Question

the terminal potential difference is 3.6 V, when the circuit is open. If the potential difference reduces to 3 V, when cell is connected to a resistance of 5Ω, the internal resistance of cell is

Answer 1

Answer:

1ohm is the correct answer

Answer 2

Answer:

The internal resistance of cell is 1Ω.

Explanation:

Given the terminal potential difference is 3.6 V, when the circuit is open.

When the circuit is open, the terminal potential difference is equal to the emf of the cell, therefore, emf, $\mathcal{E}=3.6V$

Cell is connected to a resistance, $R= 5\Omega$

When the cell is connected to an external resistance, the current through the circuit for internal resistance r of the cell is given by

$I=\dfrac{\mathcal{E}}{R+r} =\dfrac{3.6}{5+r}$

Given that there is a potential drop when connected to the external resistance, let that be $\Delta V=3 V$

Using Kirchhoff's voltage law, change in potential between the ends is given by $\Delta V=\mathcal{E}-Ir$

Substituting the values,

 $3=3.6-\dfrac{3.6}{5+r}\times r\implies 0.6=\dfrac{3.6}{5+r}\times r$

$\implies 0.6(5+r)=3.6\times r\implies 6(5+r)=36\times r$

$\implies 30+6r=36 r\implies 36 r-6r=30$

$\implies 30r=30\implies r=1\Omega$

Therefore, the internal resistance of cell is 1Ω.