A 9 V battery is connected in series with a resistor .The terminal voltage is found to be 8 V.
Current through the circuit is measured as 5 A. What is the internal resistance of the battery?

Asked by ayushmittal2424 | 9th Dec, 2018, 09:45: PM

Expert Answer:

As we know battery supplied voltage = E - Ir
Where E is the EMF of cell , I is the current passing through battery and r is internal resistance of battery . Putting all the values we get
8=9-5×r
=> r=1/5 =0.2 ohm

Answered by Ankit K | 9th Dec, 2018, 10:29: PM