A cell has an emf of 2 , V and an internal re | vedclass

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(A) Key Idea: When a cell is supplying current, the potential difference across its terminals is less than its emf due to the potential drop across it

Vedclass · Accepted Answer

$1.95$

Vedclass · Answer

(A) Key Idea: When a cell is supplying current, the potential difference across its terminals is less than its emf due to the potential drop across its internal resistance.Given:Emf of the cell, $E = 2 \, V$Internal resistance, $r = 0.1 \, \Omega$External resistance, $R = 3.9 \, \Omega$The current $i$ flowing through the circuit is given by:$i = \frac{E}{R + r} = \frac{2}{3.9 + 0.1} = \frac{2}{4.0} = 0.5 \, A$The terminal voltage $V$ across the cell is given by the formula:$V = E - ir$Substituting the values:$V = 2 - (0.5 	imes 0.1)$$V = 2 - 0.05$$V = 1.95 \, V$Thus, the terminal voltage across the cell is $1.95 \, V$.

$A$ cell has an emf of $2 \, V$ and an internal resistance of $0.1 \, \Omega$. It is connected to an external resistance of $3.9 \, \Omega$. The terminal voltage across the cell will be: (in $V$)

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