Question

Three parallel plate capacitors of capacitances \(4 \mu F\), \(6 \mu F\), and \(12 \mu F\) are first connected in series and then in parallel. The ratio of the effective capacitances in the two cases is:

• A. \(1 : 11\)
• B. \(5 : 8\)
• C. \(3 : 7\)
• D. \(4 : 9\)

Hint:

When connecting capacitors: - In series, the total capacitance decreases. - In parallel, the total capacitance increases. This is due to the distribution and combination of charges and potential differences.

Answer 1

The Correct Option is A

Capacitors in Series: When capacitors are connected in series, the total capacitance \(C_{{series}}\) is calculated as: \[ \frac{1}{C_{{series}}} = \frac{1}{C_1} + \frac{1}{C_2} + \frac{1}{C_3} \] Substituting the given capacitances: \[ \frac{1}{C_{{series}}} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12} = \frac{1}{2} \] \[ C_{{series}} = 2 \mu F \] Capacitors in Parallel: When capacitors are connected in parallel, the total capacitance \(C_{{parallel}}\) is the sum of individual capacitances: \[ C_{{parallel}} = C_1 + C_2 + C_3 = 4 + 6 + 12 = 22 \mu F \] Ratio of Capacitances: The ratio of the effective capacitances in series to parallel is: \[ {Ratio} = \frac{C_{{series}}}{C_{{parallel}}} = \frac{2}{22} = \frac{1}{11} \]