Energy dissipated

Can someone explain what part E of the question means and how to do it? Thanks!
https://isaacphysics.org/questions/two_capacitor_problem?stage=a_level

Reply 1

Eimmanuel

Study Forum Helper

Original post by Scarlett L

Can someone explain what part E of the question means and how to do it? Thanks!
https://isaacphysics.org/questions/two_capacitor_problem?stage=a_level

I thought the question was quite clear: in a typical DC circuit (no capacitor) that consists of a switch, a resistor with resistance R and an ideal cell of emf V, the switch is closed for 2 minutes and what is the energy dissipated in the resistor in 2 minutes? Energy dissipated = (V^2/R)*120 J

Returning to Part E, when S₁ is opened and S₂ is closed, charges on C₁ are transferred to C₂ and current flows through the resistor R₂ for some time, say t₁. When the current flows through the resistor R₂, energy is dissipated in a time interval of t₁ which is what the question is asking.
Next, we can find the energy dissipated in R₂, by using the principle of conservation of energy: find the initial energy stored in C₁, final energy in C₁, final energy in C₂ and hence the energy dissipated in R₂.

Reply 2

Scarlett L

I’ve solve the question now, thank you for your hint!