Question

The difference between equivalent capacitances of two identical capacitors connected in parallel to that in series is $6 \,\mu F$ . The value of capacitance of each capacitor is

• A. $2 \mu F$
• B. $3 \mu F$
• C. $4 \mu F$
• D. $6 \mu F$

Answer 1

The Correct Option is C

Let the capacitance of each capacitor be \( C \). - The equivalent capacitance of two capacitors connected in parallel is given by: \[ C_{\text{parallel}} = C + C = 2C \] - The equivalent capacitance of two capacitors connected in series is given by: \[ C_{\text{series}} = \frac{C \cdot C}{C + C} = \frac{C}{2} \] We are given that the difference between the capacitances is 6 μF: \[ C_{\text{parallel}} - C_{\text{series}} = 6 \, \mu F \] Substituting the values: \[ 2C - \frac{C}{2} = 6 \] Multiplying through by 2 to clear the fraction: \[ 4C - C = 12 \] \[ 3C = 12 \] \[ C = 4 \, \mu F \] Thus, the value of capacitance of each capacitor is 4 μF.