Three resistors, R₁ = 9.00 Ω, R₂ = 3.00 Ω, and R₃ = 1.00 Ω, are connected in parallel to a 9.00 V battery. What is the total power dissipated in the circuit? A) 180 W B) 117 W C) 99.0 W D) 80.1 W E) 18.0 W

Question

Quizplus · Accepted Answer

The total resistance in a parallel circuit is given by $1/R_{\text{total}} = 1/R_1 + 1/R_2 + 1/R_3$. Substituting the given values, $1/R_{\text{total}} = 1/9 + 1/3 + 1/1 = 1/9 + 3/9 + 9/9 = 13/9$. Therefore, $R_{\text{total}} = 9/13 \, \Omega$. The total power dissipated can be calculated using $P = V^2 / R_{\text{total}}$, where $V = 9.00 \, V$. Substituting the values, $P = 9^2 / (9/13) = 81 \times (13/9) = 9 \times 13 = 117 \, W$.

Three Resistors, R1 = 9

Three Resistors, R₁ = 9