QJEEAdv16Phe271: JEE Advanced 2016 Physics - Class 12 JEEAdv16Phe271 Solution: Consider two identical galvanometers and two identical…

Let i_g be the current that gives full deflection of the galvanometer. When all components are connected in series (see figure), effective resistance of the circuit is ${R_e} = 2R + 2{R_c}$ and maximum current allowed in the circuit is i_g.

Thus, the voltage between A and B is

${V_{AB}} = {i_g}{R_e} = 2{i_g}(R + {R_c})$ .

Consider the case when two resistors and one galvanometer are connected in series and the second galvanometer is connected in parallel.

The maximum current through each galvanometer is i_g and the maximum current through the resistors is 2i_g. Apply Kirchhoff's law to get the voltage between A and B as

$V{'_{AB}} = 2{i_g}R + 2{i_g}R + {i_g}{R_c}$

$= 2{i_g}(R + {R_c}) + 2{i_g}(R - {R_c}/2)$

$= {V_{AB}} + 2{i_g}(R - {R_c}/2)$

> {V_{AB}} ( $\because$ {R_c} < R/2 ).

Consider the case when all four components are connected in parallel.

Let i and i_g be the currents through the resistors and the galvanometers. By Kirchhoff's law, $iR = {i_g}{R_c}$ , which gives $i = {i_g}{R_c}/R$ . The current between A and B is

${I_{AB}} = 2i + 2{i_g} = 2{i_g}(1 + {R_c}/R)$ .

Consider the case when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors.

$I{'_{AB}} = {i_g} + 2i = {i_g} + 2{i_g}(2{R_c}/R)$

$= 2{i_g}(1 + {R_c}/R) - (1 - 2{R_c}/R){i_g}$

$= {I_{AB}} - (1 - 2{R_c}/R){i_g}$

< {I_{AB}} ( $\because$ {R_c} < R/2 ).