A uniform wire 8m long is and of resistance 5ohm is bent into the form of ABCD. The adjacent corners A & B are connected to a battery of emf 2V & internal resistance .5ohm. Find the current along AB.
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Since the complete wire has a resistance of 16 ohms, each side will have a resistance of 4 ohms.
You can use KCL and KVL to have the potential difference between Point_A and Point_B, but here is a quick way to do it.
Potential difference between Point_A and Point_C is 9V, since a battery of 9V is connected directly to it. Now this 9V divides equally between R2, R3 and R4 since they are all of equal resistance. So potential drop across R2, R3 and R4 will be 3 V each (9V/3).
So potential difference between Point_A and Point_B is 3+3 = 6V
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