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.|

 

 
 
 
 
 

Answers (1)
S safeer


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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