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
Answers
Answer:
0.4
Explanation:
In general, the rule for series circuit is that resistances add up in series. Let us say that we have three resistors Ra, Rb and Rc, then
R = Ra + Rb + Rc
Now if they are in parallel circuit, then net resistance reduces.
Its formula is;
1/R = 1/Ra + 1/Rb + 1/Rc
Now as per given information, when the resistors each having an error of -+0.2 are connected in series, then the maximum absolute error would be
=0.2+0.2=0.4
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
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
± (0.2 + 0.2)
= ± 0.4 Ω
R eff = 60 ± 0.4 Ω
maximum absolute error = ± 0.4 Ω
Learn More:
If the measured values of two quantities are A+- and ∆A andB+- ∆B ...
https://brainly.in/question/11689457
14. Mass of a body is (10 +-0.1) kg. Volume is (1 +- 0.1).m. Then ...
https://brainly.in/question/10682117