Calculate the voltage in a circuit

[DLeuPel]

Homework Statement

Given the following circuit, find the values of V1 and V2.

Homework Equations

The second law of Kirchhoff states that the sum of all voltajes must equal 0

The Attempt at a Solution

I found in the book that the solution is V1: 15V and V2: 14V I tried to solve V1 by adding 9 and 6 to form 1t since they’re batteries have they’re charge inverted. But I would prefer a more satisfactory explanation

 

[gneill]

The second law of Kirchhoff states that the sum of all voltajes must equal 0

Which voltages? Kirchhoff's laws are quite specific about what things must sum to zero.

I tried to solve V1 by adding 9 and 6 to form 1t since they’re batteries have they’re charge inverted.

I don't understand what you mean by inverted charge. Can you elaborate please?

 

[DLeuPel]

Which voltages? Kirchhoff's laws are quite specific about what things must sum to zero.

I don't understand what you mean by inverted charge. Can you elaborate please?

The battery simbol is inverted, that’s why. What should you add and what should you subtract? How do you solve the exercise ?

 

[gneill]

The battery simbol is inverted, that’s why.

Hmm. That doesn't mean anything to me. Are you saying that the potential difference across the voltage sources is somehow reversed? That's not correct.

What should you add and what should you subtract? How do you solve the exercise ?

Can you write out the full statement of Kichhoff's Voltage Law?

 

[DLeuPel]

Hmm. That doesn't mean anything to me. Are you saying that the potential difference across the voltage sources is somehow reversed? That's not correct.

Can you write out the full statement of Kichhoff's Voltage Law?

. Since the 6V battery and the 9V battery have oposite charges, the flow of electrons coming from both batteries sum each other in V1, leaving 15 V and satisfying Kirchhoffs second law: 15-9-6=0 That would be an attempt for a solution but what happens with the 9 V and the 6 V where they cancel out for the 3V at the right? It seems to me you don’t know how to solve it so better let someone else attempt to solve it.

 

[gneill]

. Since the 6V battery and the 9V battery have oposite charges, the flow of electrons coming from both batteries sum each other in V1, leaving 15 V and satisfying Kirchhoffs second law: 15-9-6=0 That would be an attempt for a solution but what happens with the 9 V and the 6 V where they cancel out for the 3V at the right? It seems to me you don’t know how to solve it so better let someone else attempt to solve it.

Yes. Note the word is not "charge" for the batteries. Charge refers to a quantity of electric charge with units of Coulombs. The words you might use are "voltage across", "potential across", or perhaps "EMF" for a battery.

Essentially what you've done is apply Kirchhoff's Voltage Law (KVL) around one loop. There are two more loops you can trace out which will include $V_2$ and the $3\;V$.

Don't assume because I haven't solved it for you (which would be against the rules here) that I cannot solve it.Edit: Fixed a typo.

 

[DLeuPel]

Yes. Note the word is not "charge" for the batteries. Charge refers to a quantity of electric charge with units of Coulombs. The words you might use are "voltage across", "potential across", or perhaps "EMF" for a battery.

Essentially what you've done is apply Kirchhoff's Voltage Law (KVL) around one loop. There are two more loops you can trace out which will include $V_2$ and the $3\;V$.

Don't assume because I haven't solved it for you (which would be against the rules here) that I cannot solve it.Edit: Fixed a typo.

Ok, so you can’t tell me how I can answer it directly. So what I can do so far is , assuming ( without knowing if i’m right or wrong ) that the squares measure the voltaje, the V 1 is equal to 15 V since from the top battery ( 6V )a flow of electrons is moving towards it and from the bottom battery (9V) the flow of electrons is moving away from the battery, combining both in V1 9+6 since two flow of electrons meet, 15 V is being measured. Now for the bottom one, there is where I get stuck. I assume that the squares are instruments that measure the voltaje so that the 3V means that from the top square at the right, 3V are traveling down the circuit. Now when it meets with the 8V battery, it should substructure leaving 5 V while the original 15 V is coming down the left part of the circuit. When they combine the would form 10 V but the answer says 14 V. I know It’s wrong but I don’t know of to solve it.

