Three resistors, $\mathrm{R}_{1}=9.00 \Omega, \mathrm{R}_{2}=3.00 \Omega$, and $\mathrm{R}_{3}=1.00 \Omega$, are connected in parallel to a $9.00 \mathrm{~V}$ battery. What is the total power dissipated in the circuit?
A) $117 \mathrm{~W}$
B) $80.1 \mathrm{~W}$
C) $99.0 \mathrm{~W}$
D) $180 \mathrm{~W}$
E) $18.0 \mathrm{~W}$

Question

Quizplus · Accepted Answer

The total resistance of the circuit is:$$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}=\frac{1}{9.00\Omega}+\frac{1}{3.00\Omega}+\frac{1}{1.00\Omega}=\frac{13}{9.00\Omega}$$$$R_{eq}=\frac{9.00\Omega}{13}=0.692\Omega$$The total power dissipated in the circuit is:$$P=\frac{V^2}{R_{eq}}=\frac{(9.00	ext{ V})^2}{0.692\Omega}=\boxed{117	ext{ W}}$$

Three Resistors R1=9.00Ω,R2=3.00Ω\mathrm{R}_{1}=9.00 \Omega, \mathrm{R}_{2}=3.00 \OmegaR1​=9.00Ω,R2​=3.00Ω , And R3=1.00Ω\mathrm{R}_{3}=1.00 \OmegaR3​=1.00Ω , Are Connected in Parallel to A

Three Resistors $\mathrm{R}_{1}=9.00 \Omega, \mathrm{R}_{2}=3.00 \Omega$ , And $\mathrm{R}_{3}=1.00 \Omega$ , Are Connected in Parallel to A