Question

V-I graph for four materials A, B, C \& D are shown in fig. The best material to make live wire of domestic wiring is :

• A. A
• B. B
• C. C
• D. D

Hint:

For an I-V graph (Current on y-axis, Voltage on x-axis), the slope represents the conductance ($1/R$). A steeper slope indicates lower resistance and thus a better conductor. For domestic wiring, a material with very low resistance is desired to efficiently transmit electricity.

Answer 1

The Correct Option is A

Step 1: Understand the V-I graph and Ohm's Law.
The V-I graph plots voltage (V) on the x-axis and current (I) on the y-axis. Ohm's Law states that V = IR, which means R = V/I. Therefore, the resistance (R) of a conductor is the reciprocal of the slope of the I-V graph (Current vs. Voltage graph). A steeper slope on an I-V graph means lower resistance and higher conductance. Step 2: Determine the ideal property for a live wire in domestic wiring.
For domestic wiring, a material with very low resistance is needed for the live wire. High resistance leads to significant energy loss as heat ($P = I^2R$) and a drop in voltage, which is undesirable for efficient electrical transmission. Step 3: Analyze the slopes of the V-I graphs for materials A, B, C, and D.
Looking at the provided graph:
Material A has the steepest slope, indicating the largest I/V ratio.
Material B has a less steep slope than A.
Material C has a less steep slope than B.
Material D has the least steep slope, indicating the smallest I/V ratio. Step 4: Relate the slope to resistance.
A steeper slope on an I-V graph means a higher current for a given voltage, which implies lower resistance ($R = V/I$). Therefore, material A has the lowest resistance, and material D has the highest resistance. Step 5: Select the best material.
Since the best material for domestic wiring (live wire) should have the lowest resistance to minimize energy loss and voltage drop, material A is the most suitable choice. $$(1) A$$