Voltage Calculations for Simple Circuit with Switches - VA and VB

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Homework Statement

I'm going to try drawing the circuit. Numbers are added to it for reference. I apologize for the formatting. I'm not sure how to do white space. Edit: for V_B the + is on top.

V_A and V_B are switches.

I need to figure out the voltage across V_A and VB_B when they are:
1. open, open
2. open, closed
3. closed, open
4. closed, closed

1-------(+ V_A -)-------(+ 8 V -)-------2--------------3
|_______________________________|___________|
12Ω____________________________15Ω_______(+ V_B -)
|_______________________________|___________|
6-----------------------------------------5--------------4

Homework Equations

V = IR
Kirchhoff's Laws

The Attempt at a Solution

1. When they are both open there isn't a circuit. So, V_A = 0 V and V_B = 0 V.
2. Again, no circuit.
3. There is a circuit. It has a total resistance of 27 Ω (I don't know if this matters). V_A = -8 V and V_B = 0 V.
4. I'm not sure about this one. There are two loops, but loop 1256 has a total resistance of 27 Ω and loop 1346 has a total resistance of 12 Ω. So 1256 gets 27/39 = 9/13 of the current and voltage and 1346 gets 12/39 = 4/13 of the current and voltage? So V_A = -8*9/13 V and V_B = -8*4/13 V?

 

[NascentOxygen]

1. When they are both open there isn't a circuit. So, V_A = 0 V and V_B = 0 V.

Circuit or no circuit, does not give the answer. If a voltmeter (or any device) were to be connected between the 2 terminals of the switch in question, would that form a path for current to flow through the meter? If so, what voltage would it register? That's the question you must address.

 

[NascentOxygen]

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[CWatters]

1. When they are both open there isn't a circuit. So, VA = 0 V and VB = 0 V.

You need a circuit for current to flow. You don't need a circuit for a voltage to be present. For example a 9V battery still produces 9V even when it's disconnected from the circuit.

 

[rezeew]

I would like to clarify a few things about the given circuit and proposed solutions. Firstly, the circuit shown is a series-parallel circuit, which means that the resistors are connected both in series and in parallel. This complicates the calculations as the total resistance is not simply the sum of the individual resistances. Additionally, the switches VA and VB are not labeled with their respective positions (open or closed) for each scenario, which could affect the calculations.

To accurately calculate the voltage across VA and VB, we would need to know the positions of the switches for each scenario. Assuming that the switches are labeled correctly and that the top of the circuit is connected to the positive terminal of a battery, here is a possible solution for each scenario:

1. When both switches are open, there is no complete circuit and no current flows. Therefore, the voltage across VA and VB is 0 V.

2. When VA is open and VB is closed, the circuit is a simple series circuit with a total resistance of 27 Ω. Using Ohm's law (V = IR), the current flowing through the circuit is I = V/R = (8 V)/(27 Ω) = 0.296 A. Since VA is open, there is no voltage drop across it, so VA = 0 V. For VB, the voltage drop is V = IR = (0.296 A)(15 Ω) = 4.44 V. However, since VB is connected to the positive terminal of the battery, the voltage across VB is actually 8 V - 4.44 V = 3.56 V.

3. When VA is closed and VB is open, the circuit is a simple parallel circuit with a total resistance of 12 Ω. Using Ohm's law, the current flowing through the circuit is I = V/R = (8 V)/(12 Ω) = 0.667 A. Since VB is open, there is no voltage drop across it, so VB = 0 V. For VA, the voltage drop is V = IR = (0.667 A)(12 Ω) = 8 V. However, since VA is connected to the positive terminal of the battery, the voltage across VA is actually 8 V - 8 V = 0 V.

4. When both switches are closed, the circuit is a combination of two series circuits: 1256 and 1346. The