Question

Assume that for a circuit similar to that shown in Problem 2-1,$R_1=200\Omega ,R_2=400\Omega ,R_3=2.00k\Omega ,andV=15.0V.$

(a) What is the voltage V2?

(b) What would be the power loss in resistor R2?

(c) What fraction of the total power lost by the circuit would be dissipated in resistor R2?

Short answer

The voltage across R₂ is 2.31V, the power dissipated is 0.013W and the fraction of power loss across R₂is 15%

Step-by-step solution

Given Information

$R_1=200\Omega ,R_2=400\Omega ,R_3=2.00k\Omega ,andV=15.0V.$

Part (a)

The resistances R₁,R₂ and R₃ are connected in series. So the voltage drop V₂ is:

$$\begin{gathered} V_2=\frac{R_2}{R_1+R_2+R_3}V \\ {V}_2=\frac{400}{2600}\times 15V \\ {V}_2=2.31V \end{gathered}$$

Part (b)

The power loss through the resistor R₂will be:

$$\begin{gathered} P=\frac{V^2}{R} \\ P=\frac{{(2.31)}^2}{400}W \\ P=0.013W \end{gathered}$$

Part (c)

The current in the circuit is given by:

$$\begin{gathered} I=\frac{V}{R} \\ I=\frac{15}{2600}A \\ I=5.77mA \end{gathered}$$

So the total power dissipated by the circuit is :

role="math" localid="1645444352874" $$\begin{gathered} P=VI \\ P=15\times 5.77\times {10}^{-3}W \\ P=0.086W \end{gathered}$$

Now the fraction of power dissipated by R₂is:

$=\frac{13}{86}\times 100=15\%$