A copper (Cu) rod of length 25 cm and cross-sectional area 3 mm² is joined with a similar aluminium

Q: Why are the rods considered to be in parallel and not in series?

Because both rods are connected across the exact same two terminals, A and B. The current entering at A splits into two paths, one through each rod, and recombines at B. This is the defining characteristic of a parallel circuit.

Q: Why do we need to convert the units of length and area?

The resistivity is given in $\Omega\text{-m}$ (ohm-meters), which is the standard SI unit. To ensure our final resistance is in ohms, all other measurements must also be in meters. So, cm must be converted to m, and $\text{mm}^2$ to $\text{m}^2$.

Q: How do I convert $\text{mm}^2$ to $\text{m}^2$?

Since $1\ \text{mm} = 10^{-3}\ \text{m}$, squaring both sides gives $1\ \text{mm}^2 = (10^{-3}\ \text{m})^2 = 10^{-6}\ \text{m}^2$.

Q: Is there a faster way to calculate the parallel equivalent without dealing with the $10^{-4}$ powers?

Yes! Since both resistances have a $10^{-4}$ multiplier, you can factor it out. Calculate $(14.167 \times 21.667) / (14.167 + 21.667)$ first, and then multiply the final result by $10^{-4}$.

Q: Why does the final answer $0.8566$ approximate to $0.858$ in the options?

This is likely due to intermediate rounding differences in the problem setter's calculations (e.g., rounding $14.167$ to $14.17$ or similar). Always choose the option closest to your mathematically sound result.

Question

A copper (Cu) rod of length $25	ext{ cm}$ and cross-sectional area $3	ext{ mm}^2$ is joined with a similar aluminium (Al) rod as shown in figure. Find the resistance of the combination between the ends $A$ and $B$.
(Take, resistivity of copper $= 1.7 	imes 10^{-8}\ \Omega	ext{-m}$, resistivity of aluminium $= 2.6 	imes 10^{-8}\ \Omega	ext{-m}$)

Options:

(A) $2.170	ext{ m}\Omega$

(B) $1.420	ext{ m}\Omega$

(C) $0.0858	ext{ m}\Omega$

(D) $0.858	ext{ m}\Omega$

Sigma Percentile · Accepted Answer

The problem asks us to find the equivalent resistance of a combination of two rods: one made of copper (Cu) and the other of aluminium (Al).

### Analyzing the Setup
Look closely at the diagram. Both the copper and aluminium rods are connected between the same two terminals, $A$ and $B$. This means they are connected in a **parallel combination**.

Imagine the current entering from point $A$; it will divide and flow through both rods simultaneously before recombining at point $B$.

### The Master Equation
To find the equivalent resistance, we first need to calculate the individual resistance of each rod. The resistance $R$ of any uniform conductor is given by the formula:
$$R = \frac{ho l}{A}$$
where $ho$ is the resistivity of the material, $l$ is the length, and $A$ is the cross-sectional area.

### Calculating Individual Resistances
Let's start with the copper rod. We are given:
- $ho_{	ext{Cu}} = 1.7 	imes 10^{-8}\ \Omega	ext{-m}$
- $l = 25\ 	ext{cm} = 25 	imes 10^{-2}\ 	ext{m}$
- $A = 3\ 	ext{mm}^2 = 3 	imes 10^{-6}\ 	ext{m}^2$

Substituting these values into our formula:
$$R_{	ext{Cu}} = \frac{1.7 	imes 10^{-8} 	imes 25 	imes 10^{-2}}{3 	imes 10^{-6}}$$
$$R_{	ext{Cu}} = \frac{42.5 	imes 10^{-10}}{3 	imes 10^{-6}} = 14.167 	imes 10^{-4}\ \Omega$$

Now, let's do the same for the aluminium rod. The length and area are identical; only the resistivity changes:
- $ho_{	ext{Al}} = 2.6 	imes 10^{-8}\ \Omega	ext{-m}$

$$R_{	ext{Al}} = \frac{2.6 	imes 10^{-8} 	imes 25 	imes 10^{-2}}{3 	imes 10^{-6}}$$
$$R_{	ext{Al}} = \frac{65 	imes 10^{-10}}{3 	imes 10^{-6}} = 21.667 	imes 10^{-4}\ \Omega$$

### Final Calculation
Since the rods are in parallel, the equivalent resistance $R_{	ext{eq}}$ is given by:
$$R_{	ext{eq}} = \frac{R_{	ext{Cu}} R_{	ext{Al}}}{R_{	ext{Cu}} + R_{	ext{Al}}}$$

Substituting the values we just calculated:
$$R_{	ext{eq}} = \frac{(14.167 	imes 10^{-4}) 	imes (21.667 	imes 10^{-4})}{(14.167 	imes 10^{-4}) + (21.667 	imes 10^{-4})}$$
$$R_{	ext{eq}} = \frac{306.95 	imes 10^{-8}}{35.834 	imes 10^{-4}}$$
$$R_{	ext{eq}} = 8.566 	imes 10^{-4}\ \Omega$$

The options are given in milli-ohms ($	ext{m}\Omega$). To convert ohms to milli-ohms, we multiply by $10^3$:
$$R_{	ext{eq}} = 0.8566 	imes 10^{-3}\ \Omega \approx 0.858\ 	ext{m}\Omega$$

Thus, the correct option is **(d)**.

Select Answer:

Visualized Solution

$Analyzing the Circuit$

$Resistance Formula$

$Resistance of Copper Rod$

$Calculating R_{Cu}$

$Resistance of Aluminium Rod$

$Calculating R_{Al}$

$Equivalent Resistance in Parallel$

$Substituting Values$

$Calculating R_{eq}$

$Final Answer$

The Sigma Insight: Combination of Resistors

Analyzing the Setup Look closely at the diagram

The Master Equation To find the equivalent resistance, we first need to calculate the individual resistance of each rod

Calculating Individual Resistances Let's start with the copper rod

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Analyzing the Circuit

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resistance Formula

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resistance of Copper Rod

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating RCu​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Resistance of Aluminium Rod

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating RAl​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equivalent Resistance in Parallel

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Req​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer