What can we deduce about $R$ in this graph of $I$ against $V$, where $R = \frac{V}{I}$?

Question

Why is D the correct answer to the multiple-choice question below?

For a reader unfamiliar with physics, the only physics one needs to know is that the resistance of a component is given as $R = \frac{V}{I}$.

Answer 1

First: The correct answer most be C or D because of the formula that is given $R = V / I$. Now the question is just which of the two $$ \frac{V_1}{I_1} \quad\text{ and }\quad \frac{V_2}{I_2} $$ is greatest.

Since the curve passes through the origin for any point $(V,I)$ on the curve the quantity $\frac{I}{V}$ gives the slope of the line that passes through the origin and the point $(V,I)$. It is clear then from the graph that $$ \frac{I_2}{V_2} > \frac{I_1}{V_1}. $$ That means $$ R_2 = \frac{V_2}{I_2} > \frac{V_1}{I_1} = R_2. $$

Answer 2

This is not a 'rate of change' question. To get the resistance you simply need the values of V and I, as your formula suggests. To determine that the resistance is decreasing, notice that proportionally the quotient is decreasing. Roughly $V_2$ = 1.25$V_1$ and $I_2$ = 1.5$I_1$ so $\frac{V_2}{I_2}$ = $\frac{1.25V_1}{1.5I_1}$

Answer 3

I think the assumption here is that resistance is defined in terms of Ohm's law; that is, the component's resistance changes depending on $V$ so that Ohm's law still holds. This means that $R = \frac{V_2}{I_2}$ simply by definition, and since $\frac{V_1}{I_1} > \frac{V_2}{I_2}$, resistance decreases from $V_1$ to $V_2$.

Differentiating the function with respect to $V$ and taking the reciprocal gives the correct answer if the component is purely Ohmic ($I = V/R$ for constant $R$), but in the case of a non-linear function, say $I = V^2$, you will get $\frac{dI}{dV} = 2V$ so that $\frac{V}{I} = \frac{1}{V} \neq \frac{1}{\frac{dI}{dV}} = \frac{1}{2V}$.

Answer 4

V=IR(note: here, R is a variable, for, the relation between V,I does not give a straight-line)

or, dV=R(dI)+I(dR)

or, (dI/dV)=(1/R)(1-I(dR/dV))

(dI/dV) is the slope of the curve

solving this differential-equation, we get V=IR, because we have merely retraced our path back to the starting-point.

R=V/I

so, at V=V2,I=I2, we have R=V2/I2

similarly, at V=V1,I=I1, we have R=V1/I1

from rough-sketch, you can view that (dI/dV)>0

now, (dI/dV)=(1/R)(1-I(dR/dV))

so, 1-I(dR/dV)>0 or, 1+(I/dV)(-dR)

this is possible, if dR<0, because, then -dR>0, as I,dV are positive

so, resistance decreases.