Chapter 8, Problem 1RQ

Referto Figure 8-2. Replace the values shown with the following. Solvefor all theunknownvalues.

$\begin{array}{l}{\text{I}}_{\text{T}}\text{= 0}\text{.6 A}\hfill \\ {\text{R}}_{\text{1}}\text{= 470 }\Omega \hfill \\ {\text{R}}_{\text{2}}\text{= 360 }\Omega \hfill \\ {\text{R}}_{\text{3}}\text{= 510 }\Omega \hfill \\ {\text{R}}_{\text{4}}\text{= 430 }\Omega \hfill \end{array}$

To determine

Solve for the unknown values in the figure.

Answer to Problem 1RQ

ET = 264.47 V	E1 = 149.87 V	E2 = 114.79 V	E3 = 143.48 V	E4 = 120.98 V
IT = 0.6 A	I1 = 0.3186 A	I2 = 0.3186 A	I3 = 0.2813 A	I4 = 0.2813 A
RT = 440.79 Ω	R1 = 470 Ω	R2 = 360 Ω	R3 = 150 Ω	R4 = 250Ω

Explanation of Solution

The circuit diagram is redrawn with given data;

Description:

For the given combination circuit, total resistance will be series combination of R₁ and R₂ in parallel combination with series combination of R₃ and R_4. Calculations are shown below,

$\begin{array}{l}{R}_T=\frac{1}{\frac{1}{(R_1+R_2)}+\frac{1}{(R_3+R_4)}}\\ R_T=\frac{1}{\frac{1}{(470+360)}+\frac{1}{(510+430)}}\\ R_T=440.79\Omega \end{array}$

Using the current I_T and resistance R_T, we calculate the voltage E_T

$E_T=I_T{R}_T=(0.6)(440.79)=264.47\text{ V}$

The individual voltage drops can be calculated using the voltage divider rule.

$\begin{array}{l}{E}_1=E_T\frac{R_1}{R_1+R_2}=264.47(\frac{470}{470+360})=149.87\text{ V}\\ E_1=E_T\frac{R_2}{R_1+R_2}=264.47(\frac{360}{470+360})=114.79\text{ V}\\ E_3=E_T\frac{R_3}{R_3+R_4}=264.47(\frac{510}{510+430})=143.48\text{ V}\\ E_4=E_T\frac{R_4}{R_3+R_4}=264.47(\frac{430}{510+430})=120.98\text{ V}\end{array}$

Since R₁ and R₂ are series connected, the same current flows through them. Hence, I₁=I₂

Similarly, R₃ and R₄ are series connected, therefore I₃=I₄

$\begin{array}{l}{I}_1=I_2=\frac{E_T}{R_1+R_2}=\frac{264.47}{470+360}=0.3186\text{ A}\\ I_3=I_4=\frac{E_T}{R_3+R_4}=\frac{264.47}{510+430}=0.2813\text{ A}\end{array}$

Conclusion:

The unknown values have been calculated using rules for combination circuit and ohm’s law.