Question

A cell has an EMF of 1.08v and an internal resistance of 0.5 ohm. when it is connected in series with an external resistance R .The potential difference between the terminals of cell fell to 0.96v .Find value of R​

Answer 1

Answer:

Explanation:

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 VA−VB

Now, it is given that the potential drop across the cell is 0.96V; that means this drop is due to some potential drop across the 0.5Ω internal resistance in the cell, we can say the rest i.e (1.08−0.96)=0.12V has dropped across 0.5Ω.

So, if the current flowing through the circuit is I, then we can say, I×0.5=0.12

Or, I=0.24A

Now, the voltage drop across resistor R is (1.08−0.12)=0.96V (note it is the same as VA−VB)

So,if voltage drop across R is 0.96V and Current flowing is 0.24A

Then, using Ohm's Law, we can write,

R=0.960.24=4Ω