Question 97: Units , Dimensions and Error Analysis

Question

Two resistances are given as $$\mathrm{R}_{1}=(10 \pm 0.5) \Omega$$ and $$\mathrm{R}_{2}=(15 \pm 0.5) \Omega$$. The percentage error in the measurement of equivalent resistance when they are connected in parallel is -

Accepted Answer

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 $(\Delta R / R) 	imes 100\%$.
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:
$$\frac{\Delta R}{R^2} = \frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}$$
We can then solve this equation for $\Delta R / R$:
$$\frac{\Delta R}{R} = \left(\frac{\Delta R_1}{R_1^2} + \frac{\Delta R_2}{R_2^2}ight)R$$
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:
$$\frac{\Delta R}{R} = \left(\frac{0.5}{100} + \frac{0.5}{225}ight) 	imes 6 = \frac{13}{300}$$
Finally, to convert this to a percentage, we multiply by 100, giving:
$$\frac{\Delta R}{R} 	imes 100 = \frac{13}{3} = 4.33 \%$$
This tells us that the percentage error in the equivalent resistance of the two resistances in parallel is $4.33\%$.

Answer

2.33

Answer

5.33

Answer

4.33

Answer

6.33

Units , Dimensions and Error Analysis

Two resistances are given as $\mathrm{R}_{1}=(10 \pm 0.5) \Omega and \mathrm{R}_{2}=(15 \pm 0.5) \Omega$. The percentage error in the measurement of equivalent resistance when they are connected in parallel is -

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