First-order RL Circuit analysis

In summary, the conversation discusses how to handle a circuit with both a voltage source and a current source present. It is suggested to use techniques such as Thevenin and Norton equivalence, as well as Kirchoff's and Ohm's laws, to simplify the circuit and solve for the voltages and currents. The conversation also clarifies that the current sources cannot simply be combined, and that further simplification may be necessary.
  • #1
Mohdoo
20
0

Homework Statement



CUog4.jpg


Homework Equations



I am not really sure which equations would be relevant since my question seems to be more so concepts instead of numbers, so forgive me please :(

I am familiar with how to do just about all of this problem, I am simply baffled as to how I handle the idea of a voltage source and a current source both being there at the same time. I have solved such problems only with one voltage source, or one current source, but never one of each or two of one of them.

The Attempt at a Solution



I see that there is a .5mA source, and a 30mA source. One is going to shoot current one direction into the capacitor, and one is going to shoot. I say a 30mA source because I assume that I can simply turn the 150V source into a 30mA source in parallel with the 5K resistor. Is it really as simple as subtracting .5mA from 30mA and then simply having i be negative? And then once the switch closes, the components to the left of the switch get cut off?

All my other problems on this homework assignment seemed so straight forward, this one is such a curve ball *sigh*

Many thanks! :)
 
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  • #2
Mohdoo said:
I am familiar with how to do just about all of this problem, I am simply baffled as to how I handle the idea of a voltage source and a current source both being there at the same time. I have solved such problems only with one voltage source, or one current source, but never one of each or two of one of them.
You have a couple of things you can do. You can convert voltage sources into current sources and vice versa using Thevenin and Norton equivalence. If you have two voltage sources in series, you can combine them into one, and if you have two current sources in parallel, you can combine them into one. You could continue to combine elements to simplify the circuit, or you could use Kirchoff's and Ohm's laws to solve for the voltages and currents instead.

The techniques really don't change just because you have multiple sources. What specifically about multiple sources baffles you?

I see that there is a .5mA source, and a 30mA source. One is going to shoot current one direction into the capacitor, and one is going to shoot. I say a 30mA source because I assume that I can simply turn the 150V source into a 30mA source in parallel with the 5K resistor. Is it really as simple as subtracting .5mA from 30mA and then simply having i be negative?
No, it doesn't work that way. You're going to have to work it out.

By the way, it's an inductor, not a capacitor.
 
  • #3
vela said:
You have a couple of things you can do. You can convert voltage sources into current sources and vice versa using Thevenin and Norton equivalence. If you have two voltage sources in series, you can combine them into one, and if you have two current sources in parallel, you can combine them into one. You could continue to combine elements to simplify the circuit, or you could use Kirchoff's and Ohm's laws to solve for the voltages and currents instead.

Well, I figured that I could simply combine the two current sources. But in this case do they combine or do they subtract from one another? They both point upwards, but following each of them, would they not be going different directions when crossing through the inductor? It seems like the one on the left would be pushing current from left to right, and the one on the right pushing current from right to left. Which one is it? :(
 
  • #4
You can't combine them unless they're in parallel. I don't know if you've simplified he circuit further, but they're definitely not in parallel in the handwritten drawing.
 

1. What is a first-order RL circuit?

A first-order RL circuit is a circuit that contains a resistor (R) and an inductor (L) connected in series. It is called a first-order circuit because it contains only one energy storage element, the inductor.

2. How do I analyze a first-order RL circuit?

To analyze a first-order RL circuit, you can use Kirchhoff's voltage law (KVL) and Kirchhoff's current law (KCL) to write equations for the voltage across the resistor and the current through the inductor. Then, you can solve these equations using differential equations to find the behavior of the circuit over time.

3. What is the time constant of a first-order RL circuit?

The time constant of a first-order RL circuit is the time it takes for the current in the circuit to reach 63.2% of its maximum value. It is calculated by dividing the inductance (L) by the resistance (R) in the circuit.

4. How does the current in a first-order RL circuit change over time?

At the initial moment, the current in a first-order RL circuit is equal to zero. As time passes, the current will increase towards its maximum value, which is determined by the voltage source and the resistance in the circuit. The rate of change of the current is determined by the time constant of the circuit.

5. How does the addition of a capacitor affect a first-order RL circuit?

Adding a capacitor to a first-order RL circuit will create a second energy storage element and turn the circuit into a second-order circuit. This will change the behavior of the circuit, and the analysis will require solving a second-order differential equation. The addition of a capacitor can also affect the time constant and the overall behavior of the circuit.

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