$\text { Figure } 6 \text { '1 }$ -If V_S=20 V, R₁=10 $\mathrm { k } \Omega$, R₂=50 $\mathrm { k } \Omega$ and R3=15 $\mathrm { k } \Omega$ in Figure 6-1, P_R2 equals________ A) 7.64 mW B) 2.30 mW C) 8.63 mW D) 18.6 mW

Question

Quizplus · Accepted Answer

First, we can use voltage division to find the voltage across R2: $$V_{R2} = \frac{R2}{R2+R3} \cdot V_S = \frac{15}{15+15} \cdot 20 = 10 ext{ V}$$ Then, we can use Ohm's Law to find the power dissipated by R2: $$P_{R2} = \frac{V_{R2}^2}{R2} = \frac{10^2}{50 ext{ k}\Omega} = 2.00 ext{ mW}$$ However, the resistor network between _and affects the voltage across and current through R2, so we need to take those into account as well. We can first find the total resistance between _and: $$R_{total} = R₁ + \frac{1}{\frac{1}{R₂}+\frac{1}{R2+R3}} = 10 ext{ k}\Omega + \frac{1}{\frac{1}{50 ext{ k}\Omega}+\frac{1}{50 ext{ k}\Omega}} = 60 ext{ k}\Omega$$ Using that, we can find the current through R2: $$I_{R2} = \frac{V_{_S}}{R_{total}} = \frac{20}{60 ext{ k}\Omega} = 0.333\overline{3} ext{ mA}$$ Finally, we can use the current and voltage across R2 to find the power dissipated: $$P_{R2} = I_{R2}^2 \cdot R2 = (0.333\overline{3} ext{ mA})^2 \cdot 50 ext{ k}\Omega = 2.22 ext{ mW}$$ Therefore, the answer is B.

Figure 6 ’1 \text { Figure } 6 \text { '1 } Figure 6 ’1 -If VS=20 V, R1=10

$\text { Figure } 6 \text { '1 }$ -If V_S=20 V, R₁=10