EMF & Circuits Homework: Solve for Vab

In summary, when an ideal voltmeter is connected to a voltage source, the voltmeter will read a certain voltage. If a variable resistor is connected in parallel with the voltmeter, the voltage across the resistor will be the same as the voltage across the voltmeter, but the current through the resistor will be the same as the current through the voltmeter.
  • #1
rasmusuperfan
7
0

Homework Statement



In the circuit (http://i.imgur.com/II38M.jpg),a variable resistor R is connected across the terminals of a battery, and an ideal ammeter and ideal voltmeter are also connected. In the figure, E is the internal emf of the battery, and r is its internal resistance. All the connecting cables have no appreciable resistance. The ammeter reads 3.10 A when the voltmeter reads 5.08 V, and the voltmeter reads 15.0 V when the ammeter reads 0.00 V.

Homework Equations



Vab = E - IR

The Attempt at a Solution



I thought since an ideal voltmeter has infinite resistance, so there would be no current through the resistor. I thought it should just be 15V then, but I feel like I am missing something. Any guidance would be greatly appreciated, thanks.
 
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  • #2
What, precisely, were you asked to determine? It looks to me like you're in a position to determine both E and r.
 
  • #3
I am sorry, don't know how I forgot the most important part. It asks when R = 6.00 ohms, what is the reading of the voltmeter
 
  • #4
If you knew the values of E and r, then when R is made 6.00 Ω you can calculate the voltage across it (it'll be a simple voltage divider circuit). So, can you think of how you might determine E and r from the given data?
 
  • #5
Well, V = E - Ir, would you solve for r to be 5.08V = 15V - 3.1A*r? And I am sorry, but I don't quit get what you mean by a voltage divider, they split evenly?
 
  • #6
rasmusuperfan said:
Well, V = E - Ir, would you solve for r to be 5.08V = 15V - 3.1A*r? And I am sorry, but I don't quit get what you mean by a voltage divider, they split evenly?

Yes. You can find r that way. I see that you've also decided that E = 15V (the open circuit voltage across the battery). That's correct.

A voltage divider using resistors is a series connection of two resistors, with the "output" voltage taken across one of the resistors. So, if you had resistors R1 and R2 connected to a supply voltage V, and the output was the potential across R2, then

Vout = V*R2/(R1 + R2)

This follows from the fact that the current through the series connected resistors is V/(R1 + R2), so the voltage across R2 must be R2* V/(R1 + R2).
 
  • #7
Awesome, thank you so much for all the help. That makes perfect sense, sorry I was confused. You really helped out a lot, have a great day
 

1. How do I solve for Vab in an EMF circuit?

In order to solve for Vab, you will need to use Kirchhoff's voltage law (KVL). This law states that the sum of all voltages in a closed loop must equal zero. By setting up a loop in the circuit, you can create an equation that includes all of the voltages in the loop, including Vab. Then, you can solve for Vab algebraically.

2. What is the difference between EMF and voltage in a circuit?

EMF (electromotive force) is the potential difference created by a source, such as a battery, in order to drive current through a circuit. Voltage, on the other hand, is the potential difference between two points in a circuit. In other words, EMF is the source of voltage in a circuit.

3. How does a change in resistance affect Vab in a circuit?

A change in resistance can affect Vab in a circuit by altering the current flow. This is because according to Ohm's law, V = IR, where V is voltage, I is current, and R is resistance. So, if the resistance increases, the current will decrease, and vice versa. This change in current can then affect Vab, as it is dependent on the current in the circuit.

4. Can Vab be negative in a circuit?

Yes, Vab can be negative in a circuit. This can occur when the voltage drop across a component is greater than the EMF, resulting in a negative potential difference. It is important to pay attention to the direction of current flow when calculating Vab, as this can also affect the sign of the value.

5. How does the placement of a resistor in a circuit affect Vab?

The placement of a resistor in a circuit can affect Vab by changing the amount of voltage drop. When a resistor is placed in series, it will cause a voltage drop proportional to its resistance. In parallel, however, the voltage drop will be the same across all components, including the resistor. This can affect the value of Vab, as it may be different depending on the placement of the resistor in the circuit.

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