if two resistance of R1=(2.0+-0.1) ohm and R2=(12.3+-0.2)ohm are put in (i) parllel and in (ii) serier, find the error in equvalent equation
Answers
Explanation:
Two resistors, R
1
andR
2
are connected in parallel.
We know,
R
e
1
=
R
1
1
+
R
2
1
Then R
e
=
R
1
+R
2
R
1
R
2
=
50+100
50×100
=
150
5000
=
3
100
Now, Parallel Connection error:
=
(R
1
+R
2
)
2
R
1
2
(dB)+R
2
2
(dA)
=
150
2
50
2
(3)+100
2
(2)
=
150×150
7500+20000
=
9
11
Relative Error =
(100/3)
(11/9)
=0.03666
Solution :-
As per the given data ,
- R 1 = ( 2 ± 0.1 )
- R 2 = ( 12.3 ± 0.2 )
Series Combination :-
Equivalent resistance for two resistors connected in series is given by the formula ,
➽ Rs = R 1 + R2
➽ Rs = 2 + 12.3
➽ Rs = 14.3
➽ ΔRs = ± ( ΔR1 + ΔR2 )
➽ ΔRs = ± ( 0.1 + 0.2 )
➽ ΔRs = ± 0.3
The error in equivalent resistance in series combination is ( 14.3 ± 0.3 )
Parallel Combination
Equivalent resistance for two resistors connected in series is given by the formula ,
➽ Rp = R1 R2 / R 1 + R 2
➽ Rp = 2 x 12.3 / 14.3
➽ Rp = 24.6 / 14.3
➽ Rp = 1.72
Now ,
➽ ΔRp /Rp² = ±(ΔR1 /R1² + ΔR2 /R2²)
➽ ΔRp /Rp² =±( 0.1 /4+ 0.2/151.29)
➽ ΔRp /Rp² = ±(0.025 + 0.0013 )
➽ ΔRp =±( 0.0263 x Rp²)
➽ ΔRp = ±(0.0263 x 2.95 )
➽ ΔRp = ±0.078
The error in equivalent resistance in series combination is ( 1.73 ± 0.078 )