Two resistors of resistances R1 = 100 ±3 ohm and R2 = 200 ± 4 ohm are connected (a) in series, (b) in parallel. Find the equivalent resistance of the (a) series combination, (b) parallel combination
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Answer:
Here, R
1
(100±3)Ω;R
2
=(200±4)Ω The equivalent resistance in parallel combination is
R
1
1
=
R
1
1
+
R
2
1
,
R
p
1
=
100
1
+
200
1
=
200
3
,R
p
=
3
200
=66.7Ω
The error in equivalent resistance is given by
R
p
2
ΔR
p
=
R
1
2
ΔR
1
+
R
2
2
ΔR
2
;ΔR
p
=ΔR
1
(
R
1
R
p
)
2
+ΔR
2
(
R
2
R
p
)
2
=3(
100
66.7
)
2
+4(
200
66.7
)
2
=1.8Ω
Hence, the equivalent resistance along with error in parallel combination is (66.7 ± 1.8)Ω.
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