Calculate DC current through an inductor?

Hi everyone,

Say I have a series DC circuit. The circuit contains a 100V DC source, a switch, a 1.5H inductor, and a 1nF capacitor (all in series).

If I close the switch for 2uS what is the current at that time?

Initially I thought I could rearrange V=L*di/dt to

V*Ton/L = di/dt which would be 100*.000002/.150=1.3mA???

For some reason I am thinking this is not right since the capacitor charge should also effect the current.

Can anyone help me understand and calculate this correctly?

Simon Bridge
Homework Helper
Once you have turned the circuit on it settles to a steady state very quickly - so you use V=IR - where R is the DC resistance of the inductor. An ideal inductor is just a length of wire ... R=0.

The transient response requires you solve the differential equation.

For an ideal switch closed for time T, ##v(t)=V_0\big ( u(t)-u(t-T)\big )##

Include the cap voltage too, which is 1/c times integral of i(t) dt. Basically a KVL. Take derivative to get rid of integral and get a 2nd order diff eq. Then solve for i(t)

Ok, I guess I need to go study differential equations. Never got that far in math.

Could anyone show me step by step how to solve?

Last edited:
Simon Bridge