 

[CWatters]

The squares are circuit components of unknown type. Probably resistors but you don't need to know that to solve the problem

I think you need to develop or read up on the correct process for applying KVL (and KCL). The first thing to do is mark up the original circuit...

All voltages are relative so it's meaningless to refer to "V1" without specifying which side of that component is your reference. Normally you do this by putting an arrow next to the text V1. Personally I would do this for all voltages in the circuit (eg V1, V2, V3 and the battery voltages). If you don't do this you cannot tell if V1 should be +15V or -15V.

Then before you can apply KVL you need to specify the loop or loops to which you are applying KVL nd the direction in which you are "travelling" around the loop(s). Typically you do this with another arrow labelled "loop 1", "loop 2) etc.

Then you can write your KVL equations. Personally I like to write them in the form A + B - C = 0 and not A + B = C as I find I'm less likely to make mistakes. If B is a negative voltage I have even been known to write A + (-B) + C = 0 rather than A - B + C = 0.

All this may seem unnecessary/overkill but sign errors are the biggest source of error when applying KVL. You can make shortcuts later if you like but don't say we didn't warn you :-)

 

[DLeuPel]

The squares are circuit components of unknown type. Probably resistors but you don't need to know that to solve the problem

I think you need to develop or read up on the correct process for applying KVL (and KCL). The first thing to do is mark up the original circuit...

All voltages are relative so it's meaningless to refer to "V1" without specifying which side of that component is your reference. Normally you do this by putting an arrow next to the text V1. Personally I would do this for all voltages in the circuit (eg V1, V2, V3 and the battery voltages). If you don't do this you cannot tell if V1 should be +15V or -15V.

Then before you can apply KVL you need to specify the loop or loops to which you are applying KVL nd the direction in which you are "travelling" around the loop(s). Typically you do this with another arrow labelled "loop 1", "loop 2) etc.

Then you can write your KVL equations. Personally I like to write them in the form A + B - C = 0 and not A + B = C as I find I'm less likely to make mistakes. If B is a negative voltage I have even been known to write A + (-B) + C = 0 rather than A - B + C = 0.

All this may seem unnecessary/overkill but sign errors are the biggest source of error when applying KVL. You can make shortcuts later if you like but don't say we didn't warn you :-)

Ok, so this is what I have achieved so far. The top square has as V1 15 V since the direction of the current in both batteries have both positive sings which means that the V1 must have a negative sign and a value that substrates both batteries.This value is 15 V. Then, you have 3V but it’s not a battery, then you have 8 V which is a battery. Following the signs I put, V2 equals 5 V which is not the right answer. Since there is only one battery in the bottom square, the 3V needs an extra 5 V to obey Kirchhoffs second law. This makes V2 5 V. Where have I made my mistake, also I took into an account that the top square doesn’t affect the bottom square.

 

[CWatters]

Ok, so this is what I have achieved so far. The top square has as V1 15 V since the direction of the current in both batteries have both positive sings which means that the V1 must have a negative sign and a value that substrates both batteries.This value is 15 V.

No. V1 can be positive or negative depending on your definition of V1. You cannot say V1 is positive or negative without drawing the arrow on the circuit diagram to define what you mean by a positive or negative value.

This is important because in some circuits you will have no idea which direction the current is going or if some voltages are positive or negative until after you have solved all the equations.

The correct approach is to draw arrows on the circuit. Then solve the equations. Then if a value comes out positive or negative you must refer back to your arrow for what that actually means.

 

[Apashanka]

and V2.

Applying kirchhoffs voltage law,
-3V+8V-V₂-V₁-6V=0
Similarly,
-3V+8V-V₂+9V=0
From which V₁ and V₂ is obtained.
I may be wrong!

 

[CWatters]

On the right hand side of the circuit is a component labeled 3V. Which side of that component is +ve? To find V2 you will have to assume one side is +ve.

See my earlier comment about marking up the circuit.