1. ## Determine the current

A 30-volt electromotive force is applied to an-LR series circuit in which the inductance is 0.1 henry and the resistance is 50 ohms. Find the current i(t) if i(0)=0 Determine the current as t --> infinity.

Can anybody help me to solve this question?

2. As you know the relation between the current flowing through the inductor and the voltage drop across it is (if the inductance is time invariant):

$\displaystyle L\frac{d} {{dt}}I_L (t) = V_L$

combined with Kirchoff's voltage law we get:

$\displaystyle \left\{ \begin{gathered} RI_L (t) + L\frac{d} {{dt}}I_L (t) = V(t)_{emf} \hfill \\ I_L (t = 0) = 0 \hfill \\ \end{gathered} \right.$

this is a simple first order ODE, which can be easily solved...

3. how do i relate that with t-->infinity?

4. after you'll solve the ode you'll get some form of solution:

$\displaystyle I_L (t) = f(t)$

so you just take the limit of this function when t approaches infinity which is the current in the steady state of the circuit:

$\displaystyle \mathop {\lim }\limits_{t \to \infty } \;f(t)$

Alternatively the current in the steady state can be determined without first finding the general expression of the current in the circuit. The voltage drop across the inductor is directly proportional to the rate of the change of the current that flows through it, when the circuit reaches steady state ($\displaystyle t \to \infty$) there will no longer be change in the current and thus there will be no voltage drop across the inductor, it will basically become a short circuit (assuming that it is ideal and doesn't contain parasitic resistance), and thus all the voltage will drop on the resistor, so the current in the steady state will simply be:

$\displaystyle I_L (t \to \infty ) = \frac{{\left| {V_{emf} } \right|}} {R}$

you can easily verify that you obtain the same result using the first method described above.