Consider two identical galvanometers and two identical resistors with resistance R. If the internal…

1. The maximum voltage range is obtained when all the components are connected in series
2. The maximum voltage range is obtained when the two resistors and one galvanometer are connected in series, and the second galvanometer is connected in parallel to the first galvanometer
3. The maximum current range is obtained when all the components are connected in parallel
4. The maximum current range is obtained when the two galvanometers are connected in series and the combination is connected in parallel with both the resistors.

Solution

$i=\frac{E}{2R+2R_c}$ $V_{g_1}=i{R}_c=\frac{E{R}_c}{2\left(R+R_c\right)}=\frac{E{R}_c}{2R+2R_c}$

$i_2=\frac{E}{2R+\frac{R_c}{2}}$ $V_{g_2}=\frac{i_2}{2}\times R_c$
$=\frac{1}{2}\left(\frac{E}{2R+\frac{R_c}{2}}\right)\times R_c$
$V_{g_2}=\frac{E{R}_c}{4R+R_c}$ ....(2)
Since $R_c<\frac{R}{2}$ $\left(2R_c<R\right)$
$2R+2R_c<3R$
$V_{g_1}>V_{g_2}$

 

$i_A=\frac{E}{R_c}$

Here $i=\frac{E(2R+2R_c)}{2R.2R_c}=\frac{E(R+R_c)}{2R{R}_c}$
$i_B=\frac{2R}{2R+2R_c}\times \frac{E(R+R_c)}{2R{R}_c}=\frac{E}{2R_c}$
$i_A>i_B$ ^

Asked in: JEE Advanced 2016 (Paper 2)

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