The $I-V$ graphs of two resistors,and their series combination,are shown below. Which one of these graphs represents the series combination of the other two? Give reasons for your answer.

(A) In an $I-V$ graph,the slope represents the conductance,which is the reciprocal of resistance $(1/R)$.
To determine the series combination,draw a vertical line from any point on the $V$ axis such that it intersects the graphs $1, 2,$ and $3$ at points $P, Q,$ and $R$ respectively,as shown in the provided solution image.
For a constant potential $V$,the currents corresponding to the three graphs are $I_1, I_2,$ and $I_3$.
From the graph,we observe that $I_1 < I_2 < I_3$.
Since $R = V/I$,for a fixed $V$,a smaller current implies a larger resistance. Therefore,$R_1 > R_2 > R_3$.
In a series combination,the equivalent resistance $(R_s = R_a + R_b)$ is always greater than the individual resistances ($R_a$ and $R_b$).
Since graph $1$ has the lowest current for a given potential,it corresponds to the highest resistance. Thus,graph $1$ represents the series combination of the other two resistors (graphs $2$ and $3$).