Solving Steady State-Circuits: Step by Step Example

  • Thread starter Thread starter rafterman
  • Start date Start date
  • Tags Tags
    Steady
AI Thread Summary
The discussion focuses on understanding the equation for current in an RL series circuit, specifically the part involving the exponential function. The correct equation is highlighted as I(t) = V/R * [1 - e^(-t*R/L)], emphasizing the importance of the negative sign. Participants clarify that e represents the base of the natural logarithm and is used for exponentiation in the equation. One user expresses confusion about substituting values into the equation, particularly the bracketed part. Ultimately, the conversation underscores the need for a solid grasp of exponential functions to accurately solve the problem.
rafterman
Messages
13
Reaction score
0
This is a simple question. I know the equation I am supposed to use but I can not get my head around one part. The part in brackets! I keep getting wild numbers as my answer

The equation is :-

I = V/R * (1 - e^t*R/L)

can I ask someone to give me a step by step example of how you place the relevant numbers into it.

V=6
R=10
L= 1 x 10 -4 H
t=1.0 x 10-5 s
e=2.72 (3sf)
 
Last edited:
Physics news on Phys.org
The equation describes how current changes with time in a RL series circuit fed by a battery V.
However, there is a mistake, the equation should read:

I(t)=V/R * [1-e^(-t*R/L)] (note the sign change)

My guess is that you don't understand the meaning of e^. In computer jargon ^ stands for exponentiation. Now you shouldn't have problems with this equation.
 
Hi

So are you saying that [1-e^(-t*R/L)] just means (-e) -2.72 to the power of the total -t*R/L
 
I(t) = \frac V R (1-e^{-\frac{tR}{L}})
 
Hi Borak

I know that's the equation I need, but the part in brackets is confusing me. I am not sure I am utilising it correctly.

My book has the equation but no examples of it workings, so I can not see how the bracket part can be substituted for the numbers I have worked out.
 
e^{-\frac{tR}{L}}

is an exponential function, e to

-\frac{tR}{L}

power. If you don't know what that means and how to calculate it, you should quickly get back to your math books.
 
Yeah, I guess...my physics teacher just says read the books instead of him teaching it.

Thanks anyway.
 
Don't get me wrong. What I am aiming at is that it is a simple calculation - if you have a problem with it it means your math skills are way too low for things you are expected to do. We can help you checking if your result is OK, we can point you to the fact exponential function exists, but it is up to you to learn about it.
 
Borek

Honestly thanks, I am struggling slightly because I am playing catch up. Be away from school for a long while due to injury. I have re-read my books and eventually figured it out, some of the numbers I was getting originally as the answer were just not looking right, I really started to confuse myself.

Thanks again
 
Back
Top