The magnitude and direction of the current in the circuit shown will be
Medium
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Since ${E_1}(10\,V) > {E_2}(4\,V)$
So current in the circuit will be clockwise.
Applying Kirchoff's voltage law
$ - \,1 \times i + 10 - 4 - 2 \times i - 3i = 0$ $ \Rightarrow $ $i = 1\,A$ $(a$ to $b$ via $e)$
Current $ = \frac{V}{R} = \frac{{10 - 4}}{6} = 1.0\,ampere$
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