00:01
Given is a inductor which has inductance l equal to 25 millinery, given is a resistor which has resistance equal to 8 .0 volts, and a battery which has emf 6 volt, and these are connected in series.
00:26
And so, in this rl circuit, the switch is closed at t equal to 0 seconds.
00:34
We have to find the voltage drop across the resistor.
00:41
First part is asking us to find the voltage drop of resistor at t equals to 0.
00:55
For this current in the rl circuit, at time t is given by the relation i equals to emf by r 1 minus e to the power minus t by tau.
01:08
Where tau is the time constant, t is the time, e is the emf and r is the resistance, where tau is l by r.
01:23
After this, the potential difference across the resistor, del v across the resistor is equal to r -i.
01:34
Multiplying r in this equation will give us del vr equals to r -i equals to r -i equals to r multiplied e by r 1 minus e to the power minus t by two.
01:52
And now the relation becomes del vr equals to e 1 minus e to the power minus t by two...