Kirchhoff's Law & Circuit Analysis: Q&A

In summary, Kirchhoff's current law states that the voltage across an inductor is independent of the current through the inductor. The current through the inductor can be determined from the voltage across the inductor and the inductor's inductance.
  • #1
TheRedDevil18
408
1

Homework Statement



Using Kirchhoff's current law, determine:

1) The nodal equations
2) The expression and solution for the inductor voltage
3) The expression and solution for the inductor current
4) Determine the initial and final inductor current

Circuit diagram:

WP_20150613_004.jpg

Homework Equations

The Attempt at a Solution



I'm really stuck with this question. The best I could try and do was question one which I don't think is correct but here it is

1) (9-V1)/12 = (V1-V2)/Zl + V1/6

V1 is the voltage across the 6 ohm resistor and V2 is the voltage across the inductor
 
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  • #2
TheRedDevil18 said:
V1 is the voltage across the 6 ohm resistor and V2 is the voltage across the inductor
Do you expect those voltages to be different?
Probably just a typo in the V1 definition.

TheRedDevil18 said:
(V1-V2)/Zl
Independent of the definition of V1 and V2, this term looks wrong.
 
  • #3
TheRedDevil18 said:
I'm really stuck with this question. The best I could try and do was question one which I don't think is correct but here it is

1) (9-V1)/12 = (V1-V2)/Zl + V1/6
Since you've got a reactive component in the circuit (an inductor), you should realize that you are going to end up with a differential equation. Impedance won't help unless you move to the Laplace Domain, which I'm guessing you haven't covered yet.

What do you know about the relationship between the voltage and current for an inductor? Can you write an expressions for the current as a function of the voltage and vice versa?
 
  • #4
So if the voltages are the same since they are in parallel, then from the equation

(9-V1)/12 = (V1-V2)/Zl +V1/6

since V1-V2 = 0

then, solving for V1 I get 3V which is the voltage across the 6 ohm resistor and the inductor ?
 
  • #5
If V1 and V2 are as you described, then the equation you set up for current does not make sense.
 
  • #6
You should answer the question gneill asked in post #3. Until you get that right, there is no point in proceeding with your effort.
 
  • #7
gneill said:
Since you've got a reactive component in the circuit (an inductor), you should realize that you are going to end up with a differential equation. Impedance won't help unless you move to the Laplace Domain, which I'm guessing you haven't covered yet.

What do you know about the relationship between the voltage and current for an inductor? Can you write an expressions for the current as a function of the voltage and vice versa?

Ok, I know that V = L*di/dt , so

I(t) = 1/L integral V2 dt

so, 9-V2/12 = V2/6 + 1/L integral V2 dt ?, so I just have to make an expression for V2 in terms of t ?
 
  • #8
Integrating over the voltage is not so helpful. Better look at voltages and derivatives of the current.
 
  • #9
TheRedDevil18 said:
Ok, I know that V = L*di/dt , so

I(t) = 1/L integral V2 dt

so, 9-V2/12 = V2/6 + 1/L integral V2 dt ?, so I just have to make an expression for V2 in terms of t ?
That's the idea; you write the node equation incorporating the ##I(t) = \frac{1}{L} \int V_2 dt## for the inductor branch's current. That gives you an integro-differential equation. Collect the ##V_2## terms together.

Hint: you will want to differentiate the whole thing to clear the integral and leave the equation in purely differential form.
 
  • #10
Okay thanks, I found the voltage expression to be 3*e^-800t

Also the general expression for the voltage of an inductor in an RL circuit is Vo*e^-(Rp/L)t, so to do this without integration you could have found Vo by removing the inductor from the circuit because it acts as an open at t=0 and measure the voltage across the 6 ohm resistor which you can get using the voltage divider rule. Rp would just be the 12 and 6 ohm resistors and the inductance is 5mh

Anyway I think they wanted you to derive the expression which is why they asked for the nodal equations
 

1. What is Kirchhoff's Law?

Kirchhoff's Law, also known as Kirchhoff's Circuit Laws, are two fundamental principles in electrical engineering that govern the behavior of electric circuits. They were developed by German physicist Gustav Kirchhoff in the mid-19th century and are widely used in circuit analysis.

2. What are the two laws of Kirchhoff's Law?

The two laws of Kirchhoff's Law are the Kirchhoff's Current Law (KCL) and the Kirchhoff's Voltage Law (KVL). KCL states that the sum of all currents entering a node in a circuit must be equal to the sum of all currents leaving the node. KVL states that the sum of all voltage drops around a closed loop in a circuit must be equal to the sum of all voltage rises.

3. Why are Kirchhoff's Laws important in circuit analysis?

Kirchhoff's Laws are important because they provide a systematic method for analyzing complex circuits. They allow us to determine the unknown values of currents and voltages in a circuit and understand how the circuit behaves under different conditions.

4. Can Kirchhoff's Laws be applied to both DC and AC circuits?

Yes, Kirchhoff's Laws can be applied to both DC (direct current) and AC (alternating current) circuits. However, for AC circuits, the laws must be modified to take into account the time-varying properties of AC currents and voltages.

5. Are there any limitations to Kirchhoff's Laws?

Kirchhoff's Laws are based on the assumptions of ideal circuit elements and linear circuit behavior. Therefore, they may not be accurate in circuits that contain non-ideal components such as diodes or transistors, or in circuits with nonlinear behavior. Additionally, Kirchhoff's Laws do not take into account the effects of electromagnetic interference or noise in a circuit.

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