# Circuit problem: Figuring out Voltage Drops

• joej24
In summary, the conversation discusses the role of a voltmeter in measuring potential difference and how its location in a circuit affects the measurement. The voltmeter has two leads which are connected to specific points in the circuit, and it measures the potential difference between those points. Its high resistance ensures that it does not disturb the circuit's operation. The potential difference measured would change if the location of one of the points was altered.
joej24

V = IR

## The Attempt at a Solution

Why does the voltmeter measure only the voltage drop in V3?

Is it because the Voltmeter is specifically placed between points a and b?
I originally thought that the Voltmeter would measure the voltage drop up to where it is located.

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What do you mean by "up to where it is located"?

The voltmeter has two leads (connection points); it can only tell you the potential difference that exists between those two leads. Anything else would be magic

Starting from the positive end of the battery and ending where the voltmeter is located. What are leads and where are they located? Why does the voltmeter only show the pot. diff. between these points too?

joej24 said:
Starting from the positive end of the battery and ending where the voltmeter is located. What are leads and where are they located? Why does the voltmeter only show the pot. diff. between these points too?

Have you ever seen a voltmeter in real life? It has two flexible wires (usually one is red and the other black) that end in rigid "test probes". The user touches the ends of the probes to the circuit at the points, between which, he wishes to determine the potential difference.

In your circuit the voltmeter leads are connected to points a and b, so it will measure the potential difference between points a and b.

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Oh okay. And no, I am not too familiar with voltmeters.

So the voltmeter isn't really part of the circuit? The current first passes through r (50 Ohms) from the positive terminal of the battery and then splits at J. It meets back at the point right next to R3. It then passes through R3, past b, and goes back to the negative terminal of the battery.

The current that passes through b is 6 A and that current also passes through R3. So V = IR = 6*150 = 900 V.

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How does the location of point a affect the potential difference measured by the voltmeter?
If a was located right after r (the battery's internal resistance) would the potential difference be changed in any way?

Is this correct?

V = IR
Total Resistance that is between a and b is 150 + (1/600 + 1/300) ^ -1 = 350 Ohms
V = 6 * 350 = 2100, which is the energy lost between a charge from point a to point b?

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joej24 said:
Oh okay. And no, I am not too familiar with voltmeters.

So the voltmeter isn't really part of the circuit? The current first passes through r (50 Ohms) from the positive terminal of the battery and then splits at J. It meets back at the point right next to R3. It then passes through R3, past b, and goes back to the negative terminal of the battery.

The current that passes through b is 6 A and that current also passes through R3. So V = IR = 6*150 = 900 V.

Looks good.

Voltmeters are usually designed to present a very high resistance to the circuit under test (they "look like" a very high resistance value) so that they don't disturb the operation of the circuit.

joej24 said:
How does the location of point a affect the potential difference measured by the voltmeter?
If a was located right after r (the battery's internal resistance) would the potential difference be changed in any way?

Yes, of course it would change; you would be measuring the potential from point b to a location different from a in the circuit.
Is this correct?

V = IR
Total Resistance that is between a and b is 150 + (1/600 + 1/300) ^ -1 = 350 Ohms
V = 6 * 350 = 2100, which is the energy lost between a charge from point a to point b?

Yes, that's good.

## 1. What is a voltage drop in a circuit?

A voltage drop is the difference in voltage between two points in a circuit. It occurs when the current flows through a component, such as a resistor or wire, and some of the electrical energy is converted into other forms, such as heat or light.

## 2. How do I calculate voltage drops in a circuit?

To calculate voltage drops in a circuit, you can use Ohm's Law which states that the voltage drop across a component is equal to the current flowing through it multiplied by its resistance. Alternatively, you can use Kirchhoff's Voltage Law which states that the sum of voltage drops around a closed loop in a circuit is equal to the sum of voltage sources.

## 3. What causes voltage drops in a circuit?

Voltage drops can be caused by a variety of factors, including the resistance of components, the length and thickness of wires, and the load on the circuit. Other factors such as temperature and material properties can also affect voltage drops.

## 4. How do I troubleshoot voltage drop issues in a circuit?

To troubleshoot voltage drop issues in a circuit, you can use a multimeter to measure the voltage at various points in the circuit and compare it to the expected voltage. If there is a significant difference, you can then inspect the circuit for any damaged or faulty components and replace them if necessary.

## 5. How can I minimize voltage drops in a circuit?

To minimize voltage drops in a circuit, you can use thicker wires with lower resistance, reduce the length of the wires, and use components with lower resistance values. You can also distribute the load of the circuit across multiple components to reduce the overall current and voltage drops. Additionally, keeping the circuit clean and free of any corrosion or loose connections can also help reduce voltage drops.

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