Question

Three parallel plate capacitors $ C_1 $, $ C_2 $, and $ C_3 $ each of capacitance 5 µF are connected as shown in the figure. The effective capacitance between points A and B, when the space between the parallel plates of $ C_1 $ capacitor is filled with a dielectric medium having dielectric constant of 4, is:

• A. 22.5 µF
• B. 7.5 µF
• C. 9 µF
• D. 30 µF

Hint:

When a dielectric is inserted into a capacitor, its capacitance increases by a factor equal to the dielectric constant. Series and parallel combinations of capacitors should be handled accordingly.

Answer 1

The Correct Option is C

The given problem involves calculating the effective capacitance between points A and B in a circuit of capacitors. We have three capacitors, \(C_1\), \(C_2\), and \(C_3\), each with a capacitance of 5 µF. The capacitor \(C_1\) has a dielectric medium with a dielectric constant of 4.

Step-by-step Solution:

1. The dielectric affects \(C_1\), increasing its capacitance by the factor of the dielectric constant: 

\[C_1' = \kappa \cdot C_1 = 4 \cdot 5 \, \text{µF} = 20 \, \text{µF}\]

1. Now, we have \(C_1'\) with 20 µF, \(C_2\) with 5 µF, and \(C_3\) with 5 µF.
2. \(C_1'\) and \(C_2\) are in series, so their equivalent capacitance \(C_{12}\) is calculated as follows: 

\[\frac{1}{C_{12}} = \frac{1}{C_1'} + \frac{1}{C_2} = \frac{1}{20} + \frac{1}{5} = \frac{1}{20} + \frac{4}{20} = \frac{5}{20}\]

1.  

\[C_{12} = \frac{20}{5} = 4 \, \text{µF}\]

1. \(C_{12}\) is in parallel with \(C_3\), so the total capacitance \(C_{\text{total}}\) is: 

\[C_{\text{total}} = C_{12} + C_3 = 4 \, \text{µF} + 5 \, \text{µF} = 9 \, \text{µF}\]

Conclusion:

The effective capacitance between points A and B is 9 µF. The correct answer is 9 µF.

Answer 2

1. After Dielectric is Inserted: The capacitance \( C_1 \) is modified due to the dielectric, and the new capacitance \( C_1' \) becomes: \[ C_1' = 4C_1 = 4 \times 5 \, \mu\text{F} = 20 \, \mu\text{F} \] 
2. Combination of Capacitors: - \( C_1' = 20 \, \mu\text{F} \) (with dielectric), 
- \( C_2 = C_3 = 5 \, \mu\text{F} \) (without dielectric). \( C_1' \) and \( C_2 \) are in series, and their equivalent capacitance \( C_{eq} \) is given by: \[ \frac{1}{C_{eq}} = \frac{1}{C_1'} + \frac{1}{C_2} \] Substituting the values: \[ \frac{1}{C_{eq}} = \frac{1}{20} + \frac{1}{5} = \frac{1}{20} + \frac{4}{20} = \frac{5}{20} \] Therefore: \[ C_{eq} = \frac{20}{5} = 4 \, \mu\text{F} \] Now, this equivalent capacitance \( C_{eq} \) is in parallel with \( C_3 \), so the total capacitance \( C_{total} \) is: \[ C_{total} = C_{eq} + C_3 = 4 \, \mu\text{F} + 5 \, \mu\text{F} = 9 \, \mu\text{F} \] 
Thus, the effective capacitance is \( 9 \, \mu\text{F} \), and the correct answer is (3).