1. Feb 7, 2012

### Stefenng

Hi there, this is one of the question of my assignment.

A voltmeter with a sensitivity of 1000 Ω/V has a full scale deflection of 150V. When connected across an unknown resistor in series with a milliammeter, it reads 100V. If the milliammeter reads 50mA, calculate the following:

a. The apparent resistance of the unknown resistor.
b. The actual resistance of the unknown resistor.

My attempt.

a. The apparent resistance of the unknown resistor. (2 marks)

V=IRa

100=50 X 10-3 ×Ra

Ra=100/(50 ×10-3 )

=2000Ω

b. The actual resistance of the unknown resistor. (4 marks)

〖Resistance at 100V range,R〗v=100×1000

=100kΩ

V=IR
V=I((Rv Rb)/(Rv+Rb ))

100=50 ×10-3 ((100 ×103×Rb)/(100 ×103+Rb ))

100/(50 ×10-3 )=(100 ×103 Rb)/(100 ×103+Rb )

200 ×106+2000Rb=100×103 Rb

200 ×106=98 ×103Rb

Rb=2.041kΩ

is this the correct way to solve the problem?

2. Feb 7, 2012

### Staff: Mentor

Meter is on the 150V FSD range.

3. Feb 7, 2012

### Staff: Mentor

It is most likely what your teacher intended.

That said, I would give your teacher only half marks for his question (assuming you have quoted the assignment verbatim).
My interpretation of that description is of a milliammeter in series with a resistor, and the voltmeter connected across that combination of meter + resistor such that the voltmeter measures the sum of two voltages VA + VR. This does not make for such an interesting question, but were I marking the assignment I would have to give full marks for that interpretation.

It's a case of answering the question that your teacher asked, versus answering the question that you teacher meant to ask. Well-crafted assignments should not be a mind-reading exercise.

4. Feb 10, 2012

### Stefenng

My interpretation of that description is of a milliammeter in series with a resistor, and the voltmeter connected across that combination of meter + resistor such that the voltmeter measures the sum of two voltages VA + VR. [/QUOTE]

I don't really get what you mean here. Did u mean that I should include the internal resistance of the milliammeter?

5. Feb 11, 2012

### Staff: Mentor

If it were known, since that resistance contributes to the voltage that is measured.

6. Feb 12, 2012

### Redbelly98

Staff Emeritus
Based on the given information, it seems that the intended configuration is: the voltmeter is in parallel with the unknown resistor, and the ammeter is in series with that parallel combination.

This way we know the actual voltage, but not the actual current, for the resistor; we are really measuring the sum of the resistor and voltmeter currents. But we can figure out what the voltmeter current is and calculate the actual resistor current.

The way the problem is worded -- "an unknown resistor in series with a milliammeter" -- means we know the actual current through the resistor, but not the actual voltage, because the voltmeter is measuring the sum of the resistor and milliammeter voltages. We would have to know the milliammeter's resistance (and then calculate its voltage) in order to solve the problem. Because we do not know its resistance, this shouldn't be what your teacher intended.

If your teacher goes over this problem in class, make sure that he/she draws a circuit showing the intended arrangement of resistor, voltmeter, and milliammeter.