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Question

Physics Question on Current electricity

Two resistors of resistances R1=(300±3)ΩR_{1}=\left(300 \pm3\right)\,\Omega and R2=(500±4)ΩR_{2}=\left(500 \pm4\right)\,\Omega are connected in series. The equivalent resistance of the series combination is

A

(800±1)Ω\left(800 \pm 1\right)\,\Omega

B

(800±7)Ω\left(800 \pm 7\right)\,\Omega

C

(200±7)Ω\left(200 \pm 7\right)\,\Omega

D

(200±1)Ω\left(200 \pm 1\right)\,\Omega

Answer

(800±7)Ω\left(800 \pm 7\right)\,\Omega

Explanation

Solution

The equivalent resistance of series combination is Rs=R1+R2=300Ω+500Ω=800ΩR_{s}=R_{1}+R_{2}=300\,\Omega+500\,\Omega=800\,\Omega The error in equivalent resistance is given by ΔR=(ΔR1+ΔR2)=(3+4)Ω=7Ω\Delta R=\left(\Delta R_{1}+\Delta R_{2}\right)=\left(3+4\right)\,\Omega=7\,\Omega Hence, the equivalent resistance along with error is (800±7)Ω\left(800\pm 7\right)\,\Omega.