# Calculate DC current through an inductor?

1. May 29, 2013

### hobbs125

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?

2. May 29, 2013

### Simon Bridge

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, $v(t)=V_0\big ( u(t)-u(t-T)\big )$

3. May 29, 2013

### omega_minus

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)

4. May 29, 2013

### hobbs125

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: May 29, 2013
5. May 30, 2013

### Simon Bridge

I missed the cap - right ... the solution is a damped/driven harmonic oscilator.
You can look it up. The driving function is the unput voltage as a function of time.
The u(t) function in post #2 are the Heaviside step function ... which you can look up too.

It would be a bit like pushing a pendulum off equilibrium and holding it for a bit before letting it go.

What you get is, basically, called the "impulse response" of the RCL circuit.
That's something else you can look up.

If you set it up right you can make it "ring" like a bell.