Answer to Question #266311 in Electric Circuits for Nishat

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

$R_s=R_1+R_2$

Now errors

∆R_s =(∆R₁+∆R₂ )

$∆R_s=\pm({0.05}+{0.05})=\pm0.10$

$R'_s=R_1+R_2=22+48=70\Omega$

Now

$R_{eq}=70\pm0.10$

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

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

$\frac{∆R_{eq}}{R_{eq}}=\pm(\frac{0.05}{22}+\frac{0.05}{47})$

$$\frac{∆R_{eq}}{R_{eq}}=\pm(0.00227+0.00106)=\pm(0.00303)$$

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

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

$$∆R_{eq}=14.98\times0.00303=0.04988\Omega$$

$R_{eq}=14.98\pm0.04988\Omega$