Question

72
PHYSICS->
6. Two resistances R1 = (1002) 2 and R2 = (2003) 12
are connected in series. Find the value of equivalent
resistance (R = R1 + R2) along with permissible
absolute error and percentage error.​

Answer 1

Answer:

R1 + R2 =3005.

Explanation:

A.mean=r1+r2÷2=1502.5

∆a1=à1-A.mean

=1002-1502.5

=-500.5

∆a2=à2-A.mean

=2003-1502.5

=500.5

absolute error =|a1|+|a2|÷2

=500.5+500.5÷2

=500.5

then easily u can find percentage error

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