Understanding the Problem Setup

The problem asks us to find the total current flowing from a cell when it's connected in a specific circuit. The circuit involves a wire of uniform resistance that is cut into pieces and then arranged to form a triangle. A cell with a given electromotive force (emf) and internal resistance is connected across the largest resistor in the triangle.

Calculating Individual Resistances

First, let's determine the resistance of each piece of the wire. The total resistance of the wire is given as $12 \Omega$. It is cut into three pieces with lengths in the ratio 1:2:3. For a uniform wire, resistance is directly proportional to its length. Therefore, the resistances of the three pieces ($R_1, R_2, R_3$) will also be in the ratio 1:2:3.

Let the resistances be $R_1 = x$, $R_2 = 2x$, and $R_3 = 3x$. The sum of these resistances must equal the total resistance of the wire:

$ R_1 + R_2 + R_3 = 12 \Omega $

Substituting the expressions in terms of $x$:

$ x + 2x + 3x = 12 \Omega $

$ 6x = 12 \Omega $

Solving for $x$:

$ x = \frac{12 \Omega}{6} = 2 \Omega $

Now we can find the resistance of each piece:

• $R_1 = x = 2 \Omega$
• $R_2 = 2x = 2 \times 2 \Omega = 4 \Omega$
• $R_3 = 3x = 3 \times 2 \Omega = 6 \Omega$

The three resistors are $2 \Omega$, $4 \Omega$, and $6 \Omega$. The highest resistance is $R_3 = 6 \Omega$.

Analyzing the Circuit Configuration

The three pieces ($R_1, R_2, R_3$) are connected to form a triangle. Let the vertices of the triangle be A, B, and C, such that the resistance between A and B is $R_1 = 2 \Omega$, between B and C is $R_2 = 4 \Omega$, and between C and A is $R_3 = 6 \Omega$.

A cell with emf $\mathcal{E} = 8$ V and internal resistance $r = 5 \Omega$ is connected across the highest resistor ($R_3 = 6 \Omega$). This means the cell is connected in parallel with $R_3$. The other two resistors, $R_1$ and $R_2$, are connected in series with each other across the same two points (vertices C and A) where $R_3$ and the cell are connected.

So, we have two parallel branches connected to the cell:

1. Branch 1: The highest resistor $R_3 = 6 \Omega$.
2. Branch 2: The series combination of $R_1$ and $R_2$, which is $R_{12} = R_1 + R_2 = 2 \Omega + 4 \Omega = 6 \Omega$.

Calculating the Equivalent External Resistance

The total external resistance ($R_{ext}$) of the circuit is the equivalent resistance of these two parallel branches ($R_3$ and $R_{12}$):

$ R_{ext} = \frac{R_3 \times R_{12}}{R_3 + R_{12}} $

Substituting the values:

$ R_{ext} = \frac{6 \Omega \times 6 \Omega}{6 \Omega + 6 \Omega} $

$ R_{ext} = \frac{36 \Omega^2}{12 \Omega} $

$ R_{ext} = 3 \Omega $

Calculating the Total Circuit Current

Now, we can find the total current ($I$) flowing from the cell using Ohm's law for the entire circuit. The total resistance in the circuit is the sum of the external resistance ($R_{ext}$) and the internal resistance ($r$) of the cell.

$ I = \frac{\mathcal{E}}{R_{ext} + r} $

Plugging in the values for emf, external resistance, and internal resistance:

$ I = \frac{8 \text{ V}}{3 \Omega + 5 \Omega} $

$ I = \frac{8 \text{ V}}{8 \Omega} $

$ I = 1 \text{ A} $

The current through the circuit is 1 A.