Chapter 20, Problem 1P

For the battery of bulbs (purely resistive) appearing in Fig. 20.48 :

a. Determine the total power dissipation.

b. Calculate the total reactive and apparent power.

c. Find the source current I_s.

d. Calculate the resistance of each bulb for the specified operating conditions.

e. Determine the currents I₁ and I₂.

To determine

(a)

The total power dissipation.

Answer to Problem 1P

The total power dissipated is $130\text{W}$.

Explanation of Solution

Calculation:

The given circuit diagram is shown in Figure 1.

  

The power dissipated in bulb 1 is $60\text{W}$, bulb 2 is $45\text{W}$ andbulb 3 is $25\text{W}$

The total power dissipation is given by the sum of power dissipated in individual bulbs, that is,

  $P_T=P_1+P_2+P_3$

Here,

  $P_1$ is the power dissipated in bulb 1.

  $P_2$ is the power dissipated in bulb 2.

  $P_3$ is the power dissipated in bulb 3.

Substitute $60\text{W}$ for $P_1$, $45\text{W}$ for $P_2$ and $25\text{W}$ for $P_3$ in the above equation.

  $\begin{array}{c}{P}_T=\left(60+45+25\right)\text{W}\\ =130\text{W}\end{array}$

Conclusion:

Therefore, the total power dissipated is $130\text{W}$.

To determine

(b)

The total reactive and apparent power.

Answer to Problem 1P

The reactive power dissipated in the bulbs is $0\text{VAR}$ and the apparent power is $130\text{VA}$.

Explanation of Solution

Calculation:

As the bulbs are purely resistive in nature therefore, the reactive power dissipated in bulb is zero that is,

  $Q_T=0\text{VAR}$

The apparent power is given by,

  $S_T=\sqrt{P_T{}^2+Q_T{}^2}$

Substitute $130\text{W}$ for $P_T$ and $0\text{VAR}$ for $Q_T$ in the above equation.

  $\begin{array}{c}{S}_T=\sqrt{{130}^2+0^2}\\ =\sqrt{{130}^2}\\ =130\text{VA}\end{array}$

Conclusion:

Therefore, the reactive power dissipated in bulb is $0\text{VAR}$ and apparent power is $130\text{VA}$.

To determine

(c)

The source current $I_S$.

Answer to Problem 1P

The source current $I_S$ is $0.54\text{A}$.

Explanation of Solution

Calculation:

The source voltage $\text{E}$ is $240\text{V}$.

The apparent power is given by,

  $S=E{I}_S$

Substitute $130\text{VA}$ for $S$ and $240\text{V}$ for $E$ in the above equation.

  $\begin{array}{l}130=240I_S\\ I_S=0.54\text{A}\end{array}$

Conclusion:

Therefore, the source current $I_S$ is $0.54\text{A}$.

To determine

(d)

The resistance of each bulb.

Answer to Problem 1P

The resistance of bulb 1 is $205.76\Omega$, bulb 2 is $369.11\Omega$ and bulb 3 is $664.4\Omega$.

Explanation of Solution

Calculation:

The power dissipated in first bulb is given by,

  $P_1=I_S{}^2{R}_1$

Substitute $60\text{W}$ for $P_1$ and $0.54\text{A}$ for $I_S$ in the above equation.

  $\begin{array}{l}60={0.54}^2{R}_1\\ R_1=205.76\Omega \end{array}$

The voltage $V_x$ across the first bulb is given by,

  $V_x=I_S{R}_1$

Substitute $0.54\text{A}$ for $I_S$ and $205.76\Omega$ for $R_1$ in the above equation.

  $\begin{array}{c}{V}_x=0.54\times 205.76\\ =111.11\text{V}\end{array}$

From the figure 1 it can be seen that voltage $V_2$ at bulb 2 and voltage $V_3$ at bulb 3 is equal and is given by,

  $\begin{array}{c}{V}_2=V_3\\ =E-V_x\end{array}$

Substitute $111.11\text{V}$ for $V_x$ and $240\text{V}$ for $E$ in the above equation.

  $\begin{array}{c}{V}_2=V_3\\ =\left(240-111.11\right)\text{V}\\ \text{=128}\text{.88}\text{V}\end{array}$

The power dissipated in bulb 2 is given by,

  $P_2=\frac{V_2{}^2}{R_2}$

Substitute $128.88\text{V}$ for $V_2$ and $45\text{W}$ for $P_2$ in the above equation.

  $\begin{array}{l}45=\frac{{128.88}^2}{R_2}\\ R_2=369.11\Omega \end{array}$

The power dissipated in bulb 3 is given by,

  $P_3=\frac{V_3{}^2}{R_3}$

Substitute $128.88\text{V}$ for $V_2$ and $45\text{W}$ for $P_2$ in the above equation.

  $\begin{array}{l}25=\frac{{128.88}^2}{R_2}\\ R_3=664.4\Omega \end{array}$

Conclusion:

Therefore, the resistance of bulb 1 is $205.76\Omega$, bulb 2 is $369.11\Omega$ and bulb 3 is $664.4\Omega$.

To determine

(e)

The current $I_1$ and $I_2$.

Answer to Problem 1P

The current $I_1$ is $0.349\text{A}$ and $I_2$ is $0.191\text{A}$.

Explanation of Solution

Calculation:

The value of current $I_1$ is given by,

  $I_1=\frac{V_2}{R_2}$

Substitute $128.88\text{V}$ for $V_2$ and $369.11\Omega$ for $R_2$ in the above equation.

  $\begin{array}{c}{I}_1=\frac{128.88}{369.11}\\ =0.349\text{A}\end{array}$

The current $I_2$ is given by,

  $I_2=I_S-I_1$

Substitute $0.54\text{A}$ for $I_S$ and $0.349\text{A}$ for $I_1$ in the above equation.

  $\begin{array}{c}{I}_2=0.54-0.349\\ =0.191\text{A}\end{array}$

Conclusion:

Therefore, the current $I_1$ is $0.349\text{A}$ and $I_2$ is $0.191\text{A}$.