Why Do Voltmeter Readings Differ in a Series Circuit?

[swaha]

Homework Statement

A person measures 9volts across the battery, 3 volts across the resistor Ra and 4.5volts across the resistor Rb connected in series (all three battery, Ra, Rb are connected in series) with a voltmeter of finite resistance. The resistor Ra=2M, and the battery has negligible internal resistance. Find- 1.The resistor Rb and 2. The internal resistance of the voltmeter.

[M=mega ohms]

Homework Equations

- Ohm's law and Kirchoff's laws are only possible relevant equations.

The Attempt at a Solution

- I really didnot get this. Shouldnt the voltages at the two resistors add up to give battery voltage?

[source of the question- tifr gs 2010]

 

[CWatters]

The voltages across each resistor should add up to the battery voltage BUT only if the impedance of the voltmeter is very high. If the impedance of the voltmeter isn't very high (compared to the circuit impedance) then the volt meter will behave like an additional resistor that loads up the circuit and messes with the voltages. Note that it's not just the displayed voltage, it's the actual voltage that's affected. In some cases a circuit might work fine until you attach a voltmeter and then it stops working! Same applies to scope probes.

Pretend the voltmeter in the problem is an ideal meter in parallel with an unknown resistor. The resistor represents the input impedance of the real world meter.

 

[CWatters]

For info...

Modern digital multimeters typically use an input amplifier based on a FET transistor or similar to ensure the input impedance is very high. I have one with an input impedance of 10MOhms.

When my father started out an engineer such things didn't exist. Voltmeters were just moving coils of wire and they might have an imput impedance of just 1,000 ohms. Every time you tried to measure a voltage you had to correct for the loading effect of the meter exactly as per this problem.

See also..

http://en.wikipedia.org/wiki/Multimeter#Analog_multimeters

 

[swaha]

okay thanks.

 

[Nickie]

I would approach this problem by first checking the accuracy and calibration of the voltmeter being used. It is possible that the voltmeter is faulty, leading to incorrect readings. I would also check for any loose connections or faulty wiring in the circuit that could be affecting the voltage readings.

Assuming that the voltmeter is functioning properly, we can use Ohm's law and Kirchoff's laws to solve for the unknowns. From the given information, we know that the voltage across the battery is 9 volts and the voltage across Ra is 3 volts. Using Ohm's law, we can calculate the value of Ra as 1.5M (9V/6mA). This suggests that the given value of Ra (2M) may be incorrect.

Next, we can apply Kirchoff's voltage law to the series circuit, which states that the sum of the voltages across each component in a closed loop must equal the applied voltage. In this case, we have three components (battery, Ra, and Rb) in series, so the sum of their voltages must equal 9 volts. We already know that the voltage across Ra is 3 volts, so the voltage across Rb must be 1.5 volts (9V-3V=6V).

Using Ohm's law, we can now calculate the value of Rb as 0.3M (1.5V/5mA). This means that the given value of Rb (4.5V) is also incorrect.

To solve for the internal resistance of the voltmeter, we can use Kirchoff's current law, which states that the sum of the currents entering and leaving a junction must be zero. In this case, the current entering the junction is 6mA (from the battery), and the current leaving the junction is 5mA (through Rb). This means that the remaining 1mA must be going through the voltmeter.

Using Ohm's law, we can calculate the internal resistance of the voltmeter as 9M (9V/1mA). This is a relatively high value, which may suggest that the voltmeter is not sensitive enough to accurately measure the small voltage across Rb.

In conclusion, the given values for Ra, Rb, and the voltmeter's internal resistance are incorrect. By using Ohm's law and Kirchoff's laws