Question

Resistance of 2000 ohm/v meter which is used to measure the resistance on 200 v scale

Answer 1

I think you need to rephrase your question. As is you are asking for the properties of a voltmeter when that voltmeter measures the resistance of another voltmeter set to a range sufficient to measure at a 200 Volt scale. Voltmeters don't measure resistance. Ohm meters measures resistance.

I think what you are really asking is if you had a voltmeter that has a input resistance such that 1 Volt input would cause 0.5 mA to flow into the meter and produce a reading of 1 Volt. If the range is set to 200 Volts what is the input resistance?

Let's assume that a 0.5 mA reading was 1 Volt so on the same scale a 2 Volt input would result in 1 mA flowing. R = V/I or 2000 Ohms. So for a 200 Volt input we want the same current to flow so the input resistance would have to be 200/ l mA or 200,000 Ohms.

Here I'm assuming a DMM with ranges of 2, 20, 200 Volts full scale.

If you assume a set of full scale deflections of 1, 10, 100 and 1000 Volts then you would have to place the meter on the 1000 Volt scale the input resistance would be 2,000,000 Ohms.

Answer 2

Answer:

The resistance of a meter in 200V scale is 400kΩ.

Explanation:

According to the Ohm's law, the voltage V applied across a conductor is directly proportional to the current I flowing through it.

$V\propto I\implies V=IR$

where R is the proportionality constant, called resistance.

Given a meter that is used to measure the resistance, measures the resistance 2000 Ω/V.

Then the current through the meter is given by rewriting the equation.

$I=\dfrac{V}{R}$

$\implies I=\dfrac{1}{2000}=0.0005=0.5\times 10^{-3} A$

So, to measure the resistance on a 200-volt scale, $0.5\times 10^{-3} A$ of current should flow through the circuit of the meter.

Applying the Ohm's law again,

$R=\dfrac{200}{0.5\times 10^{-3} }=400\times 10^{3}=400k\Omega$

Therefore, the resistance of a meter in 200V scale is 400kΩ.

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