C_3}} \]
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