A cell of EMF 15 volt records a potential difference of 1 volt when connected to an external resistance R such that current flowing through the circuit is 0.5 ampere calculate the internal resistance of the cell and the drop in potential also
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Answer: 2.8 ohms
Explanation:
Given an emf cell 15 V with internal resistance (r) series connected with another resistor(R).
The voltage across R is 1 V.
Current in the circuit is 0.5 A
According to KVL, 15 = v +1.
I.e., v = -14.
r= (voltage across the resistor,r)
/(current across it)
r= 14/0.2
r=2.8.
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