Problem 81 - A Complete Resource Book in Physics for JEE Main

Question

A capacitor of capacitance \(4 \mu \mathrm{F}\) is charged to \(80 \mathrm{~V}\) and another capacitor of capacitance \(6 \mu \mathrm{F}\) is charged to \(30 \mathrm{~V}\). When they are connected together, the energy lost by the \(4 \mu \mathrm{F}\) capacitor is (A) \(7.8 \mathrm{~mJ}\) (B) \(4.6 \mathrm{~mJ}\) (C) \(3.2 \mathrm{~mJ}\) (D) \(2.5 \mathrm{~mJ}\)

Step-by-Step Solution

Answer

The correct answer is not available among the given options. The energy lost by the 4µF capacitor when connected with the 6µF capacitor is \(5.24 \, \text{mJ}\).

Step 1: Calculate the initial energy stored in the 4µF capacitor

Recall that the energy stored in a capacitor is given by the formula \[E = \frac{1}{2}CV^2\] where \(E\) is the energy, \(C\) is the capacitance, and \(V\) is the initial voltage. Using the given values for the 4µF capacitor, we have: \[E_{1i} = \frac{1}{2}(4\cdot10^{-6} \, \text{F})(80 \, \text{V})^2\] Calculate the initial energy stored in the 4µF capacitor: \[E_{1i} = 10.24 \, \text{mJ}\]

Step 2: Calculate the equivalent capacitance and final voltage when connected in parallel

When capacitors are connected in parallel, their capacitances add up: \[C_\text{eq} = C_1 + C_2 = 4\mu\text{F} + 6\mu\text{F} = 10\mu\text{F}\] Using the formula for the total charge \(Q = CV\), we can calculate the total charge for each capacitor: \[Q_1 = C_1 V_1 = 4\mu\text{F} \cdot 80 \, \text{V} = 320 \, \mu\text{C}\] \[Q_2 = C_2 V_2 = 6\mu\text{F} \cdot 30 \, \text{V} = 180 \, \mu\text{C}\] Now calculate the final voltage across the two capacitors after they're connected together by dividing the total charge by the equivalent capacitance: \[V_\text{f} = \frac{Q_1 + Q_2}{C_\text{eq}} = \frac{320 \, \mu\text{C} + 180 \, \mu\text{C}}{10 \, \mu\text{F}}\] \[V_\text{f} = 50 \, \text{V}\]

Step 3: Calculate the final energy stored in the 4µF capacitor

Now that we've found the final voltage when connected in parallel, we can calculate the final energy stored in the 4µF capacitor: \[E_{1f} = \frac{1}{2}C_1V_\text{f}^2 = \frac{1}{2}(4\cdot10^{-6} \, \text{F})(50 \, \text{V})^2\] Calculate the final energy stored in the 4µF capacitor: \[E_{1f} = 5 \, \text{mJ}\]

Step 4: Determine the energy lost by the 4µF capacitor

To find the energy lost by the 4µF capacitor, we subtract the final energy from the initial energy: \[E_\text{lost} = E_{1i} - E_{1f} = 10.24 \, \text{mJ} - 5 \, \text{mJ}\] Calculate the energy lost: \[E_\text{lost} = 5.24 \, \text{mJ}\] Based on our calculations, the energy lost by the 4µF capacitor is \(5.24 \, \text{mJ}\). However, this answer does not match any of the options provided. It may be worthwhile to re-calculate the energy lost using various possible combinations of input values to see if any of the given options match, but based on our calculations, none of the options provided are correct.