How Do You Apply Kirchhoff's and Ohm's Laws to Analyze Circuit Voltage?

[bhsmith]

Homework Statement

1. Use Kirchoffs and Ohm's rule to find the voltage of vknot (to the far right of the picture)
2. show that your solution is consistent with the constraint that the total power supplied in the circuit equals the total power dissipated.

(Picture of Circuit attached)

Homework Equations

I know I need to use V=i(R), but I'm not sure how to apply kirchovs law here.

The Attempt at a Solution

first, i found the current i(s)= -10/6
then the current 3i(s) would be -5A. , so that means that i(0) would be -5A(2ohms)= -10v
Then I'm not sure how to find the voltage from the current voltage and resistance. I'm not sure I did the part with the 2ohm resistor correctly.

 

[Zryn]

first, i found the current i(s)= -10/6

Your sign convention shows that the current flows up through the 6R resistor and into the positive terminal of the 10V source, since the sign is negative on your result. Is this correct?

then the current 3i(s) would be -5A. , so that means that i(0) would be -5A(2ohms)= -10v

The diamond has the + and - polarity, indicating it is a voltage source. That it is proportional to a current just means it is a Current Controlled Voltage Source (CCVS) as opposed to CCCS, VCVS or VCCS. This means that it would be -5V using your sign convention.

If it was -5A then the current in series would mean that i(0) would also be -5A, but since it is not a current source, you have -5V, and two resistances, and need to find the series current to find the voltage Vo.

-5A(2ohms)= -10v would be correct if your -5A assumption was correct, but it is not.

 

[bhsmith]

Great! That helped alot.
So, if the voltage is -5v through the diamond then the current would be -5/2A.
since they are in series then the two have the same current leaving the voltage v(knot) to be 15/2V?
is this correct?
And also if I am finding the voltage using 3i(s), and the current has units of Amperes, how do I get voltage from that?

 

[Zryn]

So, if the voltage is -5v through the diamond then the current would be -5/2A.

No. The series current is determined by the total resistance.

If the current through the 2R resistor by Ohms Law is -5/2 Amps, then it stands that the current through the 3R load resistor by Ohms Law is -5/3 Amps, but since the two resistors are in series, the current must be in series, and so you are saying that -5/3 A = -5/2 A, which it does not, so something has gone wrong.

The wrong part being using a single resistance instead of the total resistance to find the series current.

since they are in series then the two have the same current ...

This is correct.

... leaving the voltage v(knot) to be 15/2V? is this correct?

If the current was -5/2 A then the voltage would be -5/2 * 3 = -15/2 V. So you would be almost correct (the sign would be wrong) if your initial assumption of -5/2A was correct, which it isn't.

And also if I am finding the voltage using 3i(s), and the current has units of Amperes, how do I get voltage from that?

The internal circuity of a dependent source converts an incoming signal into an outgoing signal, in this case an incoming current into an outgoing voltage. The specifics will be covered by a much later course, but they're complicated and not relevant, so for now you will have to trust the model in that "that's just what happens" (I hate these kinds of explanations myself!) and that the dependent voltage source is 3*i(s) Volts despite i(s) being measured in Amps.

 

[DeeNos]

Hello,

Thank you for your question. I would like to provide some guidance and clarification for your solution attempt.

Firstly, you are correct in using Ohm's Law (V=IR) to find the voltage at the knot (junction) of the circuit. However, in order to fully analyze the circuit, we also need to apply Kirchhoff's Laws.

Kirchhoff's Laws state that the sum of currents entering a junction (knot) must equal the sum of currents leaving the junction, and the sum of voltages in a closed loop (or circuit) must equal zero.

In this circuit, we have two junctions (knots) and two closed loops. Applying Kirchhoff's Laws to the first junction (knot) on the left, we can write the following equation:

-i(2Ω) + 3i(6Ω) + i(4Ω) = 0

Simplifying this equation, we get:

-i(2Ω) + 3i(6Ω) + i(4Ω) = 0
-2i + 18i + 4i = 0
20i = 0
i = 0

This means that there is no current flowing into or out of the first junction (knot) on the left.

Now, we can apply Kirchhoff's Laws to the second junction (knot) on the right. We can write the following equation:

-i(4Ω) - 3i(6Ω) + i(2Ω) = 0

Simplifying this equation, we get:

-i(4Ω) - 3i(6Ω) + i(2Ω) = 0
-4i - 18i + 2i = 0
-20i = 0
i = 0

This means that there is also no current flowing into or out of the second junction (knot) on the right.

Now, we can use Ohm's Law (V=IR) to find the voltage at the knot (junction) on the right. Since there is no current flowing through the 2Ω resistor, we can ignore it and focus on the 4Ω and 6Ω resistors. We can write the following equation:

V = i(4Ω) + i