# RL Circuit and di/dt

1. Dec 4, 2013

### dwn

Question 1

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

For a circuit show on Fig B1, what is the rate of change of the current in the inductor when the total current in the circuit is 0.50 A? (L = 10 mH, R(1) = 10 ohm, R(2) = 20 ohm, V = 12 V). In the figure the resistors of the circuit are in series.
Ans: 200

2. Relevant equations

$(di/dt = V/L(e(-Rt/L)))$
$(i = V/R(1-e(-Rt/L)))$

3. The attempt at a solution

I tried to find the time at which the current was at .50 A and the using that time to plug in the differential equation to find the rate of change in the current.

Last edited: Dec 4, 2013
2. Dec 4, 2013

### ShayanJ

Well,then...how can we help???

3. Dec 4, 2013

### dwn

When I plug in the time of .00046 seconds into the differential equation, my answer is incorrect. Am I the right path and my calculations are just incorrect? There really are not a lot of options to find di/dt when dealing with LR circuits, so I can't be that far off.

$30(.5)/12$=$(1-e(-30t/.010))$

t = .00046 seconds

$di/dt$-=(12/.010)e(-30*.00046)/0.01)

$di/dt=302$

sidenote: why isn't [itex] converting to math form? is the format not permitted?

4. Dec 4, 2013

### vela

Staff Emeritus
Your solution can't possibly be correct because solving for t would require taking the log of a negative number.

You've implied this is a circuit where all the elements are connected in series. Is that right? Because if it is, given the numbers you've provided, the maximum current is V/R = 0.4 A.

5. Dec 4, 2013

### dwn

I didn't understand what to do, so I just kinda of made it "work" (i.e. produce an answer). There are only a couple equations to solve LR circuits which 1. makes me feel stupid for not being able to figure it out and 2. unable to find error in solving the question using the equations.

I attached an imagine from the homework set, so it might be more helpful. The problem is P10.

Can you please offer some advice on how I should solve this. Thanks.

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6. Dec 4, 2013

### vela

Staff Emeritus
The equations you listed apply to an RL circuit where everything is in series. That's not what you have here, so you can't use them.

What you need to do is find the voltage across the inductor because that will allow you to find di/dt. A lot of times you can analyze a circuit simply by applying Kirchoff's voltage and current laws and Ohm's law over and over again.

You're given that the total current, by which they mean the current through the battery, is 0.50 A, so according to Kirchoff's current law, what's the current through R2? Once you have the current, use Ohm's law to determine the voltage drop across the resistor. Then use Kirchoff's voltage law to determine the voltage across the inductor.

7. Dec 4, 2013

### cepheid

Staff Emeritus
It's because you included superscripts and subscripts using the [noparse] and [/noparse] tags. These tags are for creating superscripts and subscripts using plain text (i.e. html) only. LaTeX is different, and uses ^ and _ for superscripts and subscripts respectively. Do not mix html equations with LaTeX markup. Choose one or the other, depending on how fancy your equation needs to be.