question_answer

In circuit shown below, the resistances are given in ohm and the battery is assumed ideal with emf equal to 3 V. The voltage across the resistance \[{{R}_{4}}\] is

A)  0.4 V                                    

B)  0.6V

C)  1.2 V                                    

D)  1.5 V

Correct Answer: A

Solution :

                 Equivalent resistance of the given network \[{{R}_{eq}}=75\,\Omega \] \[\therefore \]Total current through battery, \[i=\frac{3}{75}\]                 \[{{i}_{1}}-{{i}_{2}}=\frac{3}{75\times 2}=\frac{3}{150}\] Current through resistance \[{{R}_{4}}=\frac{3}{150}\times \frac{60}{(30+60)}=\frac{3}{150}\times \frac{60}{90}=\frac{2}{150}A\] \[{{V}_{4}}={{i}_{4}}\times {{R}_{4}}=\frac{2}{150}\times 30=\frac{2}{5}=0.4\]volt