Calculating voltage

Thread Starter

Bobby2017

Joined Dec 28, 2019
19
Hi guys i was asked this question... 20200111_223502.jpg20200111_223409.jpg

I done this for part a
20200111_223651.jpg

Which left me with this circuit

20200111_223733.jpg

My question is...is there an equation to work out the voltage across the R3? Or can someone show me how to do it please?
Thanks in advance
 

WBahn

Joined Mar 31, 2012
25,250
Voltages and currents are signed quantities, which means that in addition to knowing WHERE a current or a voltage is defined to be in the circuit, you also need to define the DIRECTION of that voltage and current.

Even more fundamental, you can't just throw out equations using I1 and I2 and not even define WHERE those currents are in your circuit. Don't make us, or the grader, or yourself guess at what you meant -- define them! Annotate them on the circuit diagram so that everyone, including you, knows exactly what you meant.

In your first KVL equation, you are defining I1 to be the current flowing left to right through the 1 kΩ resistor and I2 to be the current flowing right-to-left through the 1.5 kΩ resistor.

But in your second KVL equation, I1 to the the current flowing upward through the 680 Ω resistor and I2 to be the current flowing left-to-right through the 1.5 kΩ resistor.

By not clearly defining what I1 is and what I2 is, you've put yourself in a position so that even you have to guess what they are and you guessed wrong. Hence, your very first two equations are wrong and everything that follows is nothing but wasted effort.

When you start a problem, clearly annotate the variables you are going to use on your diagram. Then set up your equations (but don't do any work on them) and then verify that those equations make sense and are correct for the problem. Remember, this is where ALL of the circuit principles are applied and everything after this point is simply algebra. If you don't get the set up correct, no amount of correctly-performed algebra will make a bit of difference.

You also need to properly track your units throughout your work.
 

WBahn

Joined Mar 31, 2012
25,250
thanks for your help... what should i have done instead then if my equations are wrong?
I already told you, twice, what to do:

#1: "Annotate them on the circuit diagram so that everyone, including you, knows exactly what you meant."
#2: "When you start a problem, clearly annotate the variables you are going to use on your diagram."

Maybe the third time's the charm:

The first step, in most circuit analysis problems, should be to clearly annotate your circuit diagram with all of the voltages and currents you need to refer to, including their polarities. If you leave one off, either by accident or because you thought you wouldn't need it, then be sure to add it to the diagram if you do, in fact, use it. The reader, including YOU, should be able to look at any equation you write and refer back to that diagram to know precisely what each variable or parameter in that equation means.

Furthermore, I told you what the next step should always be: "Then set up your equations (but don't do any work on them) and then verify that those equations make sense and are correct for the problem."

So do just those two things -- annotate your diagram and write down your two loop equations. Post just that much and let's make sure that you've got that much correct before you proceed.
 

Thread Starter

Bobby2017

Joined Dec 28, 2019
19
its probably took you longer to type them 2 messages than it has to just give me the answer...
you can clearly see i have no idea what to do
the work i have done is to the best of my ability.... if i had any idea about what you are talking about then i would fix it
instead of just telling me i'm wrong.... HELP ME
how is my circuit diagram annotated incorrectly? what equation should i use? what do you mean by annotate the variables? what are the variables?
 

MrChips

Joined Oct 2, 2009
20,245
Stop being snarky and be grateful for the sound advice you are receiving.
Clearly, you need to draw I1 and I2 currents in your circuit diagram and show the direction of I1 and I2.
This way you will avoid making silly mistakes.
 

WBahn

Joined Mar 31, 2012
25,250
its probably took you longer to type them 2 messages than it has to just give me the answer...
If you are looking for someone to just give you the answer, you came to the wrong place.

you can clearly see i have no idea what to do
the work i have done is to the best of my ability.... if i had any idea about what you are talking about then i would fix it
instead of just telling me i'm wrong.... HELP ME
I didn't just tell you that you are wrong -- I explained what was wrong and what you need to do to fix it. You just don't want to put in the effort to try to apply that advice.

how is my circuit diagram annotated incorrectly? what equation should i use? what do you mean by annotate the variables? what are the variables?
It's not annotated at all! In your first equation you use I1 and I2 (those, by the way, are known as "variables" that you are trying to solve for). Well, if I tell you that I1 is 3.2 A, then what does that mean? That 3.2 A is flowing where in the circuit? In what direction? Put an arrow on the circuit diagram that is labeled "I1" so that people know that if I1 turns out to be 3.2 A that it means that there are 3.2 A flowing in the circuit at that point in that direction.

Look at your last imagine on which you indicated your answers. You annotated that diagram in such a way as so show that there is (assuming you did it correctly) 0.014 A of current flowing upward out of the 12 V battery and 0.017 A flowing downward out of the 5 V battery. Why did you put those currents in those places going in those directions? Did you just throw a dart at the diagram and flip a coin to determine the direction?

No, you didn't. You put them at those locations because you meant for I1 to be the current flowing upward through the 12 V battery and for I2 to be the current flowing upward through the 5 V battery. So when you got I1 = 0.014 A you annotated the drawing accordingly. When you got I2 = -0.017 A you annotated the drawing with that information, but showing the current as 0.017 A going downward due to the negative value of I2 (which you meant to define as the current flowing upward).

The problem is that, while you set up your first equation in a manner consistent with those definitions of I1 and I2, you set up your second equation with completely different definitions for I1 and I2 because you didn't clearly define what they were and so even YOU couldn't keep them consistent from one line to the next in your work.
 

MrChips

Joined Oct 2, 2009
20,245
Here is your first mistake.

If we are to assume that you drew I1 and I2 in the direction of the arrows in your second drawing, then your very first equation is clearly wrong.

This is what you wrote:

12 = 1000 I1 + 680 ( I1 + I2)

Can you spot your mistake?
 

Thread Starter

Bobby2017

Joined Dec 28, 2019
19
Here is your first mistake.

If we are to assume that you drew I1 and I2 in the direction of the arrows in your second drawing, then your very first equation is clearly wrong.

This is what you wrote:

12 = 1000 I1 + 680 ( I1 + I2)

Can you spot your mistake?
No i cant spot the mistake sorry, i have no clue about this sort of stuff and the work i managed to do (which is clearly wrong) was me trying to follow a worked example.
Its likely that i'll never see another question like this in my life but unfortunately its part of an assignment that i need to do
 

MrChips

Joined Oct 2, 2009
20,245
Here is your first step.

Resistors R1, R2, and R3 are connected to a single 3-way node.

1) Draw three currents I1, I2, and I3 flowing through the resistors. You can draw the arrows in which direction you choose. The correct direction of each will come out in the wash.

2) Apply KCL (Kirchhoff's Current Law) for this node relating I1, I2, and I3.
What is KCL?
The sum of all currents entering the node must equal the sum of all currents leaving the node.
Stated differently, the total sum of all currents at the node, with correct application of the sign of the current, must be equal to zero.

Edit: Consider how much longer it took me to write all of this when I could have just given you the answer. Capisce?
 
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