A 3 V battery with negligible internal resis | vedclass

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Question

(B) From the circuit diagram,the two $3 \ \Omega$ resistors on the right side are in series,forming an equivalent resistance of $R_1 = 3 \ \Omega + 3

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$1.5$

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(B) From the circuit diagram,the two $3 \ \Omega$ resistors on the right side are in series,forming an equivalent resistance of $R_1 = 3 \ \Omega + 3 \ \Omega = 6 \ \Omega$.This $R_1 = 6 \ \Omega$ is in parallel with the third $3 \ \Omega$ resistor.The equivalent resistance $R_{eq}$ of the circuit is given by:$\frac{1}{R_{eq}} = \frac{1}{6} + \frac{1}{3} = \frac{1 + 2}{6} = \frac{3}{6} = \frac{1}{2}$Therefore,$R_{eq} = 2 \ \Omega$.The total current $I$ in the circuit is:$I = \frac{V}{R_{eq}} = \frac{3 \ V}{2 \ \Omega} = 1.5 \ A$.

$A$ $3 \ V$ battery with negligible internal resistance is connected as shown in the figure. The current $I$ in the circuit will be ............ $A$.

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