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

Question

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

Tardigrade Admin · Accepted Answer

T.P.D ( V )= E - Ir (Remember it ) V=E-((E/R+r)) r=(E R/(R+r)) from given conditions E=2.2 when R=5 then TPD V =1.8 V therefore 1.8=(2.2 × 5/5+ r ) ⇒ r =(10/9) Ω

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

$\frac{10}{9} Ω$

$\frac{9}{10} Ω$

$\frac{11}{9} Ω$

$\frac{5}{9} Ω$

$T . P . D (V) = E$ - Ir (Remember it )
$V = E - (\frac{E}{R + r}) r = \frac{ER}{( R + r )}$
from given conditions $E = 2.2$ & when $R = 5$
then $TP D V = 1.8 V$
therefore $1.8 = \frac{2.2 \times 5}{5 + r}$
$\Rightarrow r = \frac{10}{9} Ω$

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

910​Ω

109​Ω

911​Ω

95​Ω