Question

The two resistances R1=(20+-0.1)Ω and R2=(5+-​​​​​​​0.25)Ω are connected in series in a circuit. Find the percentage error in the measurement of the equivalent resistances of the circuit.

Answer 1

The equivalent resistance is 24. 65 Ohm

Answer 2

Given :

Two resistance , $R_1=(20-0.1)\ \Omega$ and $R_2=(5-0.25)\ \Omega$ .

Both $R_1\ and \ R_2$ are in series .

To Find :

The percentage error in the measurement of the equivalent resistances of the circuit .

Solution :

The equivalent resistance in series is given by :

$R_{eq}=R_1+R_2\\R_{eq}=(20-0.1)+(5-0.25)\ \Omega \\R_{eq}=(25-0.35)\ \Omega$

Now , percentage error is given by :

$\%\ error=\dfrac{|\Delta R|}{R}\times 100\\\\\%\ error=\dfrac{0.35}{25}\times 100\\\\\%\ error=1.4\ \%$

Therefore , the  percentage error in the measurement of the equivalent resistances of the circuit is 1.4 % .

Learn More :

Errors and measurements

https://brainly.in/question/1557100