What is the Potential Difference Across the 4.50-kΩ Resistor?

In summary, a circuit with a series combination of resistors and a 50 Volt battery is given. A voltmeter, with an internal resistance of 10.0 kΩ, is used to measure the potential difference across the 4.50-kΩ resistor. The voltmeter is connected in parallel with the resistor and the battery. The true potential difference is calculated using V=IR and is found to be 22.5 V.
  • #1
Northbysouth
249
2

Homework Statement


A circuit consists of a series combination of 5.50-kΩ and 4.50-kΩ resistors connected across a 50.0 V battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the 4.50-kΩ resistor using a voltmeter having an internal resistance of 10.0 kΩ.

A) What potential difference does the voltmeter measure across the 4.50-kΩ resistor?

B) What is the true potential difference across this resistor when the meter is not present?

C) By what percentage is the voltmeter reading in error from the true potential difference?

Homework Equations



V = IR

The Attempt at a Solution



I'm stuck on part A. From what I understand, the question says there is a battery with resistors on either side of it.

So, knowing that in a loop the voltage is equal to 0, I did my calculations as follow:

0 = 50V - 5500I - 4500I
Thus this tells me that the current is 0.005 A

Then, using V = IR

V = 0.005*4500
V = 22.5 V

It says my answer is wrong and I'm not sure what else to try. Help would be appreciated.
 
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  • #2
Northbysouth said:

Homework Statement


A circuit consists of a series combination of 5.50-kΩ and 4.50-kΩ resistors connected across a 50.0 V battery having negligible internal resistance. You want to measure the true potential difference (that is, the potential difference without the meter present) across the 4.50-kΩ resistor using a voltmeter having an internal resistance of 10.0 kΩ.

A) What potential difference does the voltmeter measure across the 4.50-kΩ resistor?

B) What is the true potential difference across this resistor when the meter is not present?

C) By what percentage is the voltmeter reading in error from the true potential difference?

Homework Equations



V = IR

The Attempt at a Solution



I'm stuck on part A. From what I understand, the question says there is a battery with resistors on either side of it.

So, knowing that in a loop the voltage is equal to 0, I did my calculations as follow:

0 = 50V - 5500I - 4500I
Thus this tells me that the current is 0.005 A

Then, using V = IR

V = 0.005*4500
V = 22.5 V

It says my answer is wrong and I'm not sure what else to try. Help would be appreciated.
There are two resistors, 5.50-kΩ and 4.50-kΩ, in series with a 50 Volt ideal battery.

When you use a volt-meter to measure the voltage drop across the 4.50-kΩ resistor, how is the volt-meter connected in the circuit? ... in parallel or series? ... and with which device?
 
  • #3
There are THREE resistors in the circuit. Two in series and the one in the voltmeter (appears in parallel) which moves depending on where it is used to measure the voltage.
 

1. What is potential difference?

Potential difference is the difference in electric potential energy between two points in an electrical circuit. It is also known as voltage and is measured in volts (V).

2. How is potential difference calculated?

Potential difference is calculated by dividing the work done (energy transferred) by the charge moved. Mathematically, it is represented as V = W/Q, where V is the potential difference, W is the work done, and Q is the charge moved.

3. What is the unit of potential difference?

The unit of potential difference is volt (V). However, in some cases, it can also be measured in millivolts (mV) or kilovolts (kV).

4. How is potential difference related to current?

According to Ohm's Law, potential difference is directly proportional to the current flowing through a circuit. This means that as the potential difference increases, the current also increases, and vice versa.

5. What is the importance of potential difference?

Potential difference is crucial in understanding and analyzing electric circuits. It helps us determine the direction of current flow, the amount of energy transferred, and the overall functioning of the circuit. It is also used in various applications such as power generation, transmission, and distribution.

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