Solving for voltage in a circuit.

[CoolDude420]

Homework Statement

Find the voltage across the resistor and confirm that power in this circuit is conserved.

Homework Equations

KVL
KCL
V=IR, V=I^2R

The Attempt at a Solution

I have no idea where to start with this. We haven't learned voltage/current division yet. I am getting thrown by the i1 and i2 arrows pointing in weird directions. I don't know how to deal with this. Also we haven't done the principle of superposition nor the Thevenin stuff.

The chapter after the one this question was found in tells me about current divison and voltage division and in it it also mentions that anything in parallel with a voltage source can be ignored(as far as the circuit is concerned it isn't there), the same applies to anything in series with a current source. Knowing that, can't I just ignore the current source and just say that the voltage across the resistor 1V?

Super confused.

 

[gneill]

Hi CoolDude420, Welcome to Physics Forums.

I have no idea where to start with this. We haven't learned voltage/current division yet. I am getting thrown by the i1 and i2 arrows pointing in weird directions. I don't know how to deal with this. Also we haven't done the principle of superposition nor the Thevenin stuff.

The chapter after the one this question was found in tells me about current divison and voltage division and in it it also mentions that anything in parallel with a voltage source can be ignored(as far as the circuit is concerned it isn't there), the same applies to anything in series with a current source. Knowing that, can't I just ignore the current source and just say that the voltage across the resistor 1V?

Yes, the principle involved is one of the most basic and also one of the first things introduced in a course on electronics: Components in parallel share the same potential difference.

 

[CoolDude420]

Hi CoolDude420, Welcome to Physics Forums.
Yes, the principle involved is one of the most basic and also one of the first things introduced in a course on electronics: Components in parallel share the same potential difference.

So the answer is 1V?

 

[gneill]

So the answer is 1V?

Well, part of the answer is 1 V You still need to deal with the power conservation part of the question.

 

[CoolDude420]

Well, part of the answer is 1 V You still need to deal with the power conservation part of the question.

Oh yes. P=VI, the Power consumed by the resistor is then P(resistor) = 1V*1A. Um. these arrows are throwing me off. I don't understand where the current from the current source is going. Also I am assuming there's also another current from the voltage source too? Can you clarify where the currents are going?

 

[gneill]

Don't take the current arrows as specifying the actual directions of the currents. They're simply specifying how you are to interpret their directions. It may turn out that one or more of them will have negative values because they are actually drawn in the opposite direction to the actual current flow.

One current that you can be absolutely sure of its direction is that of the current source. It will cause its specified current to flow in the direction of its symbol's arrow no matter what. So what does that tell you about i2?

You know the potential difference across the resistor, so what is i3?

Apply KCL at the top node.

 

[CoolDude420]

Don't take the current arrows as specifying the actual directions of the currents. They're simply specifying how you are to interpret their directions. It may turn out that one or more of them will have negative values because they are actually drawn in the opposite direction to the actual current flow.

One current that you can be absolutely sure of its direction is that of the current source. It will cause its specified current to flow in the direction of its symbol's arrow no matter what. So what does that tell you about i2?

You know the potential difference across the resistor, so what is i3?

Apply KCL at the top node.

It means i2 is -1A (or just change the arrow upward and i2 will be 1A).

i3 is V/R = 1/1 = 1A.

So i2(1A) is entering the top node and i3(1A) is leaving the top node and let's assume i1 is also leaving the top node. So current into top node = current out. So that means i2=i3+i1
So 1 = 1 +i1
So i1 = 0A, meaning it consumes nor dissipates any power.

And so Power for the current source is P=VI = (1)(-1) =1W and for the resistor its consuming 1W so, overall is 0, meaning energy is conserved.

 

[gneill]

Sounds good!