Series R L Circuit: Equation for Current & Its Connections

[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?

 

[Simon Bridge]

My problem is how we get this equation?

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

why we consider dx/dt + px = c to get the above equation?

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.

how the above equation connects with steady response and the transient response of the circuit?

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.

 

[rude man]

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

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.