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Question

Physics Question on Current electricity

A resistor of 6kΩ6 \, k \Omega with tolerance 10%10\% and another resistance of 4kΩ4\, k \Omega with tolerance 10%10\% are connected in series. The tolerance of the combination is about

A

5%

B

10%

C

12%

D

15%

Answer

10%

Explanation

Solution

In series combination equivalent resistance, R=R1+R2R = R_1 + R_2 =6+4=10kΩ = 6 + 4 = 10 \, k \, \Omega Error in combination, ΔR=ΔR1+ΔR2\Delta R = \Delta R_1 + \Delta R_2 =10100×6+10100×4= \frac{10}{100 } \times 6 + \frac{10}{100} \times 4 =0.6+0.4=1= 0.6 + 0.4=1 ΔRR=110 \frac{\Delta R}{R} = \frac{1}{10} =10%= 10\%