Question

1. a 9 v battery is connected in parallel to three resistors: $2\\ \omega$, $3\\ \omega$, and $4\\ \omega$. how much power is consumed by the circuit?

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Explanation:

Step1: Find equivalent parallel resistance

For resistors in parallel, $\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$
$\frac{1}{R_{eq}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{4} = \frac{6 + 4 + 3}{12} = \frac{13}{12}$
$R_{eq} = \frac{12}{13} \ \Omega$

Step2: Calculate total circuit power

Use $P = \frac{V^2}{R_{eq}}$, substitute $V=9\ \text{V}$
$P = \frac{9^2}{\frac{12}{13}} = 81 \times \frac{13}{12}$

Step3: Simplify the expression

$P = \frac{1053}{12} = 87.75$

Answer:

87.75 W