Two ideal batteries of emf V _ 1 and V _ 2 and three resistances R...

Two ideal batteries of emf ${V}_{1}$ and ${V}_{2}$ and three resistances ${R}_{1}, {R}_{2}$ and ${R}_{3}$ are connected as shown in the figure. The current in resistance ${R}_{2}$ would be zero if

Two ideal batteries of emf V _ 1 and V _ 2 and three resistances R _ 1 , R _ 2 and R _ 3 are connected as shown in the

Options

A ${V}_{1} = {V}_{2}$ and ${R}_{1} = {R}_{2} = {R}_{3}$
B ${V}_{1} = {V}_{2}$ and ${R}_{1} = 2{R}_{2} = {R}_{3}$
C ${V}_{1} = 2{V}_{2}$ and $2{R}_{1} = 2{R}_{2} = {R}_{3}$
D $2{V}_{1} = {V}_{2}$ and $2{R}_{1} = {R}_{2} = {R}_{3}$

Correct answer

A. ${V}_{1} = {V}_{2}$ and ${R}_{1} = {R}_{2} = {R}_{3}$

Step-by-step solution

In order that the current is zero, the potential drop on ${R}_{2}$ due to ${V}_{1}$ must be nullified by that due to ${V}_{2}$ .

Accordingly,

${V}_{1} = \dfrac{{R}_{1}\left({V}_{1} + {V}_{2}\right)}{{R}_{1} + {R}_{3}} ⇒ {V}_{1}{R}_{3} = {V}_{2}{R}_{1}$
${V}_{2} = \dfrac{{R}_{3}\left({V}_{1} + {V}_{2}\right)}{{R}_{1} + {R}_{3}} ⇒ {V}_{2}{R}_{1} = {V}_{2}{R}_{3}$

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