Question

A series circuit consists of a 0.55-H inductor with internal resistance of 8.0 Ω connected in series with a 4.0-Ω resistor, an open switch, and an ideal 12-V battery. (a) When the switch is closed, what is the initial current through the 4.0-Ω resistor? (b) What is the current through the 4.0-Ω resistor a very long time after the switch is closed?

Answer 1

Answer:

Part a)

i = 0

Part b)

i = 1 A

Explanation:

Part a)

As per Lenz law we know that inductor in series circuit opposes the sudden change in current

And if the flux in the circuit will change then it will induce back EMF to induce opposite current in it

Now when we close the switch at t = 0

then initially it will induce the opposite EMF in such a way that net EMF of the circuit will be ZERO

so current = 0

Part b)

After long time the induced EMF in the circuit will be zero as the flux will become constant

so here we can say

$EMF = i(R_1 + R_2)$

$12 = i (8 + 4)$

$i = 1 A$