Answer:
power drain on an ideal battery, P = 0.017 W
Given:
[tex]R_{1} = 330\ohm [/tex]
[tex]R_{2} = 470\ohm [/tex]
[tex]R_{3} = 220\ohm [/tex]
Since, [tex]R_{2} = 470\ohm [/tex] and [tex]R_{3} = 220\ohm [/tex] are in parallel and this combination is in series with [tex]R_{1} = 330\ohm [/tex], so,
Equivalent resistance of the circuit is given by:
[tex]R_{eq} = \frac{R_{2}R_{3}}{R_{2} + R_{3}} + R_{1}[/tex]
[tex]R_{eq} = \frac{470\times 220}{470 + 220} + 330[/tex]
[tex]R_{eq} = 149.85 + 330 = 479.85 \ohm[/tex]
power drain on an ideal battery, P = [tex]\frac{V^{2}}{R_{eq}}[/tex]
P = [tex]\frac{2.9^{2}}{479.85}[/tex]
P = 0.017 W
