Question

Two parallel plate capacitors, each of capacitance 40 µF, are connected in series. The space between the plates of one capacitor is filled with a material of dielectric constant K = 4. The equivalent capacitance of the system would be:

• A. 30 µF
• B. 31 µF
• C. 32 µF
• D. 33 µF

Answer 1

The Correct Option is C

Capacitor System Solution:

Given:

• Two 40 μF capacitors in series
• One capacitor has dielectric (K=4)

Calculations:

1. Capacitor with dielectric: \( C_1 = K \times C = 4 \times 40\mu F = 160\mu F \)

2. Other capacitor: \( C_2 = 40\mu F \) (unchanged)

3. Series combination formula:

\[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{160} + \frac{1}{40} = \frac{1}{32} \]

4. Equivalent capacitance: \( C_{eq} = 32\mu F \)

Correct Answer:  \(32\  μF\)

Answer 2

\(\text{ First, calculate the capacitance of the capacitor with the dielectric inserted.}\)

\(C_1 \text{ The new capacitance with dielectric constant } K = 4 \text{ is:}\)

\(C_1 = K \times C = 4 \times 40 \, \mu\text{F} = 160 \, \mu\text{F}\)

\(\text{The second capacitor } C_2 \text{ remains unchanged at } 40 \, \mu\text{F}.\)

\(\text{The equivalent capacitance for capacitors in series is given by:}\)

\(\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2}\)

\(\text{Substituting values:}\)

\(\frac{1}{C_{\text{eq}}} = \frac{1}{160 \, \mu\text{F}} + \frac{1}{40 \, \mu\text{F}} = \frac{1}{160} + \frac{1}{40} = \frac{1 + 4}{160} = \frac{5}{160}\)

\(C_{\text{eq}} = \frac{160}{5} = 32 \, \mu\text{F}\)