# Is my method valid for solving this problem

Discussion in 'Homework Help' started by left shin, Dec 10, 2012.

1. ### left shin Thread Starter New Member

Apr 11, 2012
2
0
First, I would like to thank anyone who takes the time to help me with this problem.

In an RL circuit τ = -Leq/Req. In the following problem does this method still apply.

I ask because when I looked at the solution in the book the equations were set up different and the final answer was merely approximated.

I took Leq = to be --- 2mH || 15 mH = 1.76mH

and Req = to be--- 12 ||72Ω = 10Ω

Is this correct, or do the resistors in series with the inductors prevent me from taking this approach.

The book created 2 different differential equations for the two separate branches (with different values of tau) and then added them together. I created 1 using the equivalent values. I got a similar answer but I'm wondering if that was a coincidence.

The problem:

A short circuit is dropped between points a and b.

a) Find the initial value of the current through ab.
b) What is the final value of the current.
c) How many millisecond after the short will the current(ab) equal 210?

My solution:
a) Iab(0) = Isc - Iold ; 220- 20 = 200
b) Iab(inf) = Isc = 220
c) .122 ms; book = .123 ms

Thank you.

Last edited: Dec 10, 2012
2. ### tshuck Well-Known Member

Oct 18, 2012
3,531
675
First off, no, your inductors are not in parallel, neither are your resistors, so your can't calculate their equivalent inductance/resistance that way.... You have two parallel resistor and inductor branches, meaning that those two branches may be considered to be parallel and their equivalent impedance calculated.
Take a look here.

3. ### left shin Thread Starter New Member

Apr 11, 2012
2
0
So then it IS merely a coincidence that I arrived at a similar answer?

4. ### Ron H AAC Fanatic!

Apr 14, 2005
7,050
657
It is mostly due to the fact that one time constant is .167ms, and the other is .25ms, which are not too different. It is true that at DC, the equivalent parallel resistance is 10Ω, and at very high frequencies the apparent inductance of the parallel branches is L1 || L2, but since the time constants are different, they will discharge at different rates.
If the time constants had been radically different, your result would not be as close to the correct value.