Question

The equivalent resistance of two resistors R1 and R2 is given by 𝑅𝑒𝑞 = 𝑅1 + 𝑅2 and 𝑅𝑒𝑞 = 𝑅1𝑅2 𝑅1+𝑅2 when they are connected in series and parallel respectively. If 𝑅1 = 22.0 Ω ± 5% 𝑎𝑛𝑑 𝑅2 = 47.0 ± 5% , find the maximum error in the calculation of their equivalent resistance when they are connected in series and parallel.

Answer 1

$R'=R_1+R_2$ Series combination

Parallel combination

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}$

Case -1

Error in series combination

$∆R=\pm(∆R_1+∆R_2)$

$∆R=\pm(0.05+0.05)=\pm0.10\Omega$

Now

$R_{eq}=R'\pm∆R$

$R'=22+47=69\Omega$

$R_{eq}=(69\pm0.10)\Omega$

Parallel combination

$\frac{∆R}{R_{eq}}=\frac{∆R_1}{R_1}+\frac{∆R_2}{R_2}$

$\frac{∆R}{R_{eq}}=\frac{0.05}{22}+\frac{0.05}{47}$

$$\frac{∆R}{R_{eq}}=(0.002272+0.00106)=0.00333\Omega$$

$R_{eq}=\frac{R_1R_2}{R_1+R_2}$

$R_{eq}=\frac{22\times47}{22+47}=14.98\Omega$

$∆R=0.0499\Omega$

$R_{eq}=(14.98\pm0.0499)\Omega$