A 3,V battery with negligible internal resist | vedclass

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(C) From the circuit diagram,we can see that two $3\,\Omega$ resistors are in series,and this combination is in parallel with the third $3\,\Omega$ re

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$1.5$

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(C) From the circuit diagram,we can see that two $3\,\Omega$ resistors are in series,and this combination is in parallel with the third $3\,\Omega$ resistor.First,calculate the equivalent resistance of the two resistors in series: $R_s = 3\,\Omega + 3\,\Omega = 6\,\Omega$.Now,this $6\,\Omega$ resistor is in parallel with the remaining $3\,\Omega$ 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,$R_{eq} = 2\,\Omega$.Using Ohm's law,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 in a circuit as shown in the figure. The current $I$ in the circuit will be ............. $A$.

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