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Step-by-Step Solution
Step 1: List the given data
• Resistor 1, $R_{1} = (4 \pm 0.8)\,\Omega$
• Resistor 2, $R_{2} = (4 \pm 0.4)\,\Omega$
Step 2: Write the formula for resistors in parallel
When two resistors $R_{1}$ and $R_{2}$ are connected in parallel, the equivalent resistance $R_{\mathrm{eq}}$ is given by:
$ \frac{1}{R_{\mathrm{eq}}} \;=\; \frac{1}{R_{1}} \;+\; \frac{1}{R_{2}} \,. $
Step 3: Compute the nominal (central) value of the equivalent resistance
Substitute the nominal values $R_{1} = 4\,\Omega$ and $R_{2} = 4\,\Omega$:
$ \frac{1}{R_{\mathrm{eq}}} \;=\; \frac{1}{4} \;+\; \frac{1}{4} \;=\; \frac{1}{4} + \frac{1}{4} \;=\; \frac{2}{4} \;=\; 0.5. $
Thus,
$ R_{\mathrm{eq}} \;=\; \frac{1}{0.5} \;=\; 2\,\Omega. $
Step 4: Apply error-propagation formula
For resistors in parallel, one way to combine their uncertainties uses the relation:
$ \frac{\Delta R_{\mathrm{eq}}}{R_{\mathrm{eq}}^{2}} \;=\; \frac{\Delta R_{1}}{R_{1}^{2}} \;+\; \frac{\Delta R_{2}}{R_{2}^{2}}, $
where $R_{\mathrm{eq}}$ is the equivalent resistance, and $\Delta R_{\mathrm{eq}}$ is the absolute uncertainty in $R_{\mathrm{eq}}$. The terms $\Delta R_{1}$ and $\Delta R_{2}$ are the absolute uncertainties in $R_{1}$ and $R_{2}$, respectively.
Substitute the given values:
$ R_{\mathrm{eq}} = 2, \quad R_{1} = 4, \quad R_{2} = 4, \quad \Delta R_{1} = 0.8, \quad \Delta R_{2} = 0.4. $
So,
$ \frac{\Delta R_{\mathrm{eq}}}{(2)^{2}} \;=\; \frac{0.8}{(4)^{2}} \;+\; \frac{0.4}{(4)^{2}}
\quad\Longrightarrow\quad \frac{\Delta R_{\mathrm{eq}}}{4} \;=\; \frac{0.8}{16} + \frac{0.4}{16}. $
$ \frac{\Delta R_{\mathrm{eq}}}{4} \;=\; 0.05 + 0.025 \;=\; 0.075 \quad\Longrightarrow\quad \Delta R_{\mathrm{eq}} = 0.075 \times 4 = 0.3. $
Step 5: Write the final result for the equivalent resistance
Thus, the equivalent resistance of the parallel combination, including its uncertainty, is:
$ R_{\mathrm{eq}} \;=\; (2 \pm 0.3)\,\Omega. $
