A cell has an e.m.f of 1.08 v and an internal resistor of 0.5 ohm. when it is connected in series with resistor of R , the potential difference between the terminals fell to 0.96 v . what was the value of R?
The circuit diagram in the given case appears as follows,
Here, you can place the internal resistance as a separate resistor in the circuit and in that case potential drop across the cell will be
Now, it is given that the potential drop across the cell is
So, if the current flowing through the circuit is
Now, the voltage drop across resistor
So,if voltage drop across
Then, using Ohm's Law, we can write,