# how to solve questions on kvl in a circuit containing capacitors ???

Saurabh Singh
`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 -iRthen moving towards right we have a capacitorvoltage across a capacitor is simply q/C, so we write -q/C as it is a dropand finally at the rightmost we have our inductor so its -Ldi/dtonce we have this equation we can solve it using the fact that current is rate of change of charge with time so i=dq/dtsubstituting 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.`
`Thanks & RegardsSaurabh Singh,askIITians FacultyB.Tech.IIT Kanpur`