a 10volt cell of negligible internal resistance is connected in parallel across battery of emf 200 V and internal resistance 38 ohm find the value of current
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Arrange of cells are shown in figure , plz see attachment .
Let i current passing through circuit .
use Kirchoff's law,
-10V + i × 38 - 200V = 0 [ ∵in closed loop sum of potential difference equal = 0]
38i - 190 = 0
⇒ 38i = 190
⇒ i = 190/38 = 5A
Hence, current passing through it is 5Amperes
Let i current passing through circuit .
use Kirchoff's law,
-10V + i × 38 - 200V = 0 [ ∵in closed loop sum of potential difference equal = 0]
38i - 190 = 0
⇒ 38i = 190
⇒ i = 190/38 = 5A
Hence, current passing through it is 5Amperes
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39
Answer:
Explanation:
I=v/r=200-10/38=5A
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