Question

2 Resistors of resistance, R = (100 +- 3) Ohm, R2 = (200+-4)Ohm are connected in Parallel and Series.

Answer 1

According to what I've learnt, in any expression of multiplication or division, the percentage errors of each term are added up to find the equivalent percentage error. That is, if

y=ABC

y=ABCthen% error in y= % error in A + % error in B + % error in C

For the above problem, let Rs denote series combination. Then Rs=300±7 ohm.

Let Rp denote parallel combination.

∴Rp=R1R2/R1+R2 = R1R2/Rs

Ignoring errors, we get Rp=200/3 = 66.67 ohm

%error in R1=3, %error in R2=2, %error in Rs=7/3

Hence, %error in Rp=3+2+7/3=22/3

So, error in Rp will be 22/3% of 200/3, which is approximately 4.89.

Hence, I got Rp=66.67±4.89 ohm.

However, the book used the formula described and proved here and arrived at the answer Rp=66.67±1.8ohm.

Answer 2

The absolute error and % error in the measurement of equivalent resistance are :

Series → 7 Ohm and 2%

Parallel → 5 Ohm and 7%

Further explanation

Electrical circuits can generally be divided into two types , i.e :

1. Series Circuit

In series circuit, the electric current flowing on each resistor is always the same as the total current.

To find the total resistances you can use the following formula:

$\large {\boxed {R_s = R_1 + R_2 + R_3 + ...} }$

2. Parallel Circuit

In parallel circuits, the electrical voltage at each resistor is always the same as the source voltage.

To find the total resistances you can use the following formula:

$\large {\boxed {\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + ...} }$

Let's tackle the problem now !

Given:

$R_1 = 100 \pm 3 ~ Ohm$

$R_2 = 200 \pm 4 ~ Ohm$

Unknown:

Absolute Error = ?

% Error = ?

Solution:

In Series Circuit :

$R_s = R_1 + R_2$

$R_s = ( 100 \pm 3 ) + ( 200 \pm 4 )$

$\large {\boxed {R_s = 300 \pm 7 ~ Ohm} }$

Absolute Error = 7 Ohm

% Error = (7/300) x 100% ≈ 2%

In Parallel Circuit :

$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$

$\frac{1}{R_p} = \frac{R_2}{R_1 ~ R_2} + \frac{R_1}{R_2 ~ R_1}$

$\frac{1}{R_p} = \frac{R_1 + R_2}{R_1 ~ R_2}$

$R_p = \frac{R_1 ~ R_2}{R_1 + R_2}$

$R_p = \frac{100 \times 200}{100 + 200}$

$R_p = \frac{20000}{300}$

$R_p = \frac{200}{3} ~ Ohm$

$R_p \approx 67 ~ Ohm$

To find the % Error , we will use this following formula :

$\frac{\Delta R_p}{R_p} = \frac{\Delta R_1}{R_1} + \frac{\Delta R_2}{R_2} + \frac{\Delta (R_1 +R_2)}{R_1 + R_2}$

$\frac{\Delta R_p}{R_p} = \frac{3}{100} + \frac{4}{200} + \frac{7}{300}$

$\frac{\Delta R_p}{R_p} = \frac{11}{150}$

$\frac{\Delta R_p}{R_p} = \frac{11}{150} \times 100{\%}$

$\frac{\Delta R_p}{R_p} \approx 7{\%}$

$Absolute ~ Error = \Delta R_p$

$Absolute ~ Error = \frac{11}{150} \times \frac{200}{3}$

$Absolute ~ Error = \frac{44}{9}$

$Absolute ~ Error \approx 5 ~ Ohm$

$\large {\boxed {R_p = 67 \pm 5 ~ Ohm} }$

Learn more

• The three resistors : brainly.com/question/9503202
• A series circuit : brainly.com/question/1518810
• Compare and contrast a series and parallel circuit : brainly.com/question/539204

Answer details

Grade: High School

Subject: Physics

Chapter: Current of Electricity

Keywords: Series , Parallel , Measurement , Absolute , Error , Combination , Resistor , Resistance , Ohm