First Order RL Circuit: Finding a Current I2(0-)

In summary, the conversation discusses using mesh analysis and the node voltage method to find the currents in a linear system. The value for i2(0-) should be 15 mA according to the book, but the individual is getting -2mA. They are confused and wonder if the mesh method is the most efficient way to find i2. The conversation also mentions using Ohm's law to find initial currents and the issue of reversing the direction of i2 on the diagram.
  • #1
johnsmith7565
13
4
Homework Statement
The switch in the circuit in Fig. P7.1 has been closed for a long time before opening at t=0.

a. Finding i1(0-) and i2(0-). I am stuck on this.
Relevant Equations
Mesh equations

V1r1 + V2r2 + ... + VnRn = 0
IMG-3091.jpg
IMG-3092.jpg


I shorted the inductor and performed mesh analysis. The solutions to the linear system were done using a calculator. The book says that the value for i2(0-) should be 15 mA but I'm getting -2mA. What am I doing wrong? I'm completely confused. Maybe mesh isn't the most efficient way to find I2 but it should work, right?
 
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  • #2
Currents are flowing at t=0, before switch is opened.
Short inductor, to find the voltage on the central node ?
Remove the short from the inductor after DC solution.
Then Ohms law to find initial currents i1 and i2 ?
What is inductor current ?
Switch opens, but current through inductor continues.
What are the currents i1 and i2 now ?
 
  • #3
Baluncore said:
Currents are flowing at t=0, before switch is opened.
Short inductor, to find the voltage on the central node ?
Remove the short from the inductor after DC solution.
Then Ohms law to find initial currents i1 and i2 ?
What is inductor current ?
Switch opens, but current through inductor continues.
What are the currents i1 and i2 now ?

I’m going to do the first part of your question first.

Here is the node voltage equation:

(v-40)/500 + v/2000 + v/6000 = 0

Using a calculator I get v=30 V.

By ohms law: i2(0-) = -30/2000 = -0.0015 A. That’s right.
I1(0-) = 30/6000 = 0.005 A which is right. Why do I get the correct answers using the node voltage method but not by using the mesh method? That’s what confuses me.
 
  • #4
I think you may have reversed the direction of i2 on the diagram.
Maybe you have as much trouble reading your writing as I do.
 

FAQ: First Order RL Circuit: Finding a Current I2(0-)

What is a first order RL circuit?

A first order RL circuit is a type of electrical circuit that contains a resistor (R) and an inductor (L). It is characterized by the presence of a single energy storage element, the inductor, and is commonly used in electronic systems to control the flow of current.

How do you calculate the current (I2) in a first order RL circuit?

The current (I2) in a first order RL circuit can be calculated using the formula I2(t) = I0 + (V0/R)e^(-t/τ), where I0 is the initial current, V0 is the initial voltage, R is the resistance, t is time, and τ is the time constant of the circuit.

What is the significance of the time constant (τ) in a first order RL circuit?

The time constant (τ) in a first order RL circuit is a measure of how quickly the current in the circuit reaches a steady-state value. It is equal to the inductance divided by the resistance (τ = L/R) and determines the rate at which the current changes over time.

How does the initial current (I0) affect the current (I2) in a first order RL circuit?

The initial current (I0) in a first order RL circuit is the current that is present in the circuit at time t = 0. It affects the current (I2) by adding a constant value to the exponential decay function, resulting in a shift in the current's amplitude but not its shape.

What is the relationship between the current (I2) and the voltage (V0) in a first order RL circuit?

The current (I2) and the voltage (V0) in a first order RL circuit are directly proportional to each other. This means that as the voltage increases, the current also increases, and vice versa. However, the rate at which the current changes is dependent on the resistance and inductance of the circuit.

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