resistance of two resistors is measured as( 20+- 0.2 ) ohm and (40+-0.2) ohm . the maximum absolute error in their series combination will be
a.0.2 ohm
b. 0.1 ohm
c. 0.6 ohm
d.0.4 ohm
Answers
Answer:
When the two resistors are connected in parallel, the equivalent resistance will be
R
P
=
R
1
+R
2
R
1
R
2
Differentiating both sides ,
R
P
ΔR
P
=
R
1
ΔR
1
+
R
2
ΔR
2
+
R
1
+R
2
Δ(R
1
+R
2
)
R
P
ΔR
P
=
6
0.3
+
10
0.2
+
16
0.3+0.2
=0.05+0.02+0.03125=0.10125
% error =
R
P
ΔR
P
×100=0.10125×100=10.125%
Explanation:
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Given : resistance of two resistors is measured as( 20+- 0.2 ) ohm and (40+-0.2) ohm .
To Find : maximum absolute error in their series combination
a.0.2 ohm
b. 0.1 ohm
c. 0.6 ohm
d.0.4 ohm
Solution:
Resistance in series combination gets added
R₁ = 20 ± 0.2 Ω
R₂ = 40 ± 0.2 Ω
Reff = R₁ + R₂
= 20 + 40
= 60 Ω
Error gets added during addition or subtraction.
± (0.2 + 0.2)
= ± 0.4 Ω
R eff = 60 ± 0.4 Ω
maximum absolute error = ± 0.4 Ω
option d is correct
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