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Question

${1 \over 4}$

${1 \over 2}$

${3 \over 4}$

1

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Step-by-Step Solution Step 1: Understand the Circuit Configuration The circuit has two resistors: a fixed resistor of resistance $R$ and another resistor $r$. They are arranged such that $r$ is in parallel with a certain portion of the circuit, leading to an equivalent resistance that depends on a parameter $f$. Our end goal is to find the value of $f$ that maximizes the heat generated in resistor $r$. Step 2: Write the Equivalent Resistance The total or equivalent resistance $R_{\text{eq}}$ (as given) can be written as: $$R_{\text{eq}} = R + \frac{r \times R}{r + R}.$$ In terms of $f$ (a given dimensionless factor), this becomes: $$R_{\text{eq}} = \frac{(2f + 1)\,R}{(f + 1)}.$$ Step 3: Compute the Total Current in the Circuit Let the input voltage be $V$. Then the total circuit current $I_{\text{total}}$ is: $$I_{\text{total}} = \frac{V}{R_{\text{eq}}} = \frac{V}{\frac{(2f + 1)\,R}{(f + 1)}} = \frac{V\,(f + 1)}{R\,(2f + 1)}.$$ Step 4: Find the Current through Resistor r By the problem’s derivation, the current through the resistor $r$ (denoted by $I_r$) is a fraction of the total current. From the given steps, it turns out that: $$I_r = \frac{I_{\text{total}}}{f + 1} = \frac{V\,(f + 1)}{R\,(2f + 1)} \times \frac{1}{f + 1} = \frac{V}{R\,(2f + 1)}.$$ Step 5: Express the Heat Generated in Resistor r The power dissipated in (or heat generated by) the resistor $r$ over a given time can be related to $I_r^2\,r$. Writing $H$ for the heat in a suitable time interval, we have: $$H = I_r^2 \, r = \left(\frac{V}{R\,(2f + 1)}\right)^2 \, r.$$ According to the provided expression in the solution, this is rearranged to: $$H = \frac{V^2\,f}{R\,[\,2f + 1\,]^2}.$$ (Here $r = fR$ in the detailed derivation, hence the factor of $f$.) Step 6: Differentiate the Heat Expression w.r.t. f for Maximum To find the maximum heat, we set the derivative of $H$ with respect to $f$ to zero: $$\frac{dH}{df} = 0.$$ Substituting the expression for $H$ and simplifying eventually yields: $$2f + 1 = 4f.$$ Step 7: Solve for the Optimum Value of f From the above equation: $$2f + 1 = 4f \quad\Longrightarrow\quad 2f = 1 \quad\Longrightarrow\quad f = \frac{1}{2}.$$ Final Answer The value of $f$ that maximizes the heat generated in the resistor $r$ is $$\boxed{\frac{1}{2}}.$$

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