Verified answer

Please refer to the attachment....

Answer:

Here, r

1

=5Ω,r

2

=10Ω,Δr

1

=0.2Ω and Δr

2

=0.1Ω

The equivalent resistance, r

p

=

r

1

+r

2

r

1

r

2

=

5+10

5×10

=3.3Ω

r

p

Δr

p

=

r

1

Δr

1

+

r

2

Δr

2

+

r

1

+r

2

Δ(r

1

+r

2

)

=

5

0.2

+

10

0.1

+

5+10

0.2+0.1

=0.07

Thus, The value of equivalent resistance with limits of % error =r

p

±Δr

p

=3.3Ω±7%

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SIMILAR QUESTIONS

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Find the percentage error in specific resistance given by ρ=

l

πr

2

R

where r is the radius having value (0.2±0.02) cm, R is the resistance of (60±2)Ω and l is length of (150±0.1) cm.

Medium

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>

Two resistances are expressed as R

1

(4± 0.5 %)Ω and R

2

(12± 0.5 %)Ω. The net resistance when they are connected in series with percentage error is:

(In series R=R

1

+R

2

)

Medium

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>

In an experiment, the values of two resistances were measured as R

1

=(5.00±.2)Ω and R

2

=(10.00±.1)Ω. The percentage error in the combined resistance in parallel is :

Medium

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>

In an electrical circuit two resistances R

1

=(4±0.5) and R

2

=(12±0.5) are connected in parallel. What will be the equivalent resistance with percentage error?

( Given the effective resistance

R

1

=

R

1

1

+

R

2

1

)

Hard

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>

The resistance of metal is given by V=IR. The voltage in the resistance is V=(80.5±0.1)V and current in the resistance is I=(20.2±0.2)A, the value of resistance with its percentage error is :

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