A wire of length L and 3 identical cells of negligible internal resistances are connected in series.

Q: Why does the resistance double when the length is doubled?

Resistance is directly proportional to length ($R = \rho \frac{L}{A}$). Since the material and cross-section are the same, doubling the length doubles the resistance.

Q: Why does the mass double?

Mass is volume times density. Volume is area times length. Since the cross-section area is constant, doubling the length doubles the volume, and hence the mass.

Q: Why use $V^2/R$ instead of $I^2R$?

We are given the EMF of the cells (which acts as voltage $V$) and the resistance. While $I^2R$ is also correct, finding $I$ requires dividing $V$ by $R$ anyway, so $V^2/R$ is more direct.

Q: What happens to the specific heat capacity $s$?

The specific heat capacity $s$ is a property of the material. Since the material is the same in both cases, $s$ remains constant and cancels out when we divide the equations.

Q: Why do we ignore internal resistance?

The problem explicitly states that the cells have negligible internal resistances, so we don't need to include them in our calculations.

Question

A wire of length $L$ and 3 identical cells of negligible internal resistances are connected in series. Due to the current, the temperature of the wire is raised by $\Delta T$ in a time $t$. A number $N$ of similar cells is now connected in series with a wire of the same material and cross-section but of length $2L$. The temperature of the wire is raised by the same amount $\Delta T$ in the same time. The value of $N$ is

Options:

(A) 4

(B) 6

(C) 8

(D) 9

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you have a simple electrical circuit. You've got a wire of length $L$, and you hook it up to a battery made of 3 identical cells.

When you close the switch, current starts flowing. And what happens when current flows through a wire? It heats up! This is the beautiful phenomenon of Joule heating.

The problem tells us that this heat raises the temperature of the wire by an amount $\Delta T$ in a specific time $t$.

Now, let's define some variables to make our lives easier. Let the resistance of this wire be $R$, and its mass be $m$. Since we have 3 cells, let's say the total electromotive force (EMF) is $3E$.

### The Master Equation

We need to connect the electrical energy supplied by the battery to the thermal energy gained by the wire.

The electrical heat generated in time $t$ is given by the formula:

$$H = \frac{V^2}{R}t$$

Here, our voltage $V$ is the total EMF, which is $3E$. So, the heat generated is:

$$H = \frac{(3E)^2}{R}t$$

But where does this heat go? It raises the temperature of the wire! From thermodynamics, we know that the heat required to raise the temperature of a mass $m$ by $\Delta T$ is:

$$H = ms\Delta T$$

where $s$ is the specific heat capacity of the material. Equating the two, we get our first master equation:

$$\frac{(3E)^2}{R}t = ms\Delta T \quad \dots (i)$$

### The Second Scenario

Now, the problem throws a curveball. We change the setup.

We take a new wire of the **same** material and cross-section, but we double its length to $2L$.

What does this do to the resistance and mass?

Since resistance $R = \rho \frac{L}{A}$, doubling the length doubles the resistance. So, the new resistance is $2R$.

Similarly, doubling the length doubles the volume, which means the mass also doubles to $2m$.

We connect this new wire to $N$ identical cells. So, the new voltage is $NE$.

The problem states that the temperature rise $\Delta T$ and the time $t$ are exactly the same as before. Let's write the heat equation for this new setup:

$$\frac{(NE)^2}{2R}t = (2m)s\Delta T \quad \dots (ii)$$

### Final Calculation

We now have a beautiful system of two equations. The best way to solve them is to divide them and watch the magic happen as terms cancel out.

Let's divide equation (ii) by equation (i):

$$\frac{\frac{(NE)^2}{2R}t}{\frac{(3E)^2}{R}t} = \frac{2ms\Delta T}{ms\Delta T}$$

Notice how the time $t$, the resistance $R$, the mass $m$, the specific heat $s$, and the temperature change $\Delta T$ all gracefully cancel out!

We are left with a pure, simple algebraic relation:

$$\frac{N^2 E^2 / 2}{9 E^2} = 2$$

$$\frac{N^2}{18} = 2$$

Multiply both sides by 18:

$$N^2 = 36$$

Taking the square root, we find our final answer:

$$N = 6$$

And there you have it! To achieve the exact same temperature rise in the doubled wire, we need exactly 6 cells. Physics is beautifully consistent!

Select Answer:

Visualized Solution

$Initial Setup$

$Heat Generated in First Case$

$Second Setup$

$Heat Generated in Second Case$

$Solving for N$

$Final Value of N$

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

Analyzing the Setup

The Master Equation

The Second Scenario

Final Calculation

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Visualized Solution

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Generated in First Case

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Heat Generated in Second Case

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

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