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(C) From the circuit diagram,the $2 \, \Omega$ and $6 \, \Omega$ resistors are in parallel. Their equivalent resistance $R_p$ is:$R_p = \frac{2 	imes

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$4$

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(C) From the circuit diagram,the $2 \, \Omega$ and $6 \, \Omega$ resistors are in parallel. Their equivalent resistance $R_p$ is:$R_p = \frac{2 	imes 6}{2 + 6} = \frac{12}{8} = 1.5 \, \Omega$This combination is in series with the $1.5 \, \Omega$ resistor,giving a total resistance of $1.5 + 1.5 = 3 \, \Omega$ for this branch.This branch is in parallel with the $3 \, \Omega$ resistor connected across the battery.So,the total equivalent resistance $R_{eq}$ of the circuit is:$R_{eq} = \frac{3 	imes 3}{3 + 3} = \frac{9}{6} = 1.5 \, \Omega$The total current $I$ supplied by the battery is:$I = \frac{V}{R_{eq}} = \frac{6 \, V}{1.5 \, \Omega} = 4 \, A$

The total current supplied by the battery to the circuit is ............ $A$.

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