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(D) The correct option is $D$.From the circuit diagram,the $6 \Omega$ and $2 \Omega$ resistors are connected in parallel. Their equivalent resistance

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$1.5$

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(D) The correct option is $D$.From the circuit diagram,the $6 \Omega$ and $2 \Omega$ resistors are connected in parallel. Their equivalent resistance $R_p$ is given by:$R_p = \frac{6 	imes 2}{6 + 2} = \frac{12}{8} = 1.5 \Omega$Now,this equivalent resistance $R_p = 1.5 \Omega$ is in series with the $1.5 \Omega$ resistor and the $3 \Omega$ resistor.Therefore,the total equivalent resistance $R_{eq}$ of the circuit is:$R_{eq} = 1.5 \Omega + 1.5 \Omega + 3 \Omega = 6 \Omega$Using Ohm's law,the total current $I$ supplied by the battery is:$I = \frac{V}{R_{eq}} = \frac{9 	ext{ V}}{6 \Omega} = 1.5 	ext{ A}$

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

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