A circuit is shown in figure. $R$ is a nonzero variable but finite resistance ,e is some unknown emf with polarities. Match the columns.

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Solution

$L o o p F E D C F : e - 6 = R I_{1} - 4 I_{2}$ (i)
$L o o p A F C B A : 6 - 4 = 4 I_{2} + 2 (I_{1} + I_{2})$
or $2 = 2 I_{1} + 6 I_{2}$ (ii)
Solving them, we get
$I_{1} = \frac{3 e - 14}{4 + 3 R}, I_{2} = \frac{R + 6 - e}{4 + 3 R}$
i. $I_{2} = 0 o r e = R + 6$
$e > 6 V (∵ R \neq 0)$
ii. For current from $F$ to $C$ direction,
$I_{2} > 0 o r R + 6 > e o r e < R + 6$
Possible for any finite value of $e$ , because $R$ is finite.
iii. For current from $C$ to $F$ direction
$I_{2} < 0 o r e > R + 6$
iv. For current in $2 Ω$ from $B$ to $A$ direction,
$I_{1} + I_{2} = \frac{R - 8 + 2 e}{4 + 3 R} > 0$
$R - 8 + 2 e > 0 o r e > 4 - \frac{R}{2}$
Depending on the value of $R$ , $e$ can take any value from zero to infinity.

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(IIT-JEE 1997)