By a cell,a current of 0.9, A flows through a | vedclass

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(A) Let $E$ be the electromotive force $(EMF)$ of the cell and $r$ be its internal resistance.According to Ohm's law,the terminal voltage $V$ is given

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$0.5$

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(A) Let $E$ be the electromotive force $(EMF)$ of the cell and $r$ be its internal resistance.According to Ohm's law,the terminal voltage $V$ is given by $V = E - Ir$,where $I$ is the current.For the first case: $E = I_1(R_1 + r) = 0.9(2 + r)$.For the second case: $E = I_2(R_2 + r) = 0.3(7 + r)$.Since the $EMF$ $E$ is constant,we equate the two expressions:$0.9(2 + r) = 0.3(7 + r)$.Dividing both sides by $0.3$,we get:$3(2 + r) = 7 + r$.$6 + 3r = 7 + r$.$3r - r = 7 - 6$.$2r = 1$.$r = 0.5\, \Omega$.

By a cell,a current of $0.9\, A$ flows through a $2\, \Omega$ resistor and $0.3\, A$ flows through a $7\, \Omega$ resistor. The internal resistance of the cell is ............ $\Omega$.

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