The least count of voltmeter is 0.2 V, and least count of ammeter is 1 mA. If the potential difference is 2 V, and the current is 10 mA, what is the percentage error in measurement of resistance?

Step-by-step solution

Resistance is given by Ohm's law: R = V I The maximum percentage error in resistance is the sum of the percentage errors in voltage and current: R R 100 = ( V V + I I ) 100 Given values are: V = 2 V , V = 0.2 V I = 10 mA , I = 1 mA Substituting these values: R R 100 = ( 0.2 2 + 1 10 ) 100 R R 100 = (0.1 + 0.1) 100 R R 100 = 0.2 100 = 20 % Answer: 20 %