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marusya05 [52]
3 years ago
7

A circuit consists of one 330 ohm resistor in series with two other resistors (470 ohms and 220 ohms) connected in parallel. Wha

t is the power drain on an ideal battery with a voltage of 2.9 V connected to this combination?
Physics
1 answer:
umka21 [38]3 years ago
4 0

Answer:

power drain on an ideal battery, P = 0.017 W

Given:

R_{1} = 330\ohm

R_{2} = 470\ohm

R_{3} = 220\ohm

Since, R_{2} = 470\ohm and R_{3} = 220\ohm are in parallel and this combination is in series with R_{1} = 330\ohm, so,

Equivalent resistance of the circuit is given by:

R_{eq} = \frac{R_{2}R_{3}}{R_{2} + R_{3}} + R_{1}

R_{eq} = \frac{470\times 220}{470 + 220} + 330

R_{eq} = 149.85 + 330 = 479.85 \ohm

power drain on an ideal battery, P = \frac{V^{2}}{R_{eq}}

                                                      P = \frac{2.9^{2}}{479.85}

                                                      P = 0.017 W

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