Question 715: Current Electricity

Question

For a cell terminal potential difference is 2.2V when circuit is open and reduce to 1.8 V when cell is connected to a resistance of R = 5 $$\Omega $$. Determine internal resistance of cell (r)

Accepted Answer

Terminal potential difference is 2.2 V when circuit is open.
$$ 	herefore $$ e.m.f. of the cell = E = 2.2 volt
Now, when the cell is connected to the external resistance, circuit current I is given by
$$I = {E \over {R + r}} = {{2.2} \over {5 + r}}$$ ampere, where r is the internal resistance of the cell.
Potential difference across the cell = IR
$$ = {{2.2} \over {5 + r}} 	imes 5 = 1.8$$
$$ 	herefore $$ 5 + r = 11/1.8.
$$ 	herefore r = {{11} \over {1.8}} - 5 = {{110 - 90} \over {18}} = {{10} \over 9}\Omega $$

Answer

$${{10} \over 9}\Omega $$

Answer

$${{9} \over 10}\Omega $$

Answer

$${{11} \over 9}\Omega $$

Answer

$${{5} \over 9}\Omega $$

Current Electricity

For a cell terminal potential difference is 2.2V when circuit is open and reduce to 1.8 V when cell is connected to a resistance of R = 5 $\Omega $. Determine internal resistance of cell (r)

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