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(B) According to Ohm's Law,$V = IR$,which implies $I = \frac{1}{R} V$.Comparing this with the equation of a straight line $y = mx$,where $y = I$ and $

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$\frac{1}{\sqrt{3}} \Omega$

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(B) According to Ohm's Law,$V = IR$,which implies $I = \frac{1}{R} V$.Comparing this with the equation of a straight line $y = mx$,where $y = I$ and $x = V$,the slope $m$ of the $I-V$ graph is given by $m = 	an 	heta = \frac{1}{R}$.From the given graph,the angle $	heta = 60^{\circ}$.Therefore,$	an 60^{\circ} = \frac{1}{R}$.Since $	an 60^{\circ} = \sqrt{3}$,we have $\sqrt{3} = \frac{1}{R}$.Thus,the resistance $R = \frac{1}{\sqrt{3}} \Omega$.

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

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