Solveeit Logo

Question

Question: Two resistors of resistances \(R_{1} = (300 \pm 3)\Omega\)and \(R_{2} = (500 \pm 4)\Omega\)are conne...

Two resistors of resistances R1=(300±3)ΩR_{1} = (300 \pm 3)\Omegaand R2=(500±4)ΩR_{2} = (500 \pm 4)\Omegaare connected in series. The equivalent resistance of the series combination is

A

(800±1)Ω(800 \pm 1)\Omega

B

(800±7)Ω(800 \pm 7)\Omega

C

(200±7)Ω(200 \pm 7)\Omega

D

(200±1)Ω(200 \pm 1)\Omega

Answer

(800±7)Ω(800 \pm 7)\Omega

Explanation

Solution

The equivalent resistances of series combinations is Rs=R1+R2=300Ω+500Ω=800ΩR_{s} = R_{1} + R_{2} = 300\Omega + 500\Omega = 800\OmegaThe error in equivalent resistances is given by

ΔR=(ΔR1+ΔR2)=(3+4)Ω=7Ω\Delta R = (\Delta R_{1} + \Delta R_{2}) = (3 + 4)\Omega = 7\Omega

Hence, the equivalent resistances along with error is (800±7)Ω.(800 \pm 7)\Omega.