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LR circuit

by purduegirl
Tags: circuit
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purduegirl
#1
Jul10-08, 12:41 AM
P: 74
1. The problem statement, all variables and given/known data

An LR circuit consists of an inductor with L = 2.3 10-4 H and a resistor with R = 1.10 Ω connected in series to an oscillating source of emf. This source generates a voltage
ε = εmaxsin( ωt)
with εmax = 0.80 V and ω = 6.00 104 rad/s. What is the maximum current in the circuit?

2. Relevant equations

ε = εmaxsin(ωt)

3. The attempt at a solution

ε = (0.80V)sin( 6.0E4 rad/s*time constant)
time constant = L/R = (2.3E-4 H/1.10 Ω) = 2.091e-4 H/Ω
ε = (0.80V)sin( 6.0E4 rad/s*2.091e-4 H/Ω)
ε = (0.80V)sin( 12.546)
ε = (0.80V)(0.2172)
ε = 0.1738 A

This is the incorrect answer. Any ideas where I am going wrong?
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marcusl
#2
Jul10-08, 09:24 AM
Sci Advisor
PF Gold
P: 2,082
Hold on, the question asks for the maximum current. What is the relation between current and voltage for a circuit that includes both resistance and reactance?
Defennder
#3
Jul10-08, 10:30 AM
HW Helper
P: 2,616
Is there a way to do this without solving a DE?

alphysicist
#4
Jul11-08, 01:06 AM
HW Helper
P: 2,249
LR circuit

Quote Quote by purduegirl View Post
1. The problem statement, all variables and given/known data

An LR circuit consists of an inductor with L = 2.3 10-4 H and a resistor with R = 1.10 Ω connected in series to an oscillating source of emf. This source generates a voltage
ε = εmaxsin( ωt)
with εmax = 0.80 V and ω = 6.00 104 rad/s. What is the maximum current in the circuit?

2. Relevant equations

ε = εmaxsin(ωt)

3. The attempt at a solution

ε = (0.80V)sin( 6.0E4 rad/s*time constant)
time constant = L/R = (2.3E-4 H/1.10 Ω) = 2.091e-4 H/Ω
ε = (0.80V)sin( 6.0E4 rad/s*2.091e-4 H/Ω)
ε = (0.80V)sin( 12.546)
ε = (0.80V)(0.2172)
ε = 0.1738 A

This is the incorrect answer. Any ideas where I am going wrong?
A thing going wrong with your work here is that you have misinterpreted the equation for the voltage they gave you. The t in the voltage formula is the time, not the time constant. So the procedure you followed would find the voltage at the time t=2.091e-4 s.

Quote Quote by Defennder View Post
Is there a way to do this without solving a DE?
You can solve this problem without solving a DE. As marcusl indicated in the above post, there is a quantity that relates the maximum voltage and maximum current in this circuit (similar to the way that resistance relates voltage and current in a DC circuit); and this quantity has a standard formula for a series RLC circuit (or for a series RL AC circuit).
alphysicist
#5
Jul12-08, 09:39 PM
HW Helper
P: 2,249
purduegirl,

You mentioned in another thread a question about this one. In case you haven't figured this out, I'll answer it here.

The formula you have listed in your relevant equations

ε = εmaxsin(ωt)

is not needed at all in this problem; this problem will not involve dealing with the time at all.


Let's go back to a simpler problem that you're probably familiar with. If you have a DC circuit, with just two resistors in series and an (ideal) battery, how would you find the current? You would probably add the two resistances together to get the total equivalent resistance. Once you got the total equivalent resistance, you would use Ohm's law V = I R to find the current in the circuit.


Now here we have an AC circuit, with a voltage source and two components (resistor and inductor). The idea is that we will follow the same type of procedure in this AC circuit as we did with the simple DC circuit.

So you need to "combine" the inductor and resistor (in much the same way that you combined resistors in DC circuits). How do you do that? What quantity do you get?

Once you calculate it, you'll use Ohm's law for AC circuits. For DC circuits, Ohm's law is V=IR; for the AC circuit version of Ohm's law the resistance R is replaced by the quantity you got from combining the inductor and resistor; have you found that formula yet?


(Sorry if you've already solved this problem.)
purduegirl
#6
Jul12-08, 09:45 PM
P: 74
So for this type of circuit, the inductor and the resistor are in series. Would these be added L + R or 1/L + 1/R as in capitance?
purduegirl
#7
Jul12-08, 09:55 PM
P: 74
One other question I have is, if added the inductor and the resistor are added to together, I get units of V/A + V*s/A. How would I make the units work so that they could be added together?
alphysicist
#8
Jul12-08, 09:56 PM
HW Helper
P: 2,249
Quote Quote by purduegirl View Post
So for this type of circuit, the inductor and the resistor are in series. Would these be added L + R or 1/L + 1/R as in capitance?
No, it's not quite that straightforward. The L and R have different units, so they can't just be added together.

Instead, you'll first need to find a quantity called inductive reactance [itex]X_L[/itex]. (If there were a capacitor in your equation, you would also need to find the capacitive reactance [itex]X_C[/itex].) Do you have those formulas?

Once you have the R, the [itex]X_L[/itex] (and if you had it, the [itex]X_C[/itex]), you need to calculate the impedance Z of the circuit. The impedance Z is the quantity that I was mentioning in my last post; in a sense, it plays the role that the total equivalent resistance played in the DC resistor circuits that you probably have already done a lot. There is a standard formula for Z for a series RLC circuit; do you find that one?

The total impedance of the circuit relates the maximum voltage and maximum current in the same way the the total resistance related V and I in DC circuits.
purduegirl
#9
Jul12-08, 10:27 PM
P: 74
Thank you so much. I have been working on this problem for nearly a week!


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