Answer:
Option C
Explanation:
The circuit can be redrawn as,
45 V is drop across the branch one, so current through it
$I_{1}=\frac{45}{R_{eq1}}=\frac{45}{\frac{15}{4}}=12 A$
Now, above circuit can be redrawn as,
Since , in a series circuit, the current flows will be same,
hence, $V_{ab}=I_{ab}R_{eq}(ab)=12\left(\frac{15}{4}+4+10\right)$
$\Rightarrow$ $V_{ab}$=213 V
Hence , the correct option is (c)
