# A 10 V cell of negligible internal resistance is connected in parallel across a battery of emf 200 V and internal resistance 38 $\Omega$ as shown in the figure. Find the value of current in the circuit.|

Applying Kirchohoft's law in loop ABCDA
$200-10-38i= 0$
$190= 38i$
$i= \frac{190}{38}= 5A$
Therefore the current in the circuit is 5A

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