In the given figure potential at point 'A' is $900\ \text{volt}$ and point 'B' is earthed. What will be the potential at point 'P'?
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Use the potential divider rule for series capacitors! The voltage drop across a capacitor is inversely proportional to its capacitance. Since the parallel group ($12\ \mu\text{F}$) is exactly double the value of $C_1$ ($6\ \mu\text{F}$), the voltage drop across $C_1$ must be twice as large as the drop across the parallel group ($2 : 1$ ratio). Splitting $900\ \text{V}$ into three equal parts of $300\ \text{V}$ means $C_1$ drops $600\ \text{V}$, leaving exactly $300\ \text{V}$ at point $P$.
Step 1: Understanding the Question: Find the voltage at a point in a series capacitor network using the potential divider principle. Step 2: Key Formula or Approach: For series capacitors, voltage divides inversely to capacitance: V ∝ 1/C. The parallel combination's equivalent capacitance is compared to the single series capacitor to establish the division ratio. Step 3: Detailed Explanation: The parallel group has 12 μF, exactly double the 6 μF of C₁. Therefore, C₁ receives twice the voltage drop of the parallel group—a 2:1 split. Dividing 900 V into three equal portions of 300 V assigns 600 V across C₁ and 300 V across the parallel combination. The potential at point P is thus 300 V. This inverse-capacitance reasoning avoids setting up simultaneous equations. Step 4: Final Answer: The voltage at point P is 300 V.
