If two resistors of resistances R1=(4±0.5)Ω R2=(16±0.5)Ω are connected in (i)series (ii) parallel; find the equivalent resistance in each case with limits of percentage error

R1=(4+-0.5)​ohm
R2=1.6+-0.5ohm

for series combination:
Rs=R1+R2
=4+16=20ohm
error=0.5+0.5=+-1ohm
% error=(+-1/20)x100=5%
hence Rs=20+-1ohm=20 ohm+- 5%
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What in parallel....???
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Can ask doubt

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Parallel nhi smjh aaya
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Answer

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Can somebody explain the formula for finding error in parallel case as Bhavani.a has used ?? Her answer is correct but i am not able to understand the formula used in parallel case while finding error ?
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student-nameRanapuran asked in Physics If two resistors of resistances R1=(4±0.5)Ω R2=(16±0.5)Ω are connected in (i)series (ii) parallel; find the equivalent resistance in each case with limits of percentage error
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