In the following circuit,the terminal voltage | vedclass

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(B) The circuit consists of a cell with $EMF$ $E = 2 \ V$ and internal resistance $r = 0.1 \ \Omega$,connected in series with an external resistor $R

Vedclass · Accepted Answer

$1.95$

Vedclass · Answer

(B) The circuit consists of a cell with $EMF$ $E = 2 \ V$ and internal resistance $r = 0.1 \ \Omega$,connected in series with an external resistor $R = 3.9 \ \Omega$.First,calculate the total resistance of the circuit: $R_{total} = R + r = 3.9 \ \Omega + 0.1 \ \Omega = 4.0 \ \Omega$.Next,calculate the current $i$ flowing through the circuit using Ohm's law: $i = \frac{E}{R_{total}} = \frac{2 \ V}{4.0 \ \Omega} = 0.5 \ A$.The terminal voltage $V$ across the cell is given by the formula: $V = E - i 	imes r$.Substituting the values: $V = 2 \ V - (0.5 \ A 	imes 0.1 \ \Omega) = 2 \ V - 0.05 \ V = 1.95 \ V$.

In the following circuit,the terminal voltage across the cell is (in $V$)

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