Question

A uniform wire of resistance 12 \(\Omega\) 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 \(\Omega\) is connected across the highest of the three resistors. The current through the circuit is:

• A. 4 A
• B. 2 A
• C. 0.5 A
• D. 1 A

Hint:

When a source is connected across one side of a triangle of resistors, the other two sides are always in series with each other, and that combination is in parallel with the first side. Visualizing or redrawing the circuit can make this clearer.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a circuit analysis problem. First, we need to determine the resistance of each piece of wire. Then, we analyze the circuit formed by the triangle of resistors to find the equivalent external resistance. Finally, we use Ohm's law for the entire circuit, including the internal resistance of the cell, to find the total current.

Step 2: Key Formula or Approach:
1. For a uniform wire, resistance is proportional to length (\(R \propto L\)). 2. Resistors in series: \(R_{series} = R_1 + R_2 + \dots\) 3. Resistors in parallel: \(\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots\) 4. Ohm's law for a complete circuit: \(I = \frac{\mathcal{E}}{R_{ext} + r}\), where \(R_{ext}\) is the equivalent external resistance and \(r\) is the internal resistance.

Step 3: Detailed Explanation:
Part 1: Find the resistances of the three pieces.
Total resistance \(R_{total} = 12 \, \Omega\). The lengths are in the ratio 1:2:3. Since \(R \propto L\), the resistances will also be in the ratio 1:2:3. Let the resistances be \(R_1, R_2, R_3\). \[ R_1 = \left(\frac{1}{1+2+3}\right) \times 12 \, \Omega = \frac{1}{6} \times 12 = 2 \, \Omega \] \[ R_2 = \left(\frac{2}{1+2+3}\right) \times 12 \, \Omega = \frac{2}{6} \times 12 = 4 \, \Omega \] \[ R_3 = \left(\frac{3}{1+2+3}\right) \times 12 \, \Omega = \frac{3}{6} \times 12 = 6 \, \Omega \] The highest resistance is \(R_3 = 6 \, \Omega\). Part 2: Find the equivalent external resistance (\(R_{ext}\)).
The three resistors form a triangle. The cell is connected across the highest resistor, \(R_3\). This means that resistors \(R_1\) and \(R_2\) are in series with each other, and this series combination is in parallel with \(R_3\). Resistance of the series combination: \(R_{12} = R_1 + R_2 = 2 \, \Omega + 4 \, \Omega = 6 \, \Omega\). Now, \(R_{12}\) is in parallel with \(R_3\). The equivalent external resistance \(R_{ext}\) is: \[ \frac{1}{R_{ext}} = \frac{1}{R_{12}} + \frac{1}{R_3} = \frac{1}{6 \, \Omega} + \frac{1}{6 \, \Omega} = \frac{2}{6 \, \Omega} = \frac{1}{3 \, \Omega} \] \[ R_{ext} = 3 \, \Omega \] Part 3: Calculate the total current.
Given data for the cell:
EMF, \(\mathcal{E} = 8 \, \text{V}\).
Internal resistance, \(r = 5 \, \Omega\).
Using Ohm's law for the complete circuit: \[ I = \frac{\mathcal{E}}{R_{ext} + r} = \frac{8 \, \text{V}}{3 \, \Omega + 5 \, \Omega} = \frac{8 \, \text{V}}{8 \, \Omega} = 1 \, \text{A} \]

Step 4: Final Answer:
The current through the circuit (drawn from the cell) is 1 A.