Question

Two resistance R1 = 100±3 ohm and R2 = 2001 ohm are connected in series find the equivalent resistance? please provide with explanation​

Answer 1

Given:

$R_{1}=100 \pm 3</p><p>R_{2}=200 \pm 4$

Solution:

In the series combination, the resistance are added.

$R=R_{1}+R_{2}=100+200=300</p><p>d R=d R_{1}+d R_{2}=3+4=7$

Here, the error is added because is in addition.

R=300 \pm 7

In parallel combination, the reciprocal of the resistances are added.

$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$

$R=\frac{R_{1} \cdot R_{2}}{R_{1}+R_{2}}$

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

$\frac{1}{R}=\frac{1}{R_{1}}+\frac{1}{R_{2}}$

$R=\frac{R_{1} \cdot R_{2}}{R_{1}+R_{2}}$

Physics[/tex]

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

$\therefore R=\frac{20000}{300}=66.67 \ \Omega$

After we differentiate the parallel combination formula we get,

$\frac{d R}{R^{2}}=\frac{d R_{1}}{R_{1}^{2}}+\frac{d R_{2}}{R_{2}^{2}}$

We need to keep the error on one side and bring the rest on another side.

$d R=R^{2}(\frac{d R_{1}}{R_{1}^{2}}+\frac{d R_{2}}{R_{2}^{2}})$

$\Rightarrow d R=4444.9(\frac{3}{10000}+\frac{4}{40000})$

d R=1.789 \ \Omega

On rounding, we get,

$\Rightarrow d R=1.8 \ \Omega$

$R=(66.67 \pm 1.8) \ \Omega$