Inductive Circuit Current Growth

Discussion in 'Homework Help' started by Tommy, Feb 26, 2008.

1. Tommy Thread Starter New Member

Feb 25, 2008
4
0
Hello All

I have been given this question by my HNC lecturer and need some help understanding it better. It doesnt help that Im studying a Mechanical HNC rather than Electrical so I thought Id post it up here!

The question says that the growth of an inductive circuit is given by the equation: i(t) = I (1 -e (to power of)-Rt/L)
Where R is the resistance of the circuit in Ohms
L is the inductance in Henrys
i(t) is the current in the circuit at any time t seconds in Amps
I is the final or steady current that flows in the circuit in Amps
t is time in seconds

The ratio of L/R is the time constant of the circuit and is measured in seconds.

The question says that a particular circuit has a time constant of 10 milliseconds and a steady state current of I= 10Amps. The initial Current is zero.

Im then asked to work out:
A) The value of the current flowing 4milliseconds after the circuit is switched on.
B) The time after switch on when the current has grown to 5A
C) The value of the inductance of the circuit

Ive had a go at it but still to no avail. Ive also reffered to a smilar example given in Edwars Hughes book Electrical Technology on Pg118,119 (fourth Edition) but my lack of electrical knowledge is putting a damper on my progress.

Any Help with this would be much appreciated.

All the best

Tommy

2. Dave Retired Moderator

Nov 17, 2003
6,960
171
Hi Tommy,

Glad you sorted the posting issues out.

A) Just bang the numbers into the equation; R/L = 1/10x10^-3, I = 10, t = 4x10^-3

B) Rearrange the equation to get t as the subject of the formula, and then bang the numbers in; R/L = 1/10x10^-3, I = 10, i(t) = 5

C) I'm not quite seeing this one yet. Did they give you a voltage?

Have a go at the answers and post them up, we can check through your working for you.

Dave

3. Tommy Thread Starter New Member

Feb 25, 2008
4
0
Hi Dave
Yeh, I followed you advice and managed to sort it out With regards to the question I wasnt given a voltage for it so I may have to chase that up. Im going to try rearranging it and seeing if I can derrive anything using logs.

Ill let you know how I get on

Tommy

4. Dave Retired Moderator

Nov 17, 2003
6,960
171
I only asked about the voltage because the obvious method is to say:

i(t) = I (1 -e (to power of)-Rt/L)

Is the same as:

i(t) = V/R (1 -e (to power of)-Rt/L)

If you know the voltage you can rearrange and calculate R based of your previous values. Then you can calculate L from L/R = 10x10^-3.

There will probably be another way, but I'll have a think about it later when I get a moment.

Good luck, and feel free to post up your attempts if you want to check them.

Dave

5. Tommy Thread Starter New Member

Feb 25, 2008
4
0
Hi All

Sorry its been a while sicne I last posted up here.
Was wondering if my workings are correct?

For Question A) Calculate the value of current flowing 4ms after the circuit is switched on:

The first thing I did was work the Ratio L/R

i(t)=I(1-e^-Rt/l)

I know that L/R = 10 ms therfore 10x10-3 =0.01 seconds
R/L= 1/(L/R) = 1/10x10^-3

t=4ms = 4x10^-3 and so I substituted this back into the equation:

i(t)= I(1-e^ -1(4x10^-3)/10x10^-3

so i(t) =I (1-e^-0.4)

Therfore: i(t) = 10 (1-e^-0.4) = 3.29679954

Question B)

I know that T=L/R

so from this I did:
i = 10(1-e^-t/T)
or
5=10(1-e-t/T)

Then rearranged this to get

e^-t/T = 1-5/10

e^-t/T = 0.5

Then used natural logs to work out -t/T

e^-t/T = 0.5

ln e^-t/T = ln 0.5

-t/T = ln 0.5
-t/T = -0.69314718

T/-t = -0.69314718/1

Is this all ok so far???

Ive still yet to tackle question C? any ideas on how to go about it?

6. Tommy Thread Starter New Member

Feb 25, 2008
4
0
Has anyone spotted any errors in my working at all? Math isnt a strong point Im afraid!

7. Dave Retired Moderator

Nov 17, 2003
6,960
171
The answer to A is correct.

Everything is good until this last bit.

Remember that (-t/T) = (t/-T), so:

t = ln(0.5)*(-T)

t = ln(0.5)*(-10e-3) = 6.93147..e-3 = 6.93ms

Without the voltage I cannot see how you go about C based on the information given.

Dave