The current of 0.15 A flows through three resistors,R₁ = 20 Ohm,R₂ = 50 Ohm,R₃ = 65 Ohm,connected in series.What is the power dissipated in the R₂ resistor?

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Quizplus · Accepted Answer

The total resistance in series is $R_{total} = R_1 + R_2 + R_3 = 20\, \Omega + 50\, \Omega + 65\, \Omega = 135\, \Omega$. The total voltage across the series, $V = I \times R_{total} = 0.15\, A \times 135\, \Omega = 20.25\, V$. The voltage across $R_2$ (50 Ohm) can be found using the proportion $V_2 = V \times \frac{R_2}{R_{total}} = 20.25\, V \times \frac{50\, \Omega}{135\, \Omega} = 7.5\, V$. The power dissipated in $R_2$, $P_2 = \frac{V_2^2}{R_2} = \frac{(7.5\, V)^2}{50\, \Omega} = 1.125\, W$.

The Current of 0