QJEEM23Phc14e: JEE Main 2023 Physics - Class 11 JEEM23Phc14e Solution: Two resistances are given as R 1 = ( 10 ± 0.5 ) Ω and R 2 = (…

In the problem, we are given two resistances, $R_1$ and $R_2$ , each with a certain measurement error, $\Delta R_1$ and $\Delta R_2$ . These resistances are connected in parallel, and we are asked to find the percentage error in the equivalent resistance of this combination.

The formula for the equivalent resistance $R$ of two resistors $R_1$ and $R_2$ in parallel is:

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

We want to find the percentage error in $R$ , which is given by $(\Delta R / R) \times 100\%$ .

In order to find $\Delta R / R$ , we differentiate both sides of the above equation with respect to $R$ , $R_1$ , and $R_2$ . This gives us:

$$\frac{\mathrm{\Delta }R}{R^2}=\frac{\mathrm{\Delta }{R}_1}{R_1^2}+\frac{\mathrm{\Delta }{R}_2}{R_2^2}$$

We can then solve this equation for $\Delta R / R$ :

$$\frac{\mathrm{\Delta }R}{R}=(\frac{\mathrm{\Delta }{R}_1}{R_1^2}+\frac{\mathrm{\Delta }{R}_2}{R_2^2})R$$

Substituting the given values, $R_1 = 10 \, \Omega$ , $R_2 = 15 \, \Omega$ , $\Delta R_1 = \Delta R_2 = 0.5 \, \Omega$ , and $R = R_1R_2/(R_1+R_2) = 6 \, \Omega$ , we get:

$$\frac{\mathrm{\Delta }R}{R}=(\frac{0.5}{100}+\frac{0.5}{225})\times 6=\frac{13}{300}$$

Finally, to convert this to a percentage, we multiply by 100, giving:

$$\frac{\mathrm{\Delta }R}{R}\times 100=\frac{13}{3}=4.33\mathrm{\%}$$

This tells us that the percentage error in the equivalent resistance of the two resistances in parallel is $4.33\%$ .