Question Details

A uniform wire of resistance 12 Ω is cut into three pieces in the ratio of length 1: 2: 3. Now the three pieces are connected to form a triangle. A cell of emf 8 V and internal resistance 5 Ω is connected across the highest of the three resistors. The current through the circuit is:

Options

0.5A

Show Answer

Correct Answer :

Option D

Solution :

The correct option is 1A.

Let us break down the solution step-by-step to understand how we arrive at this answer:

Step 1: Calculate the resistance of the three cut pieces
The total resistance of the uniform wire is 12 Ω. Since the wire is uniform, its resistance is directly proportional to its length. The wire is cut into three pieces in the ratio of their lengths as 1 : 2 : 3.
Let the resistances of the three pieces be $R_{1}$ , $R_{2}$ , and $R_{3}$ .
The sum of the ratio parts is:
$1 + 2 + 3 = 6$
Using this, we can find the resistance of each piece as follows:
$R_{1} = \frac{1}{6} \times 12 Ω = 2 Ω$
$R_{2} = \frac{2}{6} \times 12 Ω = 4 Ω$
$R_{3} = \frac{3}{6} \times 12 Ω = 6 Ω$
Thus, the three resistances are 2 Ω, 4 Ω, and 6 Ω. The highest of the three resistors is $R_{3} = 6 Ω$ .

Step 2: Determine the equivalent resistance of the triangular connection
When these three pieces are connected to form a triangle, the two smaller resistors ( $R_{1} = 2 Ω$ and $R_{2} = 4 Ω$ ) are in series with each other. This series combination is connected in parallel with the largest resistor ( $R_{3} = 6 Ω$ ).
The equivalent resistance of the series path, $R_{s}$ , is:
$R_{s} = R_{1} + R_{2} = 2 Ω + 4 Ω = 6 Ω$
This series equivalent resistance $R_{s}$ is in parallel with the highest resistor $R_{3}$ . The equivalent external resistance of the triangular circuit, $R_{eq}$ , is:
$R_{eq} = \frac{R_{s} \times R_{3}}{R_{s} + R_{3}} = \frac{6 \times 6}{6 + 6} = \frac{36}{12}$

Step 3: Calculate the total current through the circuit
A cell of electromotive force (emf) $E = 8 V$ and internal resistance $r = 5 Ω$ is connected across the highest resistor.
The total resistance of the entire circuit, $R_{total}$ , is the sum of the equivalent external resistance and the internal resistance:
$R_{total} = R_{eq} + r = 3 Ω + 5 Ω = 8 Ω$
Using Ohm's law, the total current $I$ through the circuit is:
$I = \frac{E}{R_{total}} = \frac{8 V}{8 Ω} = 1 A$
Therefore, the current flowing through the circuit is 1A.