Question

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?

Answer 1

`4 Omega`

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 `V_A - V_B`

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 `0.5 Omega` internal resistance in the cell, we can say the rest i.e `(1.08-0.96)=0.12V` has dropped across `0.5 Omega`.

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

Or, `I=0.24 A`

Now, the voltage drop across resistor `R` is `(1.08-0.12)=0.96V` (note it is the same as `V_A-V_B`)

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

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

`R=0.96/0.24=4 Omega`