Question

The effective capacitance between points A and B shown in the circuit is: 

• A. 2C
• B. C
• C. C/2
• D. 5C

Answer 1

The Correct Option is A

From the circuit, \( C_4 \) is connected in parallel with the rest of the combined capacitors. So, we will draw a simplified circuit:

Here, \( C' \) is the combined capacitance of the capacitors excluding \( C_4 \).

Now, we see that the rest of the circuit is equivalent to the Wheatstone bridge network. So, when we will apply the balance condition to the Wheatstone network, then \( C_3 \) could be neglected.

Now, the resulting circuit we will get is:

Using the series combination in the upper portion, we will find the equivalent capacitance:

\[ \frac{1}{C''} = \frac{1}{C_1} + \frac{1}{C_6} \] \[ \Rightarrow \frac{1}{C''} = \frac{C_1 + C_6}{C_1 \times C_6} \] \[ \Rightarrow C'' = \frac{C_1 \times C_6}{C_1 + C_6} \]

Similarly, using the series combination in the bottom portion, we will find the equivalent capacitance:

\[ \frac{1}{C'''} = \frac{1}{C_2} + \frac{1}{C_5} \] \[ \Rightarrow \frac{1}{C'''} = \frac{C_2 + C_5}{C_2 \times C_5} \] \[ \Rightarrow C''' = \frac{C_2 \times C_5}{C_2 + C_5} \]

Since \( C_1 = C_6 = C_2 = C_5 = C \), we can simplify the expressions:

\[ C'' = \frac{C \times C}{C + C} = \frac{C^2}{2C} = \frac{C}{2} \] \[ C''' = \frac{C \times C}{C + C} = \frac{C^2}{2C} = \frac{C}{2} \]

The equivalent capacitance \( C' \) (excluding \( C_4 \)) is the parallel combination of \( C'' \) and \( C''' \):

\[ C' = C'' + C''' \] \[ C' = \frac{C}{2} + \frac{C}{2} = C \]

Finally, the effective capacitance between A and B is the parallel combination of \( C_4 \) and \( C' \):

\[ C_{eq} = C_4 + C' \] \[ C_{eq} = C + C = 2C \]

Thus, the effective capacitance between A and B will be \( 2C \).

Hence, option (A) is the correct answer.