In the figure above, initially the switch is at point a so that the current flow
ID: 1595133 • Letter: I
Question
In the figure above, initially the switch is at point a so that the current flows through the circuit. Note that the governing equation is given by Vb = IR + Vc where Vb is the voltage of power supply, I is the current, R is the resistance and Vc is potential drop across the capacitor. This can be written as Vb = IR + q/C Using I=dq/dt, and the fact that Vb, R and C are constant in time, solve the above equation to find charge q on the capacitor as a function of time. Find out the charge on the capacitor q at infinite time. integral 1/ax + b dx = log(ax + b/a) + constant Define all variables and parameters. Briefly explain each step of your proof in words. Number your equations (1), (2), (3), ... in the right hand margin, using the format given in the Formal Lab Report Requirements section of the lab manual. If possible, state whether the proposed solution is the only solution for this differential equation. Cite an authority, such as your calculus textbook. Follow the format outlined in the Formal Lab Report Requirements section of the lab manual.Explanation / Answer
assumption:
initially the capacitor is uncharged.
q(0)=0
governing equation:
Vb=I*R+(q/C)
and I=dq/dt
==>Vb=R*(dq/dt)+(q/C)
==>R*C*(dq/dt)+q=C*Vb
==>R*C*(dq/dt)=(C*Vb-q)
==>R*C*dq/(C*Vb-q)=dt...(1)
let C*Vb-q=p
taking derivative on both sides,
-dq=dp
substituing in equation 1:
-R*C*dp/p=dt
interating both sides,
-R*C*ln(p)=t+k
where k is a constant of integration
using p=C*Vb-q
-R*C*ln(C*Vb-q)=t+k
==>C*Vb-q=exp(-t/R*C)*k1
where k1=exp(-k/R*C)
==>q=C*Vb-k1*exp(-t/(R*C))...(2)
as per our assumption, q(0)=0
==>0=C*Vb-k1
==>k1=C*Vb
using this expression for k1 in equation 2,
q=C*Vb*(1-exp(-t/(R*C))
(answer)
part 2:
at t=infinite, exp(-t/(R*C))=0
hence charge=C*Vb*(1-0)=C*Vb