Heat dissipated during capacitor discharge – Rankers Physics

Charging and Discharging of Capacitors: Practice Problem & Solution

A \(5.0~\mu\text{F}\) capacitor having a charge of \(20~\mu\text{C}\) is discharged through a wire of resistance \(5.0~\Omega\). Find the heat dissipated in the wire between 25 to 50 \(\mu\text{s}\) after the connections are made.
\(40\left(\frac{1}{e^2} - \frac{1}{e^4}\right)~\mu\text{J}\)
\(40\left(\frac{1}{e} - \frac{1}{e^2}\right)~\mu\text{J}\)
\(40\left(\frac{1}{e^2} - \frac{1}{e^3}\right)~\mu\text{J}\)
None of these

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Charging and Discharging of Capacitors. Understanding the underlying formula is key to arriving at the correct answer below:

The remaining energy in the capacitor is \(U(t) = \frac{q_0^2}{2C}e^{-2t/tau}\), where \(tau = RC = 25~\mu\text{s}\). The heat dissipated is \(H = U(t_1) - U(t_2) = \frac{q_0^2}{2C}\left(e^{-2} - e^{-4}\right)\) where \(frac{q_0^2}{2C} = 40~\mu\text{J}\).

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. A parallel plate capacitor has area of each plate as \(A\), the separation between the plates as \(d\) and it is charged to potential \(V\), and then disconnected from the battery. If a dielectric slab, completely filling the capacitor is introduced, how much work will be done in doing so
2. Assertion (A): When a dielectric slab is kept near an isolated parallel plate charged capacitor, it will pull the dielectric slab between the plates. Reason (R): Energy of system decreases when dielectric slab enters between plates of charged parallel plate capacitor.  
3. Assertion (A): A parallel plate capacitor is charged to a potential difference of \( 100\text{V} \), and disconnected from the voltage source. A slab of dielectric is then slowly inserted between the plates. Compared to the energy before the slab was inserted, the energy stored in the capacitor with the dielectric is decreased. Reason (R): When we insert a dielectric between the plates of a capacitor, the induced charges tend to draw in the dielectric into the field (just as neutral objects are attracted by charged objects due to induction). We resist this force while slowly inserting the dielectric, and thus do negative work on the system, removing electrostatic energy from the system.  
4. Assertion (A): After charging a capacitor of capacitance (C) from a battery, it is connected to the same battery of potential difference (V) with reverse polarity. Loss of energy in this process is \(2CV^2\). Reason (R): Work done by the battery is equal to loss of energy in the given case.  
5. It is required to construct a \(10\text{ \mu F}\) capacitor which can be connected across a \(200\text{ V}\) battery. Capacitors of capacitance \(10\text{ \mu F}\) are available but they can withstand only \(50\text{ V}\). Design a combination which can yield the desired result.