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For a cell terminal P.D. 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 : -
A
$\dfrac{10}{9}\,\Omega$
B
$\dfrac{9}{10}\,\Omega$
C
$\dfrac{11}{9}\,\Omega$
D
$\dfrac{5}{9}\,\Omega$
Detailed Solution
In open circuit no current flows, so the terminal p.d. equals the emf: E = 2.2 V.
With a load R, the terminal p.d. is $V = E - Ir$, where $I = \dfrac{E}{R + r}$:
$V = E - \dfrac{E}{R + r}r = \dfrac{ER}{R + r}$
From the given conditions: $1.8 = \dfrac{2.2 \times 5}{5 + r}$
$\Rightarrow 9 + 1.8r = 11 \Rightarrow 1.8r = 2 \Rightarrow r = \dfrac{10}{9}\,\Omega$
With a load R, the terminal p.d. is $V = E - Ir$, where $I = \dfrac{E}{R + r}$:
$V = E - \dfrac{E}{R + r}r = \dfrac{ER}{R + r}$
From the given conditions: $1.8 = \dfrac{2.2 \times 5}{5 + r}$
$\Rightarrow 9 + 1.8r = 11 \Rightarrow 1.8r = 2 \Rightarrow r = \dfrac{10}{9}\,\Omega$