Question

A network of five capacitors is shown in the figure. The total charge stored in the network connected between A and B is:

• A. $8\ \mu\text{C}$
• B. $10\ \mu\text{C}$
• C. $9\ \text{mC}$
• D. $10\ \text{mC}$

Hint:

Always look for symmetry first! In most exam questions featuring five capacitors, the bridge is balanced, which lets you ignore the middle capacitor and solve the circuit much faster.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The given circuit diagram shows a bridge configuration of five capacitors.
We must first check if the bridge is balanced by comparing the ratios of the capacitances of the outer branches.
Step 2: Key Formula or Approach:
For a balanced Wheatstone bridge of capacitors, the condition is:
\[ \frac{C_1}{C_2} = \frac{C_3}{C_4} \]
If balanced, the central capacitor (in this case, 5 $\mu$F) can be ignored as no potential difference exists across it.
The total equivalent capacitance \( C_{eq} \) is then calculated by treating the top and bottom branches as parallel combinations of series-connected pairs.
Total charge is found using \( Q = C_{eq}V \).
Step 3: Detailed Explanation:
Let \( C_1 = 4 \, \mu\text{F} \), \( C_2 = 2 \, \mu\text{F} \), \( C_3 = 6 \, \mu\text{F} \), and \( C_4 = 3 \, \mu\text{F} \).
Check the ratio: \( \frac{C_1}{C_2} = \frac{4}{2} = 2 \) and \( \frac{C_3}{C_4} = \frac{6}{3} = 2 \).
Since \( \frac{C_1}{C_2} = \frac{C_3}{C_4} \), the bridge is balanced.
The 5 $\mu$F capacitor is redundant and can be removed from the circuit.
Now, the upper branch consists of 4 $\mu$F and 2 $\mu$F in series:
\[ C_{upper} = \frac{4 \times 2}{4 + 2} = \frac{8}{6} = \frac{4}{3} \, \mu\text{F} \]
The lower branch consists of 6 $\mu$F and 3 $\mu$F in series:
\[ C_{lower} = \frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2 \, \mu\text{F} \]
The equivalent capacitance \( C_{eq} \) is the sum of these parallel branches:
\[ C_{eq} = \frac{4}{3} + 2 = \frac{4 + 6}{3} = \frac{10}{3} \, \mu\text{F} \]
The total charge \( Q \) is:
\[ Q = C_{eq} \times V = \frac{10}{3} \times 3 = 10 \, \mu\text{C} \]
Step 4: Final Answer:
The total charge stored in the network is 10 $\mu$C.