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: Conceptualization: This is a circuit analysis task. The process involves determining individual wire resistances, calculating the equivalent external resistance of a triangular resistor network, and finally applying Ohm's law for the entire circuit, inclusive of internal resistance, to find the total current.

Step 2: Core Principles: 1. Resistance of a uniform wire is directly proportional to its length (\(R \propto L\)). 2. Series resistors sum: \(R_{series} = R_1 + R_2 + \dots\). 3. Parallel resistors: \(\frac{1}{R_{parallel}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots\). 4. Complete circuit Ohm's law: \(I = \frac{\mathcal{E}}{R_{ext} + r}\), where \(R_{ext}\) is the equivalent external resistance and \(r\) is the internal resistance.

Step 3: Detailed Breakdown: Part 1: Resistance Calculation. Total resistance \(R_{total} = 12 \, \Omega\). Lengths are in the ratio 1:2:3, thus resistances are also in the ratio 1:2:3. Let the resistances be \(R_1, R_2, R_3\). \(R_1 = \left(\frac{1}{6}\right) \times 12 \, \Omega = 2 \, \Omega\). \(R_2 = \left(\frac{2}{6}\right) \times 12 \, \Omega = 4 \, \Omega\). \(R_3 = \left(\frac{3}{6}\right) \times 12 \, \Omega = 6 \, \Omega\). The largest resistance is \(R_3 = 6 \, \Omega\). Part 2: Equivalent External Resistance (\(R_{ext}\)) Determination. The three resistors form a triangle. The cell connects across the highest resistor, \(R_3\). This configuration places \(R_1\) and \(R_2\) in series, with their combined resistance in parallel with \(R_3\). Series resistance \(R_{12} = R_1 + R_2 = 2 \, \Omega + 4 \, \Omega = 6 \, \Omega\). The equivalent external resistance \(R_{ext}\) is calculated as: \(\frac{1}{R_{ext}} = \frac{1}{R_{12}} + \frac{1}{R_3} = \frac{1}{6 \, \Omega} + \frac{1}{6 \, \Omega} = \frac{1}{3 \, \Omega}\). Thus, \(R_{ext} = 3 \, \Omega\). Part 3: Total Current Calculation. Cell Parameters: EMF, \(\mathcal{E} = 8 \, \text{V}\); Internal resistance, \(r = 5 \, \Omega\). Applying 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: Conclusion: The total current drawn from the cell is 1 A.