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.$E = 2.2 \, V$When the cell is co

Vedclass · Accepted Answer

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

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(A) When the circuit is open,the terminal potential difference is equal to the electromotive force $(E)$ of the cell.$E = 2.2 \, V$When the cell is connected to an external resistance $R = 5 \, \Omega$,the terminal potential difference $(V)$ is $1.8 \, V$.The current $(I)$ flowing through the circuit is given by Ohm's law:$I = \frac{V}{R} = \frac{1.8}{5} = 0.36 \, A$The relation between electromotive force $(E)$,terminal potential difference $(V)$,current $(I)$,and internal resistance $(r)$ is:$V = E - Ir$Substituting the known values:$1.8 = 2.2 - (0.36)r$$0.36r = 2.2 - 1.8$$0.36r = 0.4$$r = \frac{0.4}{0.36} = \frac{40}{36} = \frac{10}{9} \, \Omega$

For a cell,the terminal potential difference 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.

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