Understanding Cylindrical Wire Resistance in Series and Parallel

This problem asks us to find the ratio between the effective resistance of two cylindrical wires when connected in series and when connected in parallel. Both wires are made from the same material and have the same length, but they differ in their radii, denoted as $r_1$ and $r_2$.

Calculating Individual Wire Resistances

The resistance ($R$) of a conductor is determined by its material properties (resistivity, $\rho$), its length ($L$), and its cross-sectional area ($A$). The fundamental formula for resistance is:

$$ R = \rho \frac{L}{A} $$

For a cylindrical wire, the cross-sectional area is a circle with radius $r$. Therefore, its area is $A = \pi r^2$. Substituting this into the resistance formula, we get:

$$ R = \rho \frac{L}{\pi r^2} $$

Since both wires are made of the same material (meaning they have the same resistivity, $\rho$) and have the same length ($L$), we can simplify our expressions by defining a constant factor $k = \frac{\rho L}{\pi}$. This constant $k$ will be the same for both wires, making the calculations easier.

The resistance of the first wire (wire A) with radius $r_1$ is:

$$ R_1 = \frac{k}{r_1^2} $$

The resistance of the second wire (wire B) with radius $r_2$ is:

$$ R_2 = \frac{k}{r_2^2} $$

We can summarize these individual resistances in the following table:

Effective Resistance in Series Combination

When electrical components are connected in series, their resistances add up directly to determine the total effective resistance of the combination. The formula for the effective resistance ($R_{series}$) of two resistors $R_1$ and $R_2$ connected in series is:

$$ R_{series} = R_1 + R_2 $$

Now, we substitute the expressions we found for $R_1$ and $R_2$ into this formula:

$$ R_{series} = \frac{k}{r_1^2} + \frac{k}{r_2^2} $$

To combine these fractions, we find a common denominator, which is $r_1^2 r_2^2$:

$$ R_{series} = k \left( \frac{1}{r_1^2} + \frac{1}{r_2^2} \right) = k \left( \frac{r_2^2 + r_1^2}{r_1^2 r_2^2} \right) $$

Effective Resistance in Parallel Combination

When electrical components are connected in parallel, the reciprocal of the effective resistance is equal to the sum of the reciprocals of the individual resistances. The formula for the effective resistance ($R_{parallel}$) of two resistors $R_1$ and $R_2$ connected in parallel is:

$$ \frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} $$

Substituting the expressions for $R_1$ and $R_2$:

$$ \frac{1}{R_{parallel}} = \frac{1}{\frac{k}{r_1^2}} + \frac{1}{\frac{k}{r_2^2}} $$

Simplifying the terms on the right side:

$$ \frac{1}{R_{parallel}} = \frac{r_1^2}{k} + \frac{r_2^2}{k} $$

Combining the terms on the right side gives:

$$ \frac{1}{R_{parallel}} = \frac{r_1^2 + r_2^2}{k} $$

To find $R_{parallel}$, we take the reciprocal of both sides:

$$ R_{parallel} = \frac{k}{r_1^2 + r_2^2} $$

Calculating the Ratio of Resistances

The question requires us to find the ratio of the effective resistance in the series combination to the effective resistance in the parallel combination. This ratio is expressed as $\frac{R_{series}}{R_{parallel}}$.

We have the expressions for both:

$$ R_{series} = k \left( \frac{r_1^2 + r_2^2}{r_1^2 r_2^2} \right) $$

$$ R_{parallel} = \frac{k}{r_1^2 + r_2^2} $$

Now, we compute the required ratio by dividing $R_{series}$ by $R_{parallel}$:

$$ \frac{R_{series}}{R_{parallel}} = \frac{k \left( \frac{r_1^2 + r_2^2}{r_1^2 r_2^2} \right)}{\frac{k}{r_1^2 + r_2^2}} $$

Notice that the constant $k$ cancels out from the numerator and the denominator:

$$ \frac{R_{series}}{R_{parallel}} = \left( \frac{r_1^2 + r_2^2}{r_1^2 r_2^2} \right) \times \left( \frac{r_1^2 + r_2^2}{1} \right) $$

Multiplying these terms together, we get:

$$ \frac{R_{series}}{R_{parallel}} = \frac{(r_1^2 + r_2^2)^2}{r_1^2 r_2^2} $$

This expression can be more compactly written by taking the square root of the numerator and denominator terms separately before squaring the entire expression:

$$ \frac{R_{series}}{R_{parallel}} = \left( \frac{r_1^2 + r_2^2}{r_1 r_2} \right)^2 $$

This final expression matches option 1 provided in the question.