Ideally the resistance of a Voltmeter is infinite so that so that voltmeter does not alter circuit current. A low resistance voltmeter may give correct reading when measuring voltage in low resistance circuit but the Voltmeter produces unreliable and erroneous reading when connected in high resistance circuit. This is because, as the resistance of voltmeter is less when compared to the circuit resistance, this will act as a shunt path for the current and therefore the voltage drop across the resistor where we want to measure the voltage will be less. Because of this the reading of voltmeter will not be the actual voltage drop rather it will be lower than actually existed before the connection of the voltmeter. *This effect is known as Loading Effect.*

It is desired to measure the voltage cross the 50 kOhm resistor using two voltmeter A and B. Voltmeter A has a sensitivity of 1000 ohm/V while for voltmeter B is 20,000 ohm/V. The range of both voltmeters are 0-50 V.

First of all we will calculate the voltage drop across the 50 kOhm resistor.

Voltage drop across 50 kOhm resistor = 50×150/(100+50)

= 50 V = True value

Means before the connection of voltmeter the voltage drop across the 50 kOhm resistor is 50 V.

**Case1: When voltmeter A is used.**

Sensitivity of Voltmeter A = 1000 ohm/V

Therefore,

Resistance of Voltmeter R_{VA} = SensitivityxVoltage

= 1000×50 = 50 kOhm

Now when this voltmeter A is connected across the resistor of 50 kOhm, the equivalent resistance of the parallel combination,

Req = 50×50 / (50+50)

= 25 Kohm

Hence, voltage across the Req = 25×150 / (100+25)

= 30 V

Therefore we see that voltmeter A reads much less than the true value. It should be noted that for getting good accuracy, the resistance of voltmeter >>> circuit resistance but here in this case both are equal.

**Case2**: **When voltmeter B is used.**

Sensitivity of Voltmeter A = 20,000 ohm/V

Therefore,

Resistance of Voltmeter R_{VA}= SensitivityxVoltage

= 20,000×50 = 1000 kOhm

Mind here that the resistance of voltmeter >>> circuit resistance i.e 1000 kOhm >>>>50 kOhm so definitely we are going to get voltage reading close to true value i.e 50 V.

Now when this voltmeter A is connected across the resistor of 50 kOhm, the equivalent resistance of the parallel combination,

Req = 50×1000 / (50+1000)

= 47.6 Kohm

Hence, voltage across the Req = 47.6×150 / (100+47.6)

= 48.37 V

See how close to the true reading we get. So in second case loading effect is not prominent as the voltmeter resistance is much higher than the circuit resistance but in first case loading effect id dominating as the voltmeter resistance is comparable to circuit resistance.

*It should also be noted that meter with higher sensitivity gives reliable result. This is particularly true where voltage measurement is done in higher resistance circuit.*

Thank you!

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Awesome explanation

Thnx a lot sir

Well explained

But eg is not clear

eg is for example.