A resistor, R, a capacitor C, a switch S and a battery are in series. The switch

Question

A resistor, R, a capacitor C, a switch S and a battery are in series. The switch is closed at t = 0. At that time, the capacitor holds no charge. For t > 0:

a) Write down the differential equation for the charge Q on the lower plate of the capacitor.

b) Show, using your equation above, that the correct solution is:

c) What is the current in the circuit at time t1 (t1 > 0)?

d) How much energy is stored in the capacitor at time t1?

e) How much heat has been generated in the resistor between t = 0 and t1?

A resistor, R, a capacitor C, a switch S and a battery are in series. The switch is closed at t = 0. At that time, the capacitor holds no charge. For t > 0: Write down the differential equation for the charge Q on the lower plate of the capacitor. Show, using your equation above, that the correct solution is: Q = CV [1 - exp (- t/RC)] What is the current in the circuit at time t1 (t1 > 0)? How much energy is stored in the capacitor at time t1? How much heat has been generated in the resistor between t = 0 and t1?

Explanation / Answer

a)using KVL for the above loop,

q/C+ir-V=0

but i=dq/dt

so,

q/C + ir-V=0

b)integrATING THe above equation,

we get q=CV(1-e^(-t/RC))

c) i(t)=dq/dt

=V/R *(e^(-t/RC))

d)elergy stored=0.5q^2/C

=0.5*CV^2*(1-e^(-t/RC))^2

e)heat generated=integral of i^2r from 0 to t1

which gives us energy=(V^2C)*e^(-2*t1/RC))

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