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 \[{{R}_{4}}=\frac{3}{150}\times \frac{60}{(30+60)}=\frac{3}{150}\times \frac{60}{90}=\frac{2}{150}A\] \[{{V}_{4}}={{i}_{4}}\times {{R}_{4}}=\frac{2}{150}\times 30=\frac{2}{5}=0.4\]voltYou need to login to perform this action.
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