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?

Aug 13, 2018

$4 \Omega$

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 $0.96 V$; 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 $\left(1.08 - 0.96\right) = 0.12 V$ 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 $\left(1.08 - 0.12\right) = 0.96 V$ (note it is the same as ${V}_{A} - {V}_{B}$)

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

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

$R = \frac{0.96}{0.24} = 4 \Omega$