Saurabh Singh
Last Activity: 10 Years ago
now this is the general case .....we have a voltage source, a resistance , a capacitor and an inductor.
Kirchhoff law says that for a mesh the sum of voltages should be zero ( in a loop)
so we start from left side ,
we have a voltage source so we write V(t) -(assuming it may vary with time , this makes our equation flexible)
then we have a resistor so voltage drop will be iR , since its is a drop so we write -iR
then moving towards right we have a capacitor
voltage across a capacitor is simply q/C, so we write -q/C as it is a drop
and finally at the rightmost we have our inductor so its -Ldi/dt
once we have this equation we can solve it using the fact that current is rate of change of charge with time so i=dq/dt
substituting i in the equation we get a second order differential equation in 'q' which is easy to solve.
dear student, this is the generalized method and can be very easily extended to other cases such as only
capacitor or only inductor by setting the values of L and C zero as required.
NOW IN OUR CASE WE HAVE ONLY CAPACITOR, SO WE CAN SET L = 0 AND THEN SOLVE THE EQUATION.
Thanks & Regards
Saurabh Singh,
askIITians Faculty
B.Tech.
IIT Kanpur