calculating current in RL circuit using numerical methods

[SHANKARS]

I want to calculate the current flowing in RL series circuit using numerical methods.
How it can be done can you help me?

 

[BradtheRad]

Let us assume DC voltage is unchanging...

Amperes will eventually rise to a target value (derived as V/ R).

Amperes change at a rate according to the time constant, L/ R. The greater the Henry value, the slower the amperage rises.

By definition, after one time constant goes by, amperage reaches 63 percent of the target amount.

The conventional view is that the inductor is charged (or discharged) after 5 time constants. A similar view is applied to capacitors.

 

[SHANKARS]

Yes but what I want is by considering a DC as the supply voltage we get a transdental equation for the current in the circuit involving differentiation of current to time i.e. d(i)/dt = (V-IR)/L.
So I want to solve this equation by using suitable numerical methods to find the current flowing in the circuit at all the intervals of time assuming any small step size of the time.
Do you know which numerical method is suitable for this problem?

 

[BradtheRad]

Since you quote a formula which includes voltage, let's focus on counter-EMF generated by the inductor. It starts out the same as the applied voltage. It drops to zero eventually. (Its curve is symmetrical to the Ampere curve, which rises.)

The curve is exponential. By definition it drops by 63.2 percent at the time-point calculated as L/R, in seconds.

It is patterned after the equation:

y = 0.368 ^ x

We regard x=1 as being one time constant.
We regard y=1 as being 100 percent of the initial voltage.

The equation allows you to calculate the percentage change, at any desired time-point.

 

[Venkadesh_M]

(V - IR) / L = dI / dt

(L / (v - IR) ) di = dt

integrating both sides

∫ (L / (v - IR) ) di = ∫ dt

- ( L / R ) ln(V - IR) = t + C // C = 0

ln(V - IR) = - (R/L) t

I(t) = (V / R) (1 - e ^(- (R/L) t)) for t >= 0