A battery of emf E and internal resistance r | vedclass

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Question

(A) The potential difference $V$ across a battery is given by the equation $V = E - ir$.Comparing this with the equation of a straight line $y = mx +

Vedclass · Accepted Answer

Internal resistance of the battery is $5\,\Omega$.

Vedclass · Answer

(A) The potential difference $V$ across a battery is given by the equation $V = E - ir$.Comparing this with the equation of a straight line $y = mx + c$,we get $V = -ri + E$.From the given graph,the intercept on the $V$-axis is $E = 10\,V$.The intercept on the $i$-axis is the short-circuit current when $V = 0$,which is $i_{max} = E/r = 2\,A$.Substituting $E = 10\,V$ and $i_{max} = 2\,A$,we get $r = E / i_{max} = 10 / 2 = 5\,\Omega$.Thus,the internal resistance $r = 5\,\Omega$,the $emf$ $E = 10\,V$,and the maximum current is $2\,A$.Comparing these with the options,option $A$ is correct.

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