Load voltage with shorted load / open load

In summary: Ohm's law states that the voltage between two points is proportional to the current flowing through the between those points. So, since there is no current between A and B, the voltage between them must be equal to the voltage at the ends of the battery, or 10V.
  • #1
illidari
47
0

Homework Statement



1. R = 470 ohms. Connect a jumper wire between points A and B. Measure the voltage across the load resistor. Explain why the load voltage with a shorted load is zero using Ohm's law.

2. Remove the jumper wire and open the load resistor. Measure the voltage between the AB terminals. Why is the load voltage with an open load approximately equal to the source voltage. Use ohm's law and kirchhoff's voltage law to explain.

Homework Equations



image123424.png

***Voltage source = 10V

The Attempt at a Solution



1. I am confused how to explain this. Would it simply be because the wire has no resistance it is taking all the current up? V=IR , V=0*r V= 0

2. I'm lost on this one. Opening the load resistor would just make it a 10V battery in series with a 470 resistor. That doesn't exactly make it near the source voltage?

edit: still lost :( not sure what else to add, noticed a lot of views
 
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  • #2
I'd say your answer to 1) is correct. There's no resistance between A and B, so they must be at the same potential, by Ohm's law. For 2), now there is no current flow. So the potential at A and B must be the same as the potential at the corresponding end of the battery by the same argument, right?
 
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  • #3
Measure the voltage across the load resistor.

I think the question would make more sense if it read "Measure the voltage across the points A and B", since the load resistor is detached from the circuit.

Use Ohm's law to calculate the voltage across R when no current is flowing.
Then use Kirchhoff's voltage law to determine the voltage across A and B.
 
  • #4
lewando said:
I think the question would make more sense if it read "Measure the voltage across the points A and B", since the load resistor is detached from the circuit.

Use Ohm's law to calculate the voltage across R when no current is flowing.
Then use Kirchhoff's voltage law to determine the voltage across A and B.

I edited it to the words the book used D: my bad

Ohms law: V=IR , V=0*470 = 0V across R...?

The voltage law applies to a closed circuit :/ Not sure how I would get the sums of the voltages to come out to 0.
 
  • #5
illidari said:
I edited it to the words the book used D: my bad

Ohms law: V=IR , V=0*470 = 0V across R...?

The voltage law applies to a closed circuit :/ Not sure how I would get the sums of the voltages to come out to 0.

I would use Ohm's law on the second part as well. See post 2.
 
  • #6
Not seeing how you can make the same argument

On the 1st I understand it as:
The wire has no resistance so all the current went through it. The load resistor got no current so, V = 0 * R , the V = 0

But on the 2nd question:
No current at all anywhere, V = 0*R = 0
Yet I need to show that it equals the source voltage, 10V.
 
  • #7
illidari said:
Not seeing how you can make the same argument

On the 1st I understand it as:
The wire has no resistance so all the current went through it. The load resistor got no current so, V = 0 * R , the V = 0

But on the 2nd question:
No current at all anywhere, V = 0*R = 0
Yet I need to show that it equals the source voltage, 10V.

V is a potential difference between two points. Can you show the potential difference between point A and the positive terminal of your battery is zero? And that the potential difference between B and the negative terminal is also zero? Wouldn't that do it? Kirchoff's law is what tells you there is no current. If there is no loop, there is no current.
 
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1. What is load voltage with a shorted load?

Load voltage is the voltage measured across a load, or a device that consumes electrical energy, when it is connected to a power source. When a load is shorted, it means that there is a direct connection between the positive and negative terminals of the load, resulting in very low or zero resistance. This causes the load voltage to drop significantly or become zero.

2. How is load voltage affected by an open load?

An open load refers to a load that is disconnected or has very high resistance, causing little to no current flow. With an open load, the load voltage can become very high as there is no resistance to limit the voltage. This can potentially damage the load or the circuit if the voltage exceeds the maximum limit.

3. What are the possible causes of a shorted or open load?

A shorted load can be caused by a faulty load or wiring, accidental touching of the positive and negative terminals, or a damaged power source. An open load can be caused by a broken or disconnected wire, a damaged load, or a faulty connection.

4. How is load voltage measured in the presence of a shorted or open load?

To measure load voltage in the presence of a shorted or open load, first, disconnect the load from the circuit to avoid any potential damage. Then, use a voltmeter to measure the voltage across the load. For a shorted load, the voltage will be significantly lower or zero, while for an open load, the voltage will be very high.

5. How can a shorted or open load be fixed?

The first step in fixing a shorted or open load is to identify the cause. If it is a faulty load, it may need to be replaced. If it is a wiring issue, the connections may need to be checked and repaired. In some cases, a fuse or circuit breaker may need to be replaced. It is important to address these issues promptly to prevent any further damage to the load or the circuit.

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