Question

An uncharged capacitor of capacitance 100 mu F is connected to a battery of emf 20V at t = 0 through a resistance 10 Omega, then (ii) time at which the rate has time miximum value is

Answer 1

Explanation:

Given An uncharged capacitor of capacitance 100 mu F is connected to a battery of emf 20 V at t = 0 through a resistance 10 Ohm, then maximum rate at which the energy is stored in the capacitor is

• We need to find the maximum rate at which energy stored in the capacitor.
• So maximum rate will be power.
• So du/dt is P = maximum energy stored in capacitor / minimum time taken
•  Now minimum time = 5 x time constant
•            = 5 x R x C
•       = 5 x 10 x 100 x 10^-6
•        = 5 x 10^-3 secs
• Now maximum energy stored in the capacitor U max = 1/2 c v^2
•           = 1/2 x 100 x 10^-6 x 20 x 20
•           = 0.02 J
• So dv / dt (energy) = 0.02 / 5 x 10^-3
•     = 4 J / s

Therefore maximum rate at which energy stored in the capacitor is 4 J / s

Reference link will be

https://brainly.in/question/15427958