# Circuit problem involving sources and resistors

• Santilopez10
In summary: No, it doesn't matter which direction Id points in, as long as the sign of the answer is consistent with the direction of the current.

## Homework Statement

The currents ##I_{{a}}## and ##I_{{b}}## of the circuit have values 4A and -2A in that respective order.
A) Find ##I_{{g}}##
B) Find the power dissipated by each resistance
C) Find ##V_{{g}}## (voltage drop across the current source)
D) Show that the power delivered by the current source is equal to that dissipated by every other circuit element.
See attachment "primero" for the circuit image.

## Homework Equations

Kirchoff’s DC circuits laws only, no resistive simplification neither nodal analysis.

## The Attempt at a Solution

By assuming that the negative current means that the current is flowing in the other direction, and some time trying to try me best to figure out how current circulated across all the circuit, I got attachment "segundo".

Now to my attempt, first, I wrote 4 KCEQ and 2 KVEQ: $$N_{{1}}:4A+2A=I_{{c}}, I_{{c}}=6A$$ $$N_{{2}}:I_{{d}}+I_{{c}}=I_{{g}}$$ $$N_{{3}}:I_{{d}}+2A=I_{{e}}$$ $$N_{{4}}:I_{{g}}=I_{{e}}+4A$$ $$V_{{4214}}:-V_{{g}}+30V+100V+60V=0, V_{{g}}=190V$$ $$V_{{4234}}:-190V+I_{{d}}*30\,\Omega+I_{{e}}*4\,\Omega+I_{{e}}*16\,\Omega=0$$

As you can see, ##V_{{g}}## and ##I_{{c}}## were easy to obtain. Then using ##N_{{3}}## I sustituted it in ##V_{{4234}}## to obtain that ##I_{{d}}=3A##, then plugged in ##N_{{2}}## to get the value of ##I_{{g}}##, 9A.
Until now, A and C are done, then by doing ##P=I^2*R## in each resistance I got 1310W for answer B.
For D, I calculated the current source power generated, which was ##-I*V## that got me 1710W, which is the answer from B plus the power consumed by the voltage source, then all the exercises are correct, BUT: My real question, and problem, is if I correctly indicated all the current circulations in the second attachment, because I want to know if it really matters after all to get the correct answer!

Thanks.

#### Attachments

• primero.png
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• SEGUNDO.png
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Santilopez10 said:
Now to my attempt, first, I wrote 4 KCEQ and 2 KVEQ: $$N_{{1}}:4A+2A=I_{{c}}, I_{{c}}=6A$$ $$N_{{2}}:I_{{d}}+I_{{c}}=I_{{g}}$$ $$N_{{3}}:I_{{d}}+2A=I_{{e}}$$ $$N_{{4}}:I_{{g}}=I_{{e}}+4A$$ $$V_{{4214}}:-V_{{g}}+30V+100V+60V=0, V_{{g}}=190V$$ $$V_{{4234}}:-190V+I_{{d}}*30\,\Omega+I_{{e}}*4\,\Omega+I_{{e}}*16\,\Omega=0$$

?

CWatters
phinds said:
?
What is the problem?

Santilopez10 said:
What is the problem?
You must not be seeing what I'm seeing:

EDIT: this is my bad. My browser has stopped running JavaScript for some reason.

#### Attachments

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Last edited:
Your equations look ok to me but are hard to follow.

Firstly, it's very easy to make a sign error doing KVL and KVC so I prefer to stick rigidly to the following:

1) Define current into or out of a node as +ve and stick with that when writing KCL.
2) Write both the KCL and KVC equations to sum to zero.

Your first two eqns assume out is +ve and the third that in is +ve. You get the right answer but its a more error prone approach.

Again when writing KVL equations it's good practice to state where you are starting and which way around the loop you are going. Initially I assumed that V4214 meant you were starting at node 4 and going anti clockwise to node 2 etc. In which case the first term should be +Vg. But looking again I think you went clockwise and jump about which again makes it error prone. Despite that I think this equation is correct!

Likewise for V4234. It implies clockwise this time but you have a +(Id*30) term which means you went anti-clockwise through R8. Again I think this equation is correct.

So I think all of the equations are correct if hard to follow.

Santilopez10 said:
Until now, A and C are done, then by doing P=I2∗RP=I2∗RP=I^2*R in each resistance I got 1310W for answer B.

B asks for the "power dissipated by each resistance" not the total.

Santilopez10 said:
BUT: My real question, and problem, is if I correctly indicated all the current circulations in the second attachment, because I want to know if it really matters after all to get the correct answer!

Yes and no...

It is important that the arrows and the sign of the answer are consistent with each other (eg Id points to the right and it's value is +3A).

It is not important which direction it points. (eg An equally valid answer would be Id points left and the value is -3A.)

The trick to solving these problems is not to try and guess the direction of a current unless its obvious from the info given. For example the direction of Ib is specified in the problem, however it's not obvious at the start which way Id flows. The correct procedure is just to choose a direction for Id and mark it on the diagram, then ensure your KCL and KVL equations are consistent with that choice. Later on the polarity of Id will become clear and it will be consistent with that choice of direction.

As an exercise you could work this problem again. This time mark the arrow next to Id pointing to the left. Redo all the equations and this time you will find that Id turns out to be -3A instead of +3A.

CWatters said:
As an exercise you could work this problem again. This time mark the arrow next to Id pointing to the left. Redo all the equations and this time you will find that Id turns out to be -3A instead of +3A.

PS...

This particular problem doesn't ask you for the value of Id but if it did then the arrow you mark on the circuit diagram at the start would form part of your answer. The examiner should accept either +3A or -3A depending on which way you drew the arrow.

For this reason if you ever get a negative value for a current (implying it flows the other way to your original prediction) then do not go back and change the direction of the arrow on your drawing. That would make your drawing inconsistent with your working.

CWatters said:
PS...

This particular problem doesn't ask you for the value of Id but if it did then the arrow you mark on the circuit diagram at the start would form part of your answer. The examiner should accept either +3A or -3A depending on which way you drew the arrow.

For this reason if you ever get a negative value for a current (implying it flows the other way to your original prediction) then do not go back and change the direction of the arrow on your drawing. That would make your drawing inconsistent with your working.
got it, thanks a lot!

## What is the difference between a voltage source and a current source?

A voltage source provides a constant voltage, while a current source provides a constant current. In a circuit, a voltage source is represented by a battery, and a current source is represented by a current-controlled current source symbol.

## How do I calculate the total resistance of a circuit with multiple resistors?

The total resistance of a circuit can be calculated by using the formula Rtotal = R1 + R2 + ... + Rn, where R1, R2, etc. are the individual resistances in the circuit. If the resistors are connected in series, the total resistance is the sum of all the resistors. If the resistors are connected in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of each resistor.

## What is Kirchhoff's voltage law and how is it used in circuit analysis?

Kirchhoff's voltage law states that the sum of the voltages around a closed loop in a circuit must be equal to zero. This law is used in circuit analysis to determine unknown voltages or to confirm the accuracy of calculated voltages.

## How do I calculate the current in a circuit with multiple resistors?

The current in a circuit can be calculated by using Ohm's law, which states that current (I) is equal to voltage (V) divided by resistance (R). In a series circuit, the current is the same throughout the circuit, while in a parallel circuit, the total current is equal to the sum of the individual branch currents.

## What is the purpose of a ground in a circuit?

A ground serves as a reference point for voltage in a circuit. It is typically connected to the negative terminal of a voltage source and is used as a common reference point for all other voltages in the circuit. Grounding also helps to protect against electric shock and can provide a path for excess current to flow.