Question

In a circuit, the capacitance \( C \) is connected. The effective capacitance of the circuit can be reduced by

• A. introducing a metal plate between the plates of the capacitor
• B. introducing a dielectric slab between the plates
• C. reducing the potential difference between the plates
• D. connecting another capacitor in series with it
• E. connecting another capacitor in parallel with it

Hint:

Series → decreases, Parallel → increases capacitance.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This question asks how to decrease the total or effective capacitance of a circuit that initially contains a capacitor C. We need to analyze how different modifications affect capacitance.
Step 2: Key Formula or Approach:

Dielectric Slab: Inserting a dielectric of constant K increases capacitance: \(C' = KC\), where \(K>1\).
Metal Plate: Inserting a conducting slab increases capacitance.
Series Combination: For two capacitors \(C_1\) and \(C_2\) in series, the effective capacitance \(C_{eff}\) is given by \(\frac{1}{C_{eff}} = \frac{1}{C_1} + \frac{1}{C_2}\). This results in \(C_{eff}<\min(C_1, C_2)\).
Parallel Combination: For two capacitors in parallel, \(C_{eff} = C_1 + C_2\). This results in \(C_{eff}>\max(C_1, C_2)\).
Step 3: Detailed Explanation:
We want to reduce the effective capacitance from its initial value C.
(A) introducing a metal plate: Inserting a metal plate of thickness `t` into a capacitor with plate separation `d` is equivalent to creating two capacitors in series, each with a smaller separation. The new capacitance is \(C' = C \frac{d}{d-t}\). Since \(d>d-t\), \(C'>C\). Capacitance increases.
(B) introducing a dielectric slab: This increases the capacitance by a factor of the dielectric constant K (\(C' = KC\)). Capacitance increases.
(C) reducing the potential difference: Capacitance \(C = Q/V\) is a physical property of the capacitor's geometry and material. It does not depend on the potential difference V or charge Q applied to it. This action does not change C.
(D) connecting another capacitor in series: If we connect another capacitor \(C_{add}\) in series with C, the new effective capacitance is \(C_{eff} = \frac{C \cdot C_{add}}{C + C_{add}}\). Since \(\frac{C_{add}}{C + C_{add}}<1\), the new capacitance \(C_{eff}\) will be less than C. Capacitance is reduced.
(E) connecting another capacitor in parallel: If we add \(C_{add}\) in parallel, the new capacitance is \(C_{eff} = C + C_{add}\). This is greater than C. Capacitance increases.
Step 4: Final Answer:
The effective capacitance can be reduced by connecting another capacitor in series with it.