Question

For the circuit shown in the adjoining figure, match the entries of column I with the entries of column II.

Answer 1

$I=\frac{E}{2R}=0.5$E\R

Current drawn from the battery is least.

$I=\frac{3E}{2R}=1.5$E\R

Since the voltage across is highest, Brightness $\alpha$ voltage.

$I=\frac{2E}{R}$ Current drawn from the battery is maximum.

$I=\frac{E}{R}$

Hence correct options are:

$A)-2$

$B)-3$

$C)-1,4$

$D)-4$