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It is fine if you are not familiar with the physical concepts.

i

V

t = 0

O

L

R/nIn the circuit above, a battery applies a voltage V (measured in Volt (V)) whenever the switch is closed. But the circuit

also contains a resistor with resistance R (measured in Ohm (2)) and an inductor with inductance L (measured in Henry

(H)).

We are interested how the current I(t) (measured in Ampère (A)) changes over time (in seconds). Initially once the

circuit is closed, the current is zero.

The voltage across the resistor changes over time according to:

VR(t) = R.I(t)

and the voltage across the inductor satisfies

V₁ (t) = L dI (t)

dt

By Kirchhoff's law the directed sum of the voltages in the circuit must be zero. Hence, we obtain the following

differential equation:

RI(t) + L di(t)

dt

= V.

(a) What is the long-term behavior of I(t)?

¹https://www.electronics-tutorials.ws/dccircuits/kirchhoffs-voltage-law.html

(b) Solve the differential equation (1) using seperation of variables. Do not forget the initial condition.

(c) Using the solution from part (b), find the formulas for Vr(t) and VL(t).

(d) What is the long-term behavior of the voltages of the resistor and the inductor?

(1)/nNow let us work with specific values. Set R = 100, V = 200 V and L=25H.

(e) What is the current after 0.25 seconds?

Fig: 1

Fig: 2

Fig: 3