A cell whose e.m.f. is 2, V and internal resi | vedclass

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(C) The electromotive force $(E)$ of the cell is $2\, V$.The internal resistance $(r)$ of the cell is $0.1\,\Omega$.The external resistance $(R)$ conn

Vedclass · Accepted Answer

$1.95$

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(C) The electromotive force $(E)$ of the cell is $2\, V$.The internal resistance $(r)$ of the cell is $0.1\,\Omega$.The external resistance $(R)$ connected is $3.9\,\Omega$.The total resistance of the circuit is $R_{total} = R + r = 3.9\,\Omega + 0.1\,\Omega = 4.0\,\Omega$.The current $(I)$ flowing through the circuit is given by Ohm's law: $I = \frac{E}{R + r} = \frac{2\, V}{4.0\,\Omega} = 0.5\, A$.The terminal voltage $(V)$ across the cell is given by $V = I 	imes R$.Substituting the values: $V = 0.5\, A 	imes 3.9\,\Omega = 1.95\, V$.

$A$ cell whose e.m.f. is $2\, V$ and internal resistance is $0.1\,\Omega$ is connected with a resistance of $3.9\,\Omega$. The voltage across the cell terminal will be ................ $V$.

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