Measuring Potential Difference Across 4.5kOhm Resistor w/ Voltmeter

[physstudent1]

Homework Statement

A circuit consists of a series combination of 6.00 -kOhm and 4.50 kOhm 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 -kOhm resistor using a voltmeter having an internal resistance of 10.0 kOhm.

What potential difference does the voltmeter measure across the 4.50 -kOhm resistor?

Homework Equations

The Attempt at a Solution

I tried doing (50*10000)/V - 10,000 = 4500 and solving for V but I got 35.5 and this isn't the right answer. There are no formulas in my book for calculations with voltmeters I got this formula online so I'm not even sure if it is correct can anyone help me out?

 

[alphysicist]

Hi physstudent1,

Homework Statement

A circuit consists of a series combination of 6.00 -kOhm and 4.50 kOhm 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 -kOhm resistor using a voltmeter having an internal resistance of 10.0 kOhm.

What potential difference does the voltmeter measure across the 4.50 -kOhm resistor?

Homework Equations

The Attempt at a Solution

I tried doing (50*10000)/V - 10,000 = 4500 and solving for V but I got 35.5 and this isn't the right answer. There are no formulas in my book for calculations with voltmeters I got this formula online so I'm not even sure if it is correct can anyone help me out?

Treat the voltmeter as just a resistor being hooked in parallel to the 4.5 kOhm resistor. In this new circuit, what is the potential difference across the 4.5 kOhm resistor?

 

[thomate1]

As a scientist, it is important to accurately measure and record data in an experiment. In this case, the goal is to measure the potential difference across a 4.5kOhm resistor using a voltmeter with an internal resistance of 10.0 kOhm.

Based on the given information, the potential difference across the 4.5kOhm resistor can be calculated using Ohm's Law, which states that V = IR, where V is the potential difference, I is the current, and R is the resistance. In this case, the current flowing through the circuit can be calculated using the total resistance of the circuit and the voltage of the battery.

Using the formula V = IR, we can rearrange it to solve for V: V = IR = (50.0V)/(6.00kOhm + 4.50kOhm) = 4.55V

However, this is not the true potential difference across the 4.5kOhm resistor as the voltmeter has an internal resistance of 10.0 kOhm. This means that the voltmeter will affect the circuit and change the potential difference being measured.

To account for this, we can use the formula for total resistance in a series circuit, which is R_total = R1 + R2 + ... + Rn, where R1, R2, etc. are the individual resistances in the circuit. In this case, the total resistance of the circuit would be 6.00kOhm + 4.50kOhm + 10.00kOhm = 20.5kOhm.

Using this total resistance, we can calculate the current flowing through the circuit: I = V/R = (50.0V)/(20.5kOhm) = 2.44*10^-3 A.

Now, to calculate the potential difference across the 4.5kOhm resistor, we can use Ohm's Law again: V = IR = (2.44*10^-3 A)(4.50kOhm) = 10.98V.

Therefore, the potential difference measured by the voltmeter across the 4.5kOhm resistor would be 10.98V. It is important to note that this is not the true potential difference, as the voltmeter has affected the circuit and changed the value. To measure the