Question
A battery of emf 2 V and internal resistance 0.1 Ω is being charged with a current of 5 A. In what direction will the current flow inside the battery? What is the potential difference between the two terminals of the battery?
Solution
Given, emf of the battery, E= 2V
Internal resistance, r = 0.1
Current throught the battery, I = 5 A
As seen in the figure below, the positive terminal of the battery is connected to the positive terminal of the charger in order to charge the battery. Hence, inside the battery, the direction of the current is from the positive terminal to the negative terminal.
Internal resistance, r = 0.1
Current throught the battery, I = 5 A
As seen in the figure below, the positive terminal of the battery is connected to the positive terminal of the charger in order to charge the battery. Hence, inside the battery, the direction of the current is from the positive terminal to the negative terminal.
The emf or potential difference between the terminals of the battery is (VP – VQ).
VPQ = Vp – VQ
= 2 + 0.5
= 2.5 V
