For t>0 in this circuit. It acts as a single loop circuit. so circuit flowing through 25 ohm will be iL. Solution says
v1=

and this is my approach.
MOD: Tidied your image, for you.



If I take V1= iL * 25 , I can't get the above answer. kindly help.

 

Why do you claim it acts like a single loop circuit for t > 0?

 

Why do you claim it acts like a single loop circuit for t > 0?

After a long time , (t= infinity) the circuit will be in steady state and inductor will act as a short circuit. So, no current will flow through 100 ohm resistor. Hence , a single loop circuit

 

"After a long time" and "t > 0" are two very, very different things.

You are asked to find the voltage for t > 0, which includes when t is just barely past t = 0, when t is a fraction of a percent of one time constant, when t is one time constant, and when t is three time constants. Those are all very different than "after a long time".

If you choose to simplify the circuit down to a point that it is only valid "after a long time", do not be surprised when any answer you get as a result is ONLY valid "after a long time".

The voltage across the 25 Ω is NOT equal to that resistance times the current in the inductor because the inductor and that resistor are not one and the same -- they are NOT in series.

Also, in your work, the exponential in your equation goes from R/L to "1000" by magic.

Finally -- and I've told you this more than once -- you need to track your units properly.

 

"After a long time" and "t > 0" are two very, very different things.

You are asked to find the voltage for t > 0, which includes when t is just barely past t = 0, when t is a fraction of a percent of one time constant, when t is one time constant, and when t is three time constants. Those are all very different than "after a long time".

If you choose to simplify the circuit down to a point that it is only valid "after a long time", do not be surprised when any answer you get as a result is ONLY valid "after a long time".

The voltage across the 25 Ω is NOT equal to that resistance times the current in the inductor because the inductor and that resistor are not one and the same -- they are NOT in series.

Also, in your work, the exponential in your equation goes from R/L to "1000" by magic.

Finally -- and I've told you this more than once -- you need to track your units properly.

R = 25 || 100 = 20 ohms and exponential is R/L t . So 20/0.02t = 1000t

Next time I will keep track of my units properly.
sorry for inconvenience and thanks for your time!

 

R = 25 || 100 = 20 ohms and exponential is R/L t . So 20/0.02t = 1000t

Next time I will keep track of my units properly.
sorry for inconvenience and thanks for your time!

Hello,

Is that all you had to calculate was the exponent?

 

R = 25 || 100 = 20 ohms and exponential is R/L t . So 20/0.02t = 1000t

Next time I will keep track of my units properly.
sorry for inconvenience and thanks for your time!

I have serious doubts that you will track your units properly next time since you couldn't even be bothered to track them properly in this post!

R = 25 Ω || 100 Ω = 20 Ω

The exponent is -(R/L)t = -(20 Ω / 0.02 H) t = -1000 s^-1 · t

A better, generally more useful way to express the exponent would be - t / 1 ms