A uniform heating wire of resistance 36 ohm is connected across a potential difference of 240 V. The

Q: Why do we use $P = \frac{V^2}{R}$ instead of $P = I^2 R$?

Because the potential difference $V$ is kept constant in both cases. If we used $I^2 R$, we would have to calculate the new current for each case, which adds an extra step. Using $V^2/R$ is much faster and less prone to errors.

Q: Why does the resistance become half when the wire is cut?

Resistance of a uniform wire is given by $R = \rho \frac{L}{A}$. When you cut the wire in half, the length $L$ becomes $L/2$, while the resistivity $\rho$ and cross-sectional area $A$ remain unchanged. Thus, the resistance also becomes half.

Q: Why do we add the powers in the second case?

Power is a scalar quantity representing the rate of energy dissipation. Since both halves are dissipating energy simultaneously, the total energy dissipated per second is simply the sum of the individual powers.

Q: What if the two halves were connected in series instead of parallel?

If connected in series, the total resistance would be $18 + 18 = 36\ \Omega$, which is the same as the original wire. The power dissipation would remain exactly the same as in Case I.

Q: Does the voltage $240\text{ V}$ actually matter for the final ratio?

No, it doesn't! As you saw in the calculation, the $V^2$ term cancels out completely when we take the ratio. The ratio depends only on how the resistance changes.

Question

A uniform heating wire of resistance $36\ \Omega$ is connected across a potential difference of $240	ext{ V}$. The wire is then cut into half and potential difference of $240	ext{ V}$ is applied across each half separately. The ratio of power dissipation in first case to the total power dissipation in the second case would be $1:x$, where $x$ is .........

Sigma Percentile · Accepted Answer

### The Physics of Heating Wires

When an electric current flows through a conductor, it encounters resistance. This resistance causes some of the electrical energy to be converted into heat energy, a phenomenon known as Joule heating. The rate at which this heat is produced is called the power dissipation.

For any resistor $R$ connected across a potential difference $V$, the power dissipated is given by the master equation:

$$P = \frac{V^2}{R}$$

This specific form of the power equation is incredibly useful when the voltage $V$ is kept constant, as it allows us to see exactly how power changes when we alter the resistance. Let's dive into the two cases presented in our problem.

### Analyzing Case I: The Full Wire

Imagine you have a uniform heating wire. In the first scenario, this entire wire is connected directly across a $240	ext{ V}$ source.

We are given:
- The potential difference, $V = 240	ext{ V}$
- The resistance of the full wire, $R_1 = 36\ \Omega$

Using our power equation, the power dissipated in this first case, which we'll call $P_1$, is simply:

$$P_1 = \frac{(240)^2}{36}$$

We could calculate the exact numerical value here, but as a pro-tip for JEE physics: **never evaluate large numbers until the very end**. We are looking for a ratio, and keeping the terms in their raw form will allow for beautiful cancellations later.

### Analyzing Case II: The Cut Wire

Now, things get interesting. The wire is cut exactly in half. What happens to its resistance?

We know that the resistance of a uniform wire is directly proportional to its length ($R = ho \frac{L}{A}$). If you cut the length in half, the resistance also drops by exactly half. Therefore, the resistance of each new piece is:

$$R_2 = \frac{R_1}{2} = \frac{36}{2} = 18\ \Omega$$

### The Power of Parallel Circuits

The problem states that a potential difference of $240	ext{ V}$ is applied across *each* half separately. This is the exact definition of a parallel circuit! Both halves are experiencing the full $240	ext{ V}$.

Because power is a scalar quantity representing the rate of energy transfer, the total power dissipated in this second case ($P_2$) is simply the sum of the power dissipated by each individual half.

$$P_2 = P_{2a} + P_{2b}$$

$$P_2 = \frac{V^2}{R_2} + \frac{V^2}{R_2}$$

Substituting our values:

$$P_2 = \frac{(240)^2}{18} + \frac{(240)^2}{18}$$

$$P_2 = 2 	imes \frac{(240)^2}{18} = \frac{(240)^2}{9}$$

Notice how cutting the wire and putting it in parallel drastically increased the power output! The denominator went from $36$ down to $9$.

### Calculating the Ratio

We are asked to find the ratio of the power dissipation in the first case to the total power dissipation in the second case, $P_1 : P_2$.

Let's set up the ratio using our raw expressions:

$$\frac{P_1}{P_2} = \frac{\frac{(240)^2}{36}}{\frac{(240)^2}{9}}$$

Here is where our patience pays off. The bulky $(240)^2$ terms cancel out perfectly!

$$\frac{P_1}{P_2} = \frac{9}{36} = \frac{1}{4}$$

The problem states this ratio is $1:x$. By direct comparison:

$$x = 4$$

### The General Rule: Cutting into $n$ Parts

This problem reveals a fascinating general principle. What if we cut the wire into $n$ equal parts and connected them all in parallel across the same voltage?

1. The resistance of each part would be $\frac{R}{n}$.
2. The power of each part would be $\frac{V^2}{R/n} = n \frac{V^2}{R}$.
3. Since there are $n$ parts, the total power would be $n 	imes \left(n \frac{V^2}{R}ight) = n^2 \frac{V^2}{R}$.

This means the new power is **$n^2$ times** the original power! In our problem, $n=2$, so the new power was $2^2 = 4$ times the original power, which perfectly matches our ratio of $1:4$. Keep this shortcut in your mental toolkit for future exams!

Enter Numerical Value:

Visualized Solution

$P = \frac{V ^{2}}{R}$

$Case I: P_{1} = \frac{V ^{2}}{R _{1}}$

$Case II: R_{2} = \frac{R _{1}}{2}$

$P_{2} = P_{2 a} + P_{2 b}$

$\frac{P _{1}}{P _{2}} = \frac{1}{x}$

$x = 4$

$What if n parts?$

The Sigma Insight: Heating Effect of Current

The Physics of Heating Wires

Analyzing Case I

Analyzing Case II

The Power of Parallel Circuits

Calculating the Ratio

The General Rule

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