In circuit shown below, the resistance are given in ohms and the battery is assumed ideal with emf equal to 3 volt The voltage across the resistance `R_(4)` is

Options

A. 0.4 V
B. 0.6 V
C. 1.2 V
D. 1.5 V

Answer

A

Explanation

The equivalent circuit can be redrawn as shown in figure 1. From figure 1 it is obivious that power dissipated by `R_(1)` is maximum. Potential difference across
`R_(2)` is `=(25)/(25+50)xx3` volt `=1` volt
Therefore potential difference across `R_(3)` or
`R_(4)=(20)/(20+30)xx1` volt `=0.4` volt ltbr. the equivalent resistance of circuit across the cell
is `50+25=75` ohms
Therefore current through cell is
`(3)/(75)xx1000` mA `=40` mA