# Simple Circuit (RL circuit)

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1. Nov 26, 2014

### galaxy_twirl

1. The problem statement, all variables and given/known data

2. Relevant equations

iL = (iL(0) + Vs/R)e(-t/τ + Vs/R

3. The attempt at a solution

When the switch is closed before t=0, v(t) = 100V as the first 100Ohm resistor is shorted and L acts as a short wire in DC steady state.

When the switch opens, v(t) = 50V by voltage divider rule, since current flows through both 100Ohm resistors.

However, I have trouble putting the above in a v(t) expression, as v(t) is the voltage across R. I learnt how to write first-order transient equations for L and C but can they be applied here?

τ = L/R (τ is tau, the time constant)

iL = (iL(0) + Vs/R)e-t/τ + Vs/R

2. Nov 27, 2014

### GreenPrint

I would recommend not memorizing such equations. It really makes things complicated when presented with problems that don't fit them. There are several ways you could solve the problem. Do you know the equation for KVL for after the switch is opened?

3. Nov 27, 2014

### galaxy_twirl

Oh dear. >< I see.

Assuming i is the current in the circuit, is the KVL equation 100 = i(100) + i(100) ? By V=IR, i = 0.5A.

4. Nov 27, 2014

### GreenPrint

What about the voltage across the inductor?

5. Nov 28, 2014

### Staff: Mentor

x
Sure they can. You're already halfway to the answer because this is what you have worked out: the inductor current starts from a steady level of 1A and [once the switch is opened] it moves toward a final value of 0.5A on a 1st order exponential with a time-constant set by 200Ω and 1H.

In any general first order system, the exponential is always described by either e-t/τ or (1 - e-t/τ) and you choose which to fit the situation you are needing to describe.

I use only one general equation, preferring to perform additional algebra rather than memorize more than one.

contd on next post ......

Last edited: Nov 28, 2014
6. Nov 28, 2014

### Staff: Mentor

IL = Iinitial + (Ifinal - Iinitial) (1 - e-t/τ)

Now, just fill in the numbers. When you have it finished, and tidied up, then simply multiply by 100 to convert from resistor current to resistor voltage.

That last equation I wrote is a handy form to memorize, as it reads aloud precisely how you'd express it in words, making it easy to write 'on the fly' without error, once you get the hang of it.

Last edited: Nov 28, 2014
7. Nov 30, 2014

### rude man

You don't learn much by memorizing equations. If you don't analyze the simpler circuits you will be in big trouble with the more complex ones.

8. Nov 30, 2014

### galaxy_twirl

Sorry for the late reply. >< I was busy revising for my exams tomorrow. :(

I see. Thank you for the equation. :) I managed to solve the question by converting it to Thevenin and applying my formula, which can only be used if I have my circuits in Thevenin form.

9. Nov 30, 2014

### galaxy_twirl

Is the voltage 0? This is because at DC steady state, inductor acts as short wire and hence, $VL = 0$.

10. Nov 30, 2014

### galaxy_twirl

I think the formula my teacher gave was to prevent us by solving first order differential equations which I still don't know how to do at the moment, though I did learn a little at A levels.