For a cell, the terminal potential difference is 2.2V when circuit is open and reduces to 1.8V when cell is connected to a resistance R=5Ω . The internal resistance (.r.) of the cell is

Question

For a cell, the terminal potential difference is 2.2V when circuit is open and reduces to 1.8V when cell is connected to a resistance R=5Ω . The internal resistance (.r.) of the cell is (A) (10/9)Ω (B) (9/10)Ω (C) (11/9)Ω (D) (5/9)Ω

Tardigrade Admin · Accepted Answer

When circuit is open then it would be EMF of cell so

E = 2.2 V

When connected with an external resistance

5Ω

we will measure terminal potential difference so
Solution

I = \frac{E}{R + r}

V_{A} - V_{B} = E - I (r) = I (R) = \frac{ER}{R + r}

.....(i)
So ,

V_{A} - V_{B} = 1.8 V o ltR = 5Ω

E = 2.2 V

from equation (i)

1.8 = \frac{( 2.2 ) ( 5 )}{5 + r}

9 + 1.8 r = 11

1.8 r = 2

r = \frac{20}{18} = \frac{10}{9} Ω

Q. For a cell, the terminal potential difference is $2.2 V$ when circuit is open and reduces to $1.8 V$ when cell is connected to a resistance $R = 5Ω$ . The internal resistance $(r)$ of the cell is

$\frac{10}{9} Ω$

$\frac{9}{10} Ω$

$\frac{11}{9} Ω$

$\frac{5}{9} Ω$

Q. For a cell, the terminal potential difference is 2.2V when circuit is open and reduces to 1.8V when cell is connected to a resistance R=5Ω . The internal resistance (r) of the cell is

910​Ω

109​Ω

911​Ω

95​Ω

Q. For a cell, the terminal potential difference is $2.2 V$ when circuit is open and reduces to $1.8 V$ when cell is connected to a resistance $R = 5Ω$ . The internal resistance $(r)$ of the cell is

$\frac{10}{9} Ω$

$\frac{9}{10} Ω$

$\frac{11}{9} Ω$

$\frac{5}{9} Ω$