A conducting wire of length l, area of cross-section A and electric resistivity rho is connected bet

Q: Why doesn't the resistivity $\rho$ change when the wire's dimensions change?

Resistivity is an intrinsic property of the material itself (like copper or aluminum) and depends on factors like temperature and atomic structure. It does not depend on the macroscopic shape or size of the wire. As long as the material and temperature are constant, $\rho$ remains constant.

Q: I get confused when dividing fractions. How did $\frac{2l}{A/2}$ become $4 \frac{l}{A}$?

When you divide by a fraction, it is mathematically equivalent to multiplying by its reciprocal. So, $\frac{2l}{A/2}$ becomes $2l \times \frac{2}{A}$, which simplifies to $\frac{4l}{A}$.

Q: Does the voltage of the battery change when we change the wire?

No, an ideal battery maintains a constant potential difference $V$ across its terminals regardless of the circuit it is connected to. The current will adjust itself based on the resistance, but the voltage source remains steady.

Q: What if the question said the radius was halved instead of the area?

If the radius $r$ is halved ($r' = r/2$), the new area would be $A' = \pi (r/2)^2 = \frac{\pi r^2}{4} = A/4$. The area would become one-fourth, which would make the resistance increase by a factor of 8 (if length was also doubled)!

Q: Is the volume of the wire conserved in this problem?

Yes! Even though the problem explicitly gave us the new dimensions, let's check: Initial Volume = $A \cdot l$. Final Volume = $(A/2) \cdot (2l) = A \cdot l$. The volume is indeed conserved, which is exactly what happens when a wire is physically stretched.

Question

A conducting wire of length $l$, area of cross-section $A$ and electric resistivity $\rho$ is connected between the terminals of a battery. A potential difference $V$ is developed between its ends, causing an electric current. If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be

Options:

(A) $\frac{1}{4} \frac{VA}{\rho l}$

(B) $\frac{3}{4} \frac{VA}{\rho l}$

(C) $\frac{1}{4} \frac{\rho l}{VA}$

(D) $4 \frac{VA}{\rho l}$

Sigma Percentile · Accepted Answer

### The Beauty of Ohm's Law and Resistance

Imagine you are standing inside a conducting wire. All around you, free electrons are zooming past, colliding with the massive, vibrating lattice ions of the metal. These collisions are the physical manifestation of **electrical resistance**.

When we apply a potential difference $V$ across the ends of this wire, we create an electric field that pushes these electrons, creating a steady flow known as the electric current $I$. The relationship between this push and the resulting flow is elegantly captured by Ohm's Law:

$$I = \frac{V}{R}$$

But what exactly determines the resistance $R$ of our wire? It depends on three fundamental factors: the material it's made of (its resistivity $ho$), its length $l$, and its cross-sectional area $A$. The formula is beautifully intuitive:

$$R = ho \frac{l}{A}$$

Think of it like a hallway. A longer hallway (greater $l$) means more obstacles to bump into, increasing resistance. A wider hallway (greater $A$) means more space for people to walk side-by-side, decreasing resistance.

### Analyzing the Setup

In our problem, we start with a wire of length $l$, area $A$, and resistivity $ho$. Let's call its initial resistance $R_1$.

$$R_1 = ho \frac{l}{A}$$

This wire is connected to a battery providing a voltage $V$. The initial current $I_1$ flowing through the circuit is simply $V / R_1$.

### The Transformation

Now, the problem introduces a twist. We are given a new scenario where the length of the wire is doubled, and its cross-sectional area is halved.

Let's define our new parameters:
- New length, $l' = 2l$
- New area, $A' = \frac{A}{2}$

Crucially, the problem states that the wire is made of the **same material**. This means the intrinsic property of the metal, its resistivity $ho$, remains absolutely constant.

Let's calculate the new resistance, $R_2$, by substituting these new dimensions into our resistance formula:

$$R_2 = ho \frac{l'}{A'}$$

$$R_2 = ho \frac{2l}{\frac{A}{2}}$$

When we simplify this fraction, the $2$ in the denominator of the denominator flips up to multiply with the numerator:

$$R_2 = 4 \left( ho \frac{l}{A} ight)$$

Notice the term in the parentheses? That is exactly our initial resistance $R_1$. So, we have discovered that:

$$R_2 = 4 R_1$$

By doubling the length and halving the area, we have made the wire four times more resistive!

### Final Calculation

We are asked to find the resultant current, let's call it $I_2$, flowing through this new, highly resistive wire. The battery hasn't changed, so the voltage $V$ remains the same. We apply Ohm's Law once more:

$$I_2 = \frac{V}{R_2}$$

Substitute the expression we found for $R_2$:

$$I_2 = \frac{V}{4 ho \frac{l}{A}}$$

To make this look mathematically pristine, we rearrange the fraction. The area $A$ moves to the numerator:

$$I_2 = \frac{1}{4} \frac{VA}{ho l}$$

And there we have it! The new current is exactly one-fourth of the initial current. This makes perfect physical sense: since the resistance increased by a factor of four, the current must decrease by a factor of four to maintain the balance dictated by Ohm's Law.

### The Way Forward: A Word of Caution

Before we wrap up, let's discuss a classic trap that examiners love to set in JEE and NEET.

In this specific problem, we were explicitly told that the new length is $2l$ and the new area is $A/2$. However, what if the problem had simply said, *"The wire is stretched to double its length"*?

When a wire is **stretched**, its total volume must remain constant. Since Volume = Area $	imes$ Length ($V = A \cdot l$), if you double the length ($l' = 2l$), the area must automatically halve ($A' = A/2$) to keep the product constant.

In our problem, the dimensions were given directly, but the underlying physics of volume conservation is a powerful tool to keep in your arsenal. Always read the wording carefully: is the wire *replaced* with a different one, or is it *stretched*? The physics changes entirely based on that single word!

Select Answer:

Visualized Solution

$R_{1} = ρ \frac{l}{A}$

$l^{'} = 2 l, A^{'} = \frac{A}{2}$

$R_{2} = ρ \frac{l ^{'}}{A ^{'}}$

$R_{2} = 4 ρ \frac{l}{A}$

$I = \frac{V}{R}$

$I_{2} = \frac{V}{R _{2}}$

$I_{2} = \frac{1}{4} \frac{V A}{ρl}$

What if volume is constant?

The Sigma Insight: Ohm's Law, Resistance and Electrical Power

The Beauty of Ohm's Law and Resistance

Analyzing the Setup

The Transformation

Final Calculation

The Way Forward

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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; } R1​=ρAl​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } l′=2l,A′=2A​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​=ρA′l′​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R2​=4ρAl​

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​=R2​V​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } I2​=41​ρlVA​