A voltmeter having least count 0.1 V and an ammeter having least count 0.2 A are used to measure the potentail difference across the ends of a wire and current flowing through the wire respectively. If the reading of voltmeter is 4.4 V and reading of ammeter is 2.2 A, then find (i) the resistance of wire with maximum permissible error and (ii) maximum percentage error.

A voltmeter having least count 0.1 V and an ammeter having least count

1.	A voltmeter having least count 0.1 V and an ammeter having least count 0.2 A are used to measure the potentail difference across the ends of a wire and current flowing through the wire respectively. If the reading of voltmeter is 4.4 V and reading of ammeter is 2.2 A, then find (i) the resistance of wire with maximum permissible error and (ii) maximum percentage error.
Answer» Here, `V = 4.4 V, I = 2.2 A, Delta V = 0.1 V, DeltaI = 0.2 A` `R = (V)/(I) = (4.4)/(2.2) = 2 ohm` Maximum permissible error `(DeltaR)/(R ) = (DeltaV)/(V) +(DeltaI)/(I) = (0.1)/(4.4) +(0.2)/(2.2) = 0.023 + 0.091 = 0.114` `DeltaR= 0.114xxR = 0.114xx2 = 0.228` `DeltaR = 0.2 Omega`(rounding off to one decimal place) :. Resistance of wire with max. permissible error, `R = (2.0 +- 0.2)` ohm Max. percentage error `= (DeltaR)/(R )xx100 = 0.114xx100= 11.4%`