In the given diagram,an ideal voltmeter V rea | vedclass

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(B) The circuit consists of two cells connected in a loop. The current $I$ flowing in the circuit is given by the net electromotive force divided by t

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$r_1 = 3r_2$

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(B) The circuit consists of two cells connected in a loop. The current $I$ flowing in the circuit is given by the net electromotive force divided by the total resistance:$I = \frac{1.5\,V - 1.3\,V}{r_1 + r_2} = \frac{0.2}{r_1 + r_2}$The voltmeter $V$ is connected across the cell with $1.3\,V$ emf. Since the current flows out of the $1.5\,V$ cell and into the $1.3\,V$ cell,the $1.3\,V$ cell is being charged. The terminal voltage $V$ across the $1.3\,V$ cell is given by:$V = E + Ir_1$$1.45 = 1.3 + I r_1$$0.15 = I r_1$Substituting the value of $I$:$0.15 = \left( \frac{0.2}{r_1 + r_2} \right) r_1$$0.15(r_1 + r_2) = 0.2 r_1$$0.15 r_1 + 0.15 r_2 = 0.2 r_1$$0.15 r_2 = 0.05 r_1$$3 r_2 = r_1$Thus,the relation is $r_1 = 3r_2$.

In the given diagram,an ideal voltmeter $V$ reads $1.45\,V$. Then the relation between $r_1$ and $r_2$ is:

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