question_answer

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

A)  \[0.4V\]           

B)  \[0.6V\]

C)  \[1.2\text{ }V\]           

D)  \[1.5V\]

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 \[{{R}_{4}}=\frac{3}{150}\times \frac{60}{(30+60)}=\frac{3}{150}\times \frac{60}{90}=\frac{2}{150}A\] \[{{V}_{4}}={{i}_{4}}\times {{R}_{4}}=\frac{2}{150}\times 30=\frac{2}{5}=0.4\,volt\]