Question

A \(5 \mu \mathrm{F}\) capacitor is charged by a \(220 \mathrm{v}\) supply. It is then disconnected from the supply and is connected to another uncharged $2.5 \mu \mathrm{F}$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation ? (A) \(0.02 \mathrm{~J}\) (B) \(0.121 \mathrm{~J}\) (C) \(0.04 \mathrm{~J}\) (D) \(0.081 \mathrm{~J}\)

Short answer

Final Voltage = 146.67 V #tag_title#Step 3: Calculate the final energy stored in the connected capacitors#tag_content#Now that we have the final voltage across the connected capacitors, we can calculate the final energy stored in them using the energy stored in a capacitor formula: Final Energy = \(\frac{1}{2}\) × Total Capacitance × Final Voltage^2 Final Energy = \(\frac{1}{2}\) × 7.5 × 10^{-6} F × (146.67 V)^2 #tag_title#Step 4: Determine the energy lost#tag_content#Finally, we can find the energy lost in the process by subtracting the final energy stored in the connected capacitors from the initial energy stored in the charged capacitor: Energy Lost = Initial Energy - Final Energy Energy Lost = \(\frac{1}{2} \times 5 \times 10^{-6} \text{F} \times (220 \text{V})^2 - \frac{1}{2} \times 7.5 \times 10^{-6} \text{F} \times (146.67 \text{V})^2\) Energy Lost = 0.121 J The energy lost in the form of heat and electromagnetic radiation during the process is (B) 0.121 J.

Step-by-step solution

Calculate the initial energy stored in the charged capacitor

The initial energy stored in the 5 μF capacitor charged by a 220 V supply can be calculated using the formula for the energy stored in a capacitor: Energy = \(\frac{1}{2}\) × Capacitance × Voltage^2 Let's plug in the given values and calculate the initial energy. Energy = \(\frac{1}{2}\) × 5 × 10^{-6} F × (220 V)^2

Determine the final voltage across the connected capacitors

When the capacitors are connected, the capacitor's charge will be shared, and the voltage across both capacitors will be the same. The total charge will remain constant, but the total capacitance will change. We can use the formula for capacitors connected in parallel to find the total capacitance: Total Capacitance = Capacitance1 + Capacitance2 Total Capacitance = 5 × 10^{-6} F + 2.5 × 10^{-6} F = 7.5 × 10^{-6} F Now, we can use the constant charge formula (Q = C × V) to find the final voltage across the connected capacitors: Initial Charge = Initial Capacitance × Initial Voltage Final Charge = Total Capacitance × Final Voltage As the charge is constant: Initial Charge = Final Charge Initial Capacitance × Initial Voltage = Total Capacitance × Final Voltage Final Voltage = \(\frac{\text{Initial Capacitance} \times \text{Initial Voltage}}{\text{Total Capacitance}}\) Final Voltage = \(\frac{5 \times 10^{-6} \text{F} \times 220 \text{V}}{7.5 \times 10^{-6} \text{F}}\)