The current-voltage (I-V) graphs for a given | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) From Ohm's law,$V = IR$,which implies $I = \frac{1}{R} V$.Comparing this with the equation of a straight line $y = mx$,the slope of the $I-V$ grap

Vedclass · Accepted Answer

$T_1 < T_2$

Vedclass · Answer

(B) From Ohm's law,$V = IR$,which implies $I = \frac{1}{R} V$.Comparing this with the equation of a straight line $y = mx$,the slope of the $I-V$ graph is given by $	ext{slope} = \frac{dI}{dV} = \frac{1}{R}$.From the given graph,the slope for temperature $T_1$ is greater than the slope for temperature $T_2$.Therefore,$\frac{1}{R_1} > \frac{1}{R_2}$,which implies $R_1 For a metallic wire,resistance increases with an increase in temperature $(R \propto T)$.Since $R_1 < R_2$,it follows that $T_1 < T_2$.

The current-voltage $(I-V)$ graphs for a given metallic wire at two different temperatures $T_1$ and $T_2$ are shown in the figure. The temperatures $T_1$ and $T_2$ are related as

Similar Questions

$A$ $V-i$ graph for a conductor is given. The resistance is proportional to which of the following?

Which of the following statements is correct?

The variation of current $(i)$ and voltage $(V)$ in a conductor is shown in the diagram below. The resistance of the conductor is ........... $\Omega$.

The current flowing through a circuit is shown in the figure. What is the resistance of the circuit?

The resistance of a discharge tube is

Vedclass Test Series

Exam Paper Generator

Online Exam Module