For a cell,the terminal potential difference | vedclass

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(A) When the circuit is open,the terminal potential difference is equal to the electromotive force $(E)$ of the cell. Thus,$E = 2.2 \, V$.When the cel

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$\frac{10}{9} \, \Omega$

Vedclass · Answer

(A) When the circuit is open,the terminal potential difference is equal to the electromotive force $(E)$ of the cell. Thus,$E = 2.2 \, V$.When the cell is connected to an external resistance $R$,the terminal potential difference $(V)$ is given by the formula: $V = E - Ir$,where $I = \frac{E}{R+r}$.Substituting $I$,we get $V = E - (\frac{E}{R+r})r = E(\frac{R}{R+r})$.Given $V = 1.8 \, V$,$E = 2.2 \, V$,and $R = 5 \, \Omega$:$1.8 = 2.2 	imes \frac{5}{5+r}$$1.8(5+r) = 11$$9 + 1.8r = 11$$1.8r = 2$$r = \frac{2}{1.8} = \frac{20}{18} = \frac{10}{9} \, \Omega$.

For a cell,the terminal potential difference $(P.D.)$ is $2.2 \, V$ when the circuit is open and reduces to $1.8 \, V$ when the cell is connected to an external resistance of $R = 5 \, \Omega$. Determine the internal resistance of the cell $(r)$.

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