If voltage V = (100 ± 5) V and current I = (10 ± 0.2) A, the percentage error in resistance R is

Given, voltage V=(100±5)V Current I=(10±0.2)AAccording to Ohm’s law, V=IR or R=V/ITaking log of both sides,log⁡R=log⁡V−log⁡IDifferentiating, we getΔRR=ΔVV−ΔIIFor maximum error, ΔRR=ΔVV+ΔlIMultiplying both sides by 100 for taking percentage, we getΔRR×100=ΔVV×100+ΔIl×100Percentage error in resistance R=ΔVV×100+ΔlI×100=5100×100+0.210×100=7%.