Question

The emf of a cell is 1.2V . When it is connected across a 4 ohm resistor, the p.d across the terminals is
found to be 0.8V. The internal resistance of the cell is
a) 1ohm.
b) 3 ohm.
c) 0.5 ohm.
d) 2 ohm.

give the answer with full explanation and derivation.​

Answer 1

Answer:Option d

Explanation:

Refer to page 123 of NCERT 12th Physics Ch- 3 Current Electricity

Eqn- 3.94a and 3.94b

Answer 2

Answer:

The internal resistance of the cell is 2ohms.

Explanation:

Given, emf of a cell, $\big \varepsilon =1.2V$

           resistance of a resistor, $R= 4 \Omega$

           potential difference across the terminals $V=0.8V$

From the Ohm's law, $V=IR$

                                   $\implies I=\frac{V}{R}$                                               ...(1)

In terms of internal resistance($r$), emf of a cell is given by,

                                   $\varepsilon =I(R+r)\implies I=\frac{\varepsilon}{R+r}$                    ...(2)

Equating both the equations, we get

                                  $\frac{V}{R}=\frac{\varepsilon}{R+r}$

Substituting the given values,

                               $\frac{0.8}{4}=\frac{1.2}{4+r}$

                    $\implies {4+r}=\frac{1.2}{0.8}\times4$

                          $\implies r=6-4=2\Omega$

Therefore, the internal resistance of the cell is 2ohms.

Hence, the correct option is 4.