The problem asks for the maximum percentage error in the equivalent resistance ($R_{eq}$) of two parallel resistors, $R_1$ and $R_2$, each having a maximum percentage error of $5\%$.

1. Define Nominal Values and Errors

• Nominal Resistances: $R_{1n} = 100 \ \Omega$, $R_{2n} = 900 \ \Omega$.
• Maximum Fractional Error: $\epsilon_1 = \epsilon_2 = 0.05$.

The equivalent resistance formula is $R_{eq} = \frac{R_1 R_2}{R_1 + R_2}$.

2. Method 1: Calculation using the Differential Error Propagation

For parallel resistors, the sensitivity of $R_{eq}$ to small changes in $R_1$ and $R_2$ is often simplified by considering the fractional errors ($\epsilon_{eq} = \frac{\Delta R_{eq}}{R_{eq}}$). The maximum fractional error is the weighted sum of the input fractional errors:

$$\epsilon_{eq, \max} = \left| \frac{\partial R_{eq}}{\partial R_1} \frac{R_1}{R_{eq}} \right| \epsilon_1 + \left| \frac{\partial R_{eq}}{\partial R_2} \frac{R_2}{R_{eq}} \right| \epsilon_2$$

The weighting coefficients (sensitivity factors) are:

$$W_1 = \frac{\partial R_{eq}}{\partial R_1} \frac{R_1}{R_{eq}} = \left(\frac{R_2}{R_1 + R_2}\right)^2 \cdot \frac{R_1}{R_{eq}}$$

However, a much simpler form exists derived from the reciprocal relationship $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2}$.

If the fractional errors are equal ($\epsilon_1 = \epsilon_2 = \epsilon$), the maximum fractional error $\epsilon_{eq, \max}$ is simply equal to the input fractional error $\epsilon$:

$$\epsilon_{eq, \max} = \epsilon_1 \cdot \left( \frac{R_2}{R_1 + R_2} + \frac{R_1}{R_1 + R_2} \right) = \epsilon_1 \cdot (1) = \epsilon_1$$

Substitute the nominal values and errors:

$$\epsilon_{eq, \max} = 0.05$$ $$\text{Max \% Error} = 0.05 \times 100\% = 5\%$$

3. Method 2: Calculation using Worst-Case Values

The maximum equivalent resistance ($R_{eq, \max}$) occurs when both $R_1$ and $R_2$ are maximized:

• $R_{1, \max} = 100 \cdot 1.05 = 105 \ \Omega$
• $R_{2, \max} = 900 \cdot 1.05 = 945 \ \Omega$

Nominal equivalent resistance $R_{eq, n} = 90 \ \Omega$.

$$R_{eq, \max} = \frac{105 \times 945}{105 + 945} = \frac{99225}{1050} = 94.5 \ \Omega$$

Maximum Percentage Error:

$$\text{Max \% Error} = \frac{R_{eq, \max} - R_{eq, n}}{R_{eq, n}} \times 100\%$$ $$\text{Max \% Error} = \frac{94.5 - 90}{90} \times 100\% = \frac{4.5}{90} \times 100\% = 5\%$$

4. Final Result

The maximum percentage error is $5\%$. Rounding off to the nearest integer value gives 5.