The charge on each capacitor when a voltage source of $15 \text{ V}$ is connected in the circuit as shown, is
Show Hint
In a parallel circuit, every single branch experiences the exact same voltage drop ($V = 15 \text{ V}$). You can completely skip calculating the total combination charge! Just calculate the charge directly using a single component: $Q = C \cdot V = 5 \; \mu\text{F} \times 15 \text{ V} = 75 \; \mu\text{C}$.
Step 1: Understanding the Question:
The problem asks for the electric charge stored on each individual capacitor within a network containing four identical capacitors, given that each individual capacitor has a capacitance of $5 \; \mu\text{F}$ and the system is connected across a $15 \text{ V}$ power supply.
Step 2: Key Formula or Approach:
For capacitors connected together in a parallel combination, the equivalent capacitance $C_{\text{eq}}$ is simply the sum of all individual capacitances:
$$C_{\text{eq}} = C_1 + C_2 + C_3 + C_4$$
The total charge $Q_{\text{total}}$ delivered by the voltage source $V$ to the parallel group is:
$$Q_{\text{total}} = C_{\text{eq}} \cdot V$$
Since the four parallel capacitors are entirely identical, the total charge divides evenly among them:
$$Q_{\text{each}} = \frac{Q_{\text{total}}}{4}$$
Step 3: Detailed Explanation:
First, find the total equivalent capacitance of the four $5 \; \mu\text{F}$ parallel units:
$$C_{\text{eq}} = 5 \; \mu\text{F} + 5 \; \mu\text{F} + 5 \; \mu\text{F} + 5 \; \mu\text{F} = 20 \; \mu\text{F}$$
Next, compute the total charge stored by this entire circuit combination:
$$Q_{\text{total}} = C_{\text{eq}} \cdot V = 20 \; \mu\text{F} \times 15 \text{ V} = 300 \; \mu\text{C}$$
Finally, because all branches are identical and share the same potential drop, calculate the charge on an individual capacitor:
$$Q_{\text{each}} = \frac{300 \; \mu\text{C}}{4} = 75 \; \mu\text{C}$$
Step 4: Final Answer:
The charge accumulated on each individual capacitor is $75 \; \mu\text{C}$, matching option (A).
