Chapter 7, Problem 4PP

Using the rules for parallel circuits and Ohm’slaw, solve for the missing values.

$\begin{array}{lllll}{\text{E}}_{\text{T}}\hfill & {\text{E}}_{\text{1}}\hfill & {\text{E}}_{\text{2}}\hfill & {\text{E}}_{\text{3}}\hfill & {\text{E}}_{\text{4}}\hfill \\ {\text{I}}_{\text{T}}\hfill & {\text{l}}_{\text{1}}\hfill & {\text{I}}_{\text{2}}\hfill & {\text{l}}_{\text{3}}\hfill & {\text{l}}_{\text{4}}\hfill \\ {\text{R}}_{\text{T}}\hfill & {\text{R}}_{\text{1}}\text{82 k}\Omega \hfill & {\text{R}}_{\text{2}}\text{75 k}\Omega \hfill & {\text{R}}_{\text{3}}\text{56 k}\Omega \hfill & {\text{R}}_{\text{4}}\text{62 k}\Omega \hfill \\ {\text{P}}_{\text{T}}\text{3}\text{.436 W}\hfill & {\text{P}}_{\text{1}}\hfill & {\text{P}}_{\text{2}}\hfill & {\text{P}}_{\text{3}}\hfill & {\text{P}}_{\text{4}}\hfill \end{array}$

To determine

The missing values in the given table using rules for parallel circuit and ohm’s law.

Answer to Problem 4PP

ET = 240.24 V	E1 = 240.24 V	E2 = 240.24 V	E3 = 240.24 V	E4 = 240.24 V
IT = 14.30 mA	I1 = 2.92 mA	I2 = 3.20 mA	I3 = 4.29 mA	I4 = 3.87 mA
RT= 16.80 kΩ	R1 = 82kΩ	R2 = 75kΩ	R3 = 56kΩ	R4 = 62 kΩ
PT = 3.436W	P1 = 0.6991 W	P2 = 0.768 W	P3 = 1.0306 W	P4 = 0.9285 W

Explanation of Solution

Description:

In a parallel circuit, the total resistance is equal to the reciprocal of the sum of reciprocal resistances of the individual branches.

$R_T=\frac{1}{\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\frac{1}{R_4}}$

Substitute the values and compute the unknown resistor R_T

$\begin{array}{l}{R}_T=\frac{1}{\frac{1}{82k}+\frac{1}{75k}+\frac{1}{56k}+\frac{1}{62k}}\\ R_T=\frac{1}{\frac{65100+71176+95325+86100}{5338200k}}\\ R_T=16802.6\Omega \\ R_T=16.80k\Omega \end{array}$

Using the values for total power dissipation and total resistance, we can calculate the total current I_T

$\begin{array}{l}{P}_T=I_T^2{R}_T\\ 3.436={(}^{I_T}.(16.80\text{ x }{10}^3)\\ {(}^{I_T}=\frac{3.436}{(16.80\text{ x }{10}^3)}=2.04{\text{ x 10}}^{-4}\\ I_T=\sqrt{2.04{\text{ x 10}}^{-4}}\text{ = 0}\text{.01430 A}\\ I_T=\text{ 14}\text{.30 mA}\end{array}$

Using the Total current and Total resistance, we calculate the Total voltage E_T

$E_T=I_T{R}_T=(14.30\text{ x }{10}^{-3})(16.80{\text{ x 10}}^3)=240.24\text{ V}$

By rule of parallel circuits, the total voltage drop across any branch of a parallel circuit is the same as the applied voltage. Therefore,

$E_1=E_2=E_3=E_4=E_T=240.24\text{ V}$

The current flowing in individual branches can be calculated from the voltage and resistor values. Therefore,

$\begin{array}{l}{I}_1=\frac{E_1}{R_1}=\frac{240.24}{82\text{x}{10}^3}=2.92\text{ mA}\\ I_2=\frac{E_2}{R_2}=\frac{240.24}{75\text{x}{10}^3}=3.20\text{ mA}\\ I_3=\frac{E_3}{R_3}=\frac{240.24}{56\text{x}{10}^3}=4.29\text{ mA}\\ I_4=\frac{E_4}{R_4}=\frac{240.24}{62\text{x}{10}^3}=3.\text{ 87 mA}\end{array}$

The power consumed by the resistors is given as,

$\begin{array}{l}{P}_1=I_1^2{R}_1={(}^{2.92}.(82{\text{ x 10}}^3)=\text{0}\text{.6991W}\\ P_2=I_2^2{R}_2={(}^{3.20}.(75{\text{ x 10}}^3)=\text{0}\text{.768 W}\\ P_3=I_3^2{R}_3={(}^{4.29}.(56{\text{ x 10}}^3)=\text{ 1}\text{.0306 W}\\ P_4=I_4^2{R}_4={(}^{3.87}.(62{\text{ x 10}}^3)=\text{ 0}\text{.9285 W}\end{array}$

Conclusion:

The missing values in the given table have been calculated using rules for parallel circuit and ohm’s law.