A resistor of 6 Ω with a tolerance of 10% and another of 4 Ω with a tolerance of 10% are connected in series. The tolerance of combination is about:
Correct answer: C. 10%
- A. 5%
- B. 12%
- C. 10%
- D. 13%
Explanation
the resistances are R1=6Ω and R2=4Ω. Tolerance in the resistances: ΔR1=.6Ω ΔR2 =.4Ω The total resistance of the combination Req=R1+R2=6+4=10Ω Total tolerance in the equivalent resistance:ΔReq=ΔR1+ΔR2=.6+.4=1Ω Hence percentage tolerance of combination = 1Ω/10Ω × 100= 10 %
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