Question

A circuit consists of one 330 ohm resistor in series with two other resistors (470 ohms and 220 ohms) connected in parallel. What is the power drain on an ideal battery with a voltage of 2.9 V connected to this combination?

Answer 1

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