The I-V graphs for two different electrical a | vedclass

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(B) According to Ohm's law,$V = IR$,which can be written as $I = \frac{1}{R}V$.Comparing this with the equation of a straight line $y = mx$,the slope

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$R_P > R_Q$

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(B) According to Ohm's law,$V = IR$,which can be written as $I = \frac{1}{R}V$.Comparing this with the equation of a straight line $y = mx$,the slope of the $I-V$ graph is $\frac{I}{V} = \frac{1}{R}$.Thus,the slope of the $I-V$ graph is inversely proportional to the resistance $(Slope \propto \frac{1}{R})$.In the given diagram,the slope of line $Q$ is greater than the slope of line $P$ (i.e.,$Slope_Q > Slope_P$).Therefore,$\frac{1}{R_Q} > \frac{1}{R_P}$.This implies that $R_P > R_Q$.

The $I-V$ graphs for two different electrical appliances $P$ and $Q$ are shown in the diagram. If $R_P$ and $R_Q$ are the resistances of the devices,then:

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