Question

A 200 volt battery is connected across the series combination of two capacitors 4 \(\mu\)F and 6 \(\mu\)F. The amount of energy stored in this series combination is:

• A. \(38 \times 10^{-2} { J}\)
• B. \(48 \times 10^{-2} { J}\)
• C. \(3.8 \times 10^{-2} { J}\)
• D. \(4.8 \times 10^{-2} { J}\)

Hint:

For capacitors in series, the equivalent capacitance reduces, and the voltage across each depends on its capacitance.

Answer 1

The Correct Option is D

Step 1: To calculate the energy stored in the series combination of capacitors, we first need to calculate the equivalent capacitance of the two capacitors in series. The formula for the equivalent capacitance \( C_{eq} \) for capacitors in series is given by: \[ \frac{1}{C_{eq}} = \frac{1}{C_1} + \frac{1}{C_2} \] where \(C_1 = 4 \, \mu F\) and \(C_2 = 6 \, \mu F\). Therefore: \[ \frac{1}{C_{eq}} = \frac{1}{4} + \frac{1}{6} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \] \[ C_{eq} = \frac{12}{5} = 2.4 \, \mu F. \] Step 2: Now, we calculate the energy stored in the series combination of capacitors using the formula for energy stored in a capacitor: \[ E = \frac{1}{2} C_{eq} V^2 \] where \(V = 200 \, {V}\) is the voltage across the capacitors. Substituting the values: \[ E = \frac{1}{2} \times 2.4 \times 10^{-6} \times (200)^2 = \frac{1}{2} \times 2.4 \times 10^{-6} \times 40000 = 4.8 \times 10^{-2} \, {J}. \]