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If voltage remains constant in parallel combination then why doesn't the voltmeter with low resistance give correct reading if there is only one resistor?

If there are two resistors then the one on which we are calculating voltage would be less but if there is only one resistor in the circuit and we are not seeing it through the current perspective then please tell whether I am correct or not.

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  • $\begingroup$ See if there are two resistors then the one on which we are calculating voltage would be less but if there is only one resistor in the circuit and we are not seeing it through the current perspective then please tell whether I am correct or not $\endgroup$ – Anju Sabharwal Jan 9 at 16:20
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You would be right if your circuit looked like the first picture (that is, the voltage source is indeed a "pure voltage source" with zero internal resistance).

enter image description here

Any "real" voltage source has some internal impedance which I indicate in the second diagram (the dotted line is "the real voltage source" which consists of an ideal voltage source and a series resistance). The additional current that is drawn through the internal impedance by the presence of a low-impedance volt meter will reduce the voltage reading.

A high impedance volt meter minimizes the disturbance of the circuit and is always preferable.

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Voltage in a parallel combination is same across all device in the combination. If the voltmeter has low resistance it will act like a "load" It will draw current itself leading to a change in current passing through resistance . Let me explain mathematically, In a parallel combination current passing through each component is I1 = V /R1 where I1 is current through 1 component and R1 it's resistance . If for voltmeter R1 is not very high then current will be high leading to a change in current passing through the resistance . Hence , voltmeter with high resistance should be used .

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  • $\begingroup$ The question needs a circuit diagram. A low-impedance Volt meter won't have any effect on the circuit at all if the meter is parallel with a Voltage source, but it can have a significant effect if connected anywhere else. It's not clear from the OP's question where the meter is connected. $\endgroup$ – Solomon Slow Jan 9 at 16:10
  • $\begingroup$ Please someone see my comment and please reply $\endgroup$ – Anju Sabharwal Jan 9 at 16:21

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