PhysicsHard156×since 2002Q8315Two resistances are given as R1=(10±0.5)Ω\mathrm{R}_{1}=(10 \pm 0.5) \OmegaR1=(10±0.5)Ω and R2=(15±0.5)Ω\mathrm{R}_{2}=(15 \pm 0.5) \OmegaR2=(15±0.5)Ω. The percentage error in the measurement of equivalent resistance when they are connected in parallel is -A2.33B5.33C4.33D6.33Check answerSkip