# Series R L circuit

1. Jun 25, 2013

### phydis

1. In a series R L circuit, we get this equation for the current i : i(t) = E/R[1-e^(-Rt/L)] where R: Resistance, L: inductance, E: emf

2. My problem is how we get this equation?, why we consider dx/dt + px = c to get the above equation? how the above equation connects with steady response and the transient response of the circuit?

2. Jun 25, 2013

### Simon Bridge

Usually from a mesh analysis - written in differential form, then you solve the differential equation.
A good text book should show you this.

Because that is the form the DE takes.
This way you get to use someone elses work as a shortcut. You could always just solve it yourself of course.

That should be clear from the definitions of "steady state" and "transient" response.
Note: to have a response you have to have something to respond to.

The circuit behaves a bit like a damped and driven harmonic oscillator ... so there will be a short-lived component and a long-lived one.

3. Jul 6, 2013

### rude man

This current is the response to a step input of voltage. No other input will yield this response. This IS the transient response. The steady-state response is i = E/R. In this case V = step input voltage V0 U(t).

The basic differential equation is V = Ri + L di/dt, based on the simple fact that for an inductorr V = L di/dt and for a resistor V = iR.

If a sinusoidal voltage V = V0 sin(wt) is applied at t=0 there is a transient as well as a steady-state response. The above diffrerential equation allows solving for both.