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The end of a battery is 4 volt and it`s internal resistance is 1.5 ohm . It`s potential difference is measured by a voltmeter of resistance 1000ohm. Calculate the percentage error in reading of emf shown by voltmeter.

The end of a battery is 4 volt and it`s internal resistance is 1.5 ohm . It`s potential difference is measured by a voltmeter of resistance 1000ohm. Calculate the percentage error in reading of emf shown by voltmeter.

Grade:12

2 Answers

Vikas TU
14149 Points
3 years ago
The net resistance of the voltmeter is 1.5+1000 = 1001.5 
Current = 4/1001.5 = 0.004 A 
Drop crosswise over 1.5 ohm = 0.004x 1.5 = 0.006 V 
Voltage perusing = 4-0.006 v 
= 3.994V 
Rate mistake = voltage perusing/voltage x 100 
=0.9985 x 100 
= 99.8% 
Ruhi
12 Points
3 years ago
Here emf is 4 volts ,r=1.5ohm ,R =1000ohm Current=e/(R+r),=4/1001.5 =0.00399. The p.d. across the voltmeter is. V=.00399/1000=3.99. Percentage error =e-v/e×100 =4-3.99/4×100 =0.25 percent

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