Question

two resistance (5.1±0.003) ohm and (2.6±0.02) ohm are connected in series in a circuit calculate the total resistance of the circuit with error limits?​

Answer 1

Answer:

Explanation:

R =r1+r2

= (3+-0.03+6+-0.06)

=(9+-0.09)

Answer 2

Answer:

Correct option is

B

(2.0±0.1) Ω

Let R is the equivalent resistance for parallel combination and ΔR is the error(in its measurement respective to the error in the values of resistors)

Hence,

R

1

=

R

1

1

+

R

1

R

1

=

R

1

+R

2

R

1

R

2

R=

3+6

3×6

∴R=2Ω

Now,

R

2

ΔR

=

(R

1

)

2

ΔR

1

+

(R

2

)

2

ΔR

2

R

2

ΔR

=

(3)

2

0.1

+

(6)

2

0.5

R

2

ΔR

=

4

0.1

ΔR=0.1

Hence the resistance of the combination is

R=(2.00±0.1)Ω