# Circuits: What does the voltmeter read?

• Andy13
In summary, a 12.0V battery with an internal resistance of 1.80Ohm is connected to two 8.80kilo-Ohm resistors in series. An ammeter with an internal resistance of 0.410Ohm measures the current exiting the battery. A voltmeter with an internal resistance of 17.7kilo-Ohm measures the voltage across one of the 8.80kilo-Ohm resistors in the circuit. To find the voltmeter reading, you can use Kirchhoff's loop laws to find the current through the 8.80kilo-Ohm resistor and then use Ohm's law (V=IR) to find the voltage
Andy13

## Homework Statement

A E = 12.0V battery with an internal resistance r = 1.80Ohm is connected to two 8.80kilo-Ohm resistors in series. An ammeter of internal resistance 0.410Ohm measures the current exiting the battery and at the same time a voltmeter with internal resistance 17.7kilo-Ohm measures the voltage across one of the 8.80kilo-Ohm resistors in the circuit. What does the ammeter read? What does the voltmeter read?

## Homework Equations

V = IR
Kirchhoff's Loop laws
Resistors in parallel: ((1/R1)+(1/R2)+...)^-1 = R(equivalent)

## The Attempt at a Solution

I drew the circuit described above and successfully found the ammeter reading of the current out of the battery ( or the total current). In my diagram (which should be right, given that I got I right), the battery, the ammeter, and the resistors are in series, while the voltmeter is in parallel with one of the resistors.

My I(battery) was 0.000817 A, which I found using I = V/R after finding the total equivalent resistance.

I'm having trouble finding the voltmeter reading, which should be measuring the voltage across one of the 8.8 kΩ resistors. I've done two separate calculations, but neither has been correct:

1. V(through 8.8kΩ resistor) = I(battery)*8.8 ---> from V = IR
2. Since the voltmeter has some internal resistance, some current could split along that path. I did Kirchhoff's loop equations to find the current through the resistor and used that current times the resistance to find V.

Method 2 was wrong-- is this because I messed up my math, or because the theory was wrong and no current goes through the voltmeter?

Method 2: Current goes through the 17.7 k and the 8.8 k in parallel so
V = I*R = .000817*5878 = 4.80 V.

Andy13 said:

## Homework Statement

A E = 12.0V battery with an internal resistance r = 1.80Ohm is connected to two 8.80kilo-Ohm resistors in series. An ammeter of internal resistance 0.410Ohm measures the current exiting the battery and at the same time a voltmeter with internal resistance 17.7kilo-Ohm measures the voltage across one of the 8.80kilo-Ohm resistors in the circuit. What does the ammeter read? What does the voltmeter read?

## Homework Equations

V = IR
Kirchhoff's Loop laws
Resistors in parallel: ((1/R1)+(1/R2)+...)^-1 = R(equivalent)

## The Attempt at a Solution

I drew the circuit described above and successfully found the ammeter reading of the current out of the battery ( or the total current). In my diagram (which should be right, given that I got I right), the battery, the ammeter, and the resistors are in series, while the voltmeter is in parallel with one of the resistors.

My I(battery) was 0.000817 A, which I found using I = V/R after finding the total equivalent resistance.

I'm having trouble finding the voltmeter reading, which should be measuring the voltage across one of the 8.8 kΩ resistors. I've done two separate calculations, but neither has been correct:

1. V(through 8.8kΩ resistor) = I(battery)*8.8 ---> from V = IR
2. Since the voltmeter has some internal resistance, some current could split along that path. I did Kirchhoff's loop equations to find the current through the resistor and used that current times the resistance to find V.

Method 2 was wrong-- is this because I messed up my math, or because the theory was wrong and no current goes through the voltmeter?

This voltmeter has a resistance only about twice the value of the resistor, so connecting it in parallel with one of them with reduce the effective resistance of that pair.

You will effectively have a a battery in Series with an ammeter, in series with an 8.8 kΩ resistor in series with an effective (approx) 5.9 kΩ resistor.

What is the PD across the 5.9 kΩ resistor then? That should be what the Voltmeter reads.

[that 5.9 kΩ value I quote is approximate, you should be working with what you really get when a 8.8 kΩ and 17.7 kΩ resistor are connected in parallel.]

Thanks to all!

I would first double check my calculations and make sure I didn't make any mistakes. If my calculations were correct, then I would analyze the circuit and determine if the voltmeter is indeed measuring the voltage across the 8.8kΩ resistor. If it is, then I would consider the internal resistance of the voltmeter and how it may affect the reading. I would also consider the fact that the voltmeter is in parallel with the resistor, which means it may not be measuring the full voltage drop across the resistor. I would also check if there are any other resistors or components in parallel with the voltmeter that may be affecting the reading. Overall, I would carefully analyze the circuit and take into account all factors that may affect the voltmeter reading before coming to a conclusion.

## 1. What is a voltmeter?

A voltmeter is a measuring instrument used to measure the voltage or electrical potential difference between two points in an electrical circuit. It is typically connected in parallel with the circuit component whose voltage is being measured.

## 2. How does a voltmeter work?

A voltmeter works by using a sensitive galvanometer, which is a device that measures small electrical currents, and a series of resistors. The voltage being measured creates a small current that flows through the galvanometer, causing it to move. The movement of the galvanometer is then translated into a numerical reading on the voltmeter display.

## 3. What is the difference between a voltmeter and an ammeter?

A voltmeter measures voltage, which is the potential difference between two points in a circuit, while an ammeter measures current, which is the rate of flow of electric charge. Voltmeters are connected in parallel with the circuit, while ammeters are connected in series.

## 4. When should a voltmeter be used?

A voltmeter should be used whenever you need to measure the voltage across a component or circuit. This can be helpful in troubleshooting electrical issues, testing the performance of a circuit, or determining the voltage supplied by a power source.

## 5. What are the units of measurement for a voltmeter?

The units of measurement for a voltmeter are volts, abbreviated as "V". Some voltmeters may also display millivolts (mV) or kilovolts (kV) for smaller or larger voltage readings, respectively.

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