Simple R/L Circuit Current Problem

In summary, the conversation discusses a circuit with a switch and inductor, and poses questions about the current at different times. The summary provides equations and calculations to find the current at t=2ms and the time it takes for the inductor to reach maximum current. It also raises a valid concern about the question regarding the time it takes for the inductor to reach a specific current, as it never truly reaches that current.
  • #1
cmathis
6
0

Homework Statement



http://img202.imageshack.us/img202/8355/screenshot20130330at234.png

a) The switch closes at time t=0. What is the current flowing down through the inductor at time t=2ms?

b) What time will it take for the inductor to reach maximum current?

c) How long will it take for the inductor to reach current 2mA?



Homework Equations



i(t) = Ifinal + (Iinitial - Ifinal)[itex]e^{-t/\tau}[/itex]



The Attempt at a Solution



Current before the switch closes at t=0 is obviously 0A, so that means Iinitial = 0A.

After the switch closes and the circuit has come to steady state, you can represent the inductor by a short-circuit. The short circuit being in parallel with the two 3k[itex]\Omega[/itex] (equivalent to a 1.5k[itex]\Omega[/itex] resistor) effectively cancels out the two parallel 3k[itex]\Omega[/itex] resistors. So you're left with basically a single-loop circuit, so Ifinal would be equal to the current through the 3k[itex]\Omega[/itex] resistor in series with the voltage source, so Ifinal=0.001A.

[itex]\tau[/itex] = [itex]\frac{L}{R}[/itex] = [itex]\frac{1 H}{3000 \Omega}[/itex] = 3.33 x 10-4

That gives you equation:
i(t) = 0.001 - 0.001[itex]e^{-t/3.33 x 10^{-4}}[/itex]

Solving the equation at t=2ms give you i(2ms) = 0.998mA.

Am I correct so far?

For part b, wouldn't it take t=∞ amount of time since the current never really reaches 0.001A? The professor seems to expect a definite answer though...

For part c, is this a trick question? If I've done everything right, the inductor current never reaches 2mA.
 
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  • #2
R2 and R3 will influence the value of the time constant. Try turning the voltage source and resistor network into its Thevenin equivalent, with the inductor serving as the 'load'.

Your query about part c has merit.
 
  • #3
(oops... ignore this post. I'll fix it soon.)
 
  • #4
I'd get:

http://img145.imageshack.us/img145/5180/screenshot20130330at324.png

right?

So the above equation is right, but now [itex]\tau[/itex] = 0.001.

So i(2ms) = 0.865mA.
 
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  • #5
Yup. That looks good.
 
  • #6
Ok great. But what about part b? The limiting current in the inductor is 1mA, but it never really reaches that current, so t=infinity.
 
  • #7
cmathis said:
Ok great. But what about part b? The limiting current in the inductor is 1mA, but it never really reaches that current, so t=infinity.

True. (Although as an engineering rule of thumb, it's common practice to assume that all the excitement is essentially over after five time constants have elapsed)
 

1. What is a simple R/L circuit?

A simple R/L circuit is an electrical circuit that contains a resistor (R) and an inductor (L). It is a type of series circuit, meaning that the components are connected in a single loop, and the same current flows through each component.

2. What is the purpose of solving a simple R/L circuit current problem?

The purpose of solving a simple R/L circuit current problem is to determine the current that flows through the circuit. This information can be used to analyze the behavior of the circuit and make predictions about its performance.

3. How do you calculate the current in a simple R/L circuit?

The current in a simple R/L circuit can be calculated using Ohm's law, which states that current (I) is equal to voltage (V) divided by resistance (R). In a series circuit, the total resistance is the sum of the individual resistances. The inductor's resistance (known as reactance) can be calculated using the formula Xl = 2πfL, where f is the frequency in hertz and L is the inductance in henries. Once the total resistance is known, the current can be calculated using Ohm's law.

4. What factors affect the current in a simple R/L circuit?

The current in a simple R/L circuit is affected by the resistance and reactance of the components, as well as the voltage applied to the circuit. The frequency of the alternating current also plays a role, as higher frequencies will result in higher reactance and lower current.

5. How can the current in a simple R/L circuit be adjusted?

The current in a simple R/L circuit can be adjusted by changing the values of the components, such as using a different resistor or inductor with different resistance or inductance. The frequency of the applied current can also be adjusted, which will affect the reactance and therefore the current. Additionally, the voltage applied to the circuit can be changed, which will also affect the current.

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