CBSE Class 11-science Answered
a resistor of 6 ohm with a tolerance of 10 % and another of 4 ohm with a tolerance of 10% are connected in series the tolewrance of combination is about?
Asked by samia.rubbabkazmi | 27 Jul, 2019, 07:17: PM
Expert Answer
Let Resistance of resistor-A = ( 6 ± ΔRA ) Ω
Let Resistance of resistor-B = ( 4 ± ΔRB ) Ω
When connected in series, equivalent resistance = [ 10 ±( ΔRA + ΔRB ) ] Ω
tolerence in percentage for the equivalent resistance = [ ( ΔRA + ΔRB )/( RA + RB ) ]× 100 ................(1)
In eqn.(1), RA = 6Ω and RB = 4Ω
If resistor-A has tolerence 10%, then we have, ΔRA / RA = 0.1 or ΔRA = 0.6 Ω
If resistor-B has tolerence 10%, then we have, ΔRB / RB = 0.1 or ΔRB = 0.4 Ω
if we substitute the relevant values in eqn.(1), we get the tolerence of resitors connected in series as,
tolerence in percentage = [ ( 0.6 + 0.4 )/( 6 + 4 ) ]× 100 = 10 %
Answered by Thiyagarajan K | 27 Jul, 2019, 07:57: PM
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