A) 0.4 V
B) 0.6 V
C) 1.2 V
D) 1.5 V
Correct Answer: A
Solution :
Equivalent resistance of the given network \[{{R}_{eq}}=75\,\Omega \] \ Total current through battery \[i=\frac{3}{75}\] \[{{i}_{1}}={{i}_{2}}=\frac{3}{75\times 2}=\frac{3}{150}\] Current through \[{{R}_{4}}=\frac{3}{150}\times \frac{60}{(30+60)}\]\[=\frac{3}{150}\times \frac{60}{90}=\frac{2}{150}A\] \[{{V}_{4}}={{i}_{4}}\times {{R}_{4}}=\frac{2}{150}\times 30=\frac{2}{5}V=0.4\,V\]You need to login to perform this action.
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