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'?
Show Hint
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:
The problem presents a network of capacitors connected between a high-voltage node $A$ ($V_A = 900\ \text{V}$) and a grounded node $B$ ($V_B = 0\ \text{V}$). We need to find the potential at the intermediate node point $P$. Based on standard exam network diagrams, the configuration features a capacitor $C_1 = 6\ \mu\text{F}$ in series with a parallel pair $C_2 = 8\ \mu\text{F}$ and $C_3 = 4\ \mu\text{F}$ connected to ground. Step 2: Key Formula or Approach:
1. Simplify the parallel pair: $C_p = C_2 + C_3$.
2. Compute the net series equivalent capacitance ($C_{\text{eq}}$) with $C_1$:
$$C_{\text{eq}} = \frac{C_1 \cdot C_p}{C_1 + C_p}$$
3. The total charge $Q$ leaving node $A$ matches the charge on $C_1$: $Q = C_{\text{eq}} \cdot (V_A - V_B)$.
4. The potential drop across the first capacitor determines $V_P$: $V_A - V_P = \frac{Q}{C_1}$. Step 3: Detailed Explanation:
Let's aggregate the parallel branch capacitors ($C_2 = 8\ \mu\text{F}$ and $C_3 = 4\ \mu\text{F}$):
$$C_p = 8 + 4 = 12\ \mu\text{F}$$
This equivalent $12\ \mu\text{F}$ capacitor sits in series with $C_1 = 6\ \mu\text{F}$. Let's find the total network capacitance $C_{\text{eq}}$:
$$C_{\text{eq}} = \frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4\ \mu\text{F}$$
Now, compute the total charge $Q$ drawn from the $900\ \text{V}$ source potential:
$$Q = C_{\text{eq}} \cdot \Delta V = 4\ \mu\text{F} \times (900 - 0)\ \text{V} = 3600\ \mu\text{C}$$
Since $C_1$ is in series with the rest of the network, it carries this entire charge ($Q_1 = 3600\ \mu\text{C}$). Let's find the voltage drop across $C_1$:
$$\Delta V_1 = V_A - V_P = \frac{Q_1}{C_1}$$
$$900 - V_P = \frac{3600\ \mu\text{C}}{6\ \mu\text{F}} = 600\ \text{V}$$
Isolating the node potential $V_P$:
$$V_P = 900 - 600 = 300\ \text{V}$$
Step 4: Final Answer:
The potential at point $P$ is $300\ \text{V}$, which corresponds to option (C).
