Chapter 8, Problem

${\text{E}}_{\text{T}}$		${\text{E}}_{\text{1}}$	${\text{E}}_{\text{2}}$		${\text{E}}_{\text{3}}$			${\text{E}}_{\text{4}}\text{1}\text{.248 V}$
${\text{I}}_{\text{T}}$		${\text{I}}_{\text{1}}$	${\text{I}}_{\text{2}}$		${\text{I}}_{\text{3}}$			${\text{I}}_{\text{4}}$
${\text{R}}_{\text{T}}$		${\text{R}}_{\text{1}}$	${\text{R}}_{\text{2}}$		${\text{R}}_{\text{3}}$			${\text{R}}_{\text{4}}$
${\text{P}}_{\text{T}}\text{0}\text{.576 W}$		${\text{P}}_{\text{1}}\text{0}\text{.0806 W}$	${\text{P}}_{\text{2}}\text{0}\text{.0461 W}$		${\text{P}}_{\text{3}}\text{0}\text{.00184 W}$			${\text{P}}_{\text{4}}$
${\text{E}}_{\text{5}}$	${\text{E}}_{\text{6}}$			${\text{E}}_{\text{7}}$		${\text{E}}_{\text{8}}$	${\text{E}}_{\text{9}}$
${\text{I}}_{\text{5}}$	${\text{I}}_{\text{6}}$			${\text{I}}_{\text{7}}$		${\text{I}}_{\text{8}}$	${\text{I}}_{\text{9}}$
${\text{R}}_{\text{5}}$	${\text{R}}_{\text{6}}$			${\text{R}}_{\text{7}}$		${\text{R}}_{\text{8}}$	${\text{R}}_{\text{9}}$
${\text{P}}_{\text{5}}\text{0}\text{.0203 W}$	${\text{P}}_{\text{6}}\text{0}\text{.00995 W}$			${\text{P}}_{\text{7}}\text{0}\text{.0518 W}$		${\text{P}}_{\text{8}}\text{0}\text{.0726 W}$	${\text{P}}_{\text{9}}\text{0}\text{.288 W}$

FIGURE 8-26 A combination circuit.

To determine

The unknown values in the circuit.

Answer

ET = 24.04 V	E1 = 3.4V	E2 = 4.8 V	E3 = 0.478 V	E4 = 1.248 V
IT = 24.02mA	I1 = 24.02mA	I2 = 9.614 mA	I3 = 3.854 mA	I4 = 3.854 mA
RT = 942 Ω	R1 = 144 Ω	R2 = 499 Ω	R3 = 124Ω	R4 = 324 Ω
PT = 0.576 W	P1 = 0.0806W	P2 = 0.0461W	P3 = 0.00184W	P4 = 0.00481W

E5 = 2.11 V	E6 = 1.726V	E7 = 3.6 V	E8 = 5.04V	E9 = 12 V
I5 = 9.614 mA	I6 = 5.76 mA	I7 = 14.41 mA	I8 = 14.41 mA	I9 = 24.02 mA
R5 = 220 Ω	R6 = 300Ω	R7 = 86Ω	R8 = 360Ω	R9 = 500 Ω
P5 = 0.0203 W	P6 = 0.00995 W	P7 = 0.0518W	P8 = 0.0726W	P9 = 0.288 W

Explanation of Solution

In the given question,

• Current flowing through R1 is denoted by I₁

• Current flowing through R2is denoted by I₂

• Current flowing through R3 and R4 is denoted by I₃(= I₄)

• Current flowing through R5 is denoted by I₅

• Current flowing through R6 is denoted by I₆

• Current flowing through R7 and R8 is denoted by I₇ (= I₈)

• Current flowing through R9is denoted by I₉

The total power consumed by the circuit is equal to sum of power consumed by individual resistors. Hence,

$\begin{array}{l}{P}_T=P_1+P_2+P_3+P_4+P_5+P_6+P_7+P_8+P_9\\ P_4=P_T-\left(P_1+P_2+P_3+P_5+P_6+P_7+P_8+P_9\right)\\ =0.576-\left(0.0806+0.0461+0.00184+0.0203+0.00995+0.0518+0.0726+0.288\right)\\ =0.576-\left(0.5712\right)\\ =0.00481\text{ W}\end{array}$

Using the voltage E₄ and Power P_4, we calculate resistance R₄ and current I₄

$\begin{array}{l}{I}_4=\frac{P_4}{E_4}=\frac{0.00481}{1.248}=3.854\text{mA}\\ \\ R_4=\frac{{\left(E_4\right)}^2}{P_4}=\frac{{\left(1.248\right)}^2}{0.00481}=324\Omega \end{array}$

Since R₃ and R₄ are in series, the same current flows through both. Therefore, I₃=I₄=3.854 mA

$R_3=\frac{P_3}{{\left(I_3\right)}^2}=\frac{0.00184}{{\left(3.854\text{x}{10}^{-3}\right)}^2}=124\Omega$

Using the current I₃ and resistance R₃ find the voltage E₃

$E_3=I_3{R}_3=\left(3.854{\text{ x 10}}^{\text{-3}}\right)\left(124\right)=0.478\text{ V}$

Since R6 is in parallel with series combination of R3 and R4. Hence,

E₆= 0.478 + 1.248 =1.726 V

Using the voltage E₆ and Power P_6, we calculate resistance R₆ and current I₆

$\begin{array}{l}{I}_6=\frac{P_6}{E_6}=\frac{0.00995}{1.726}=5.76\text{mA}\\ \\ R_6=\frac{{\left(E_6\right)}^2}{P_6}=\frac{{\left(1.726\right)}^2}{0.00995}=300\Omega \end{array}$

At node B,

$\begin{array}{l}{I}_2=I_3+I_6\\ =\left(\text{3}\text{.854 + 5}\text{.76}\right)\text{x}{10}^{-3}\\ =9.614\text{ mA}\end{array}$

Using P₂ and I₂, we calculate R₂

$R_2=\frac{P_2}{{\left(I_2\right)}^2}=\frac{0.0461}{{\left(9.614\text{ x }{10}^{-3}\right)}^2}=499\Omega$

Using P₂ and I₂, we calculate E₂

$E_2=\frac{P_2}{I_2}=\frac{0.0461}{9.614\text{x}{10}^{-3}}=4.8\text{V}$

Since the incoming current in a network is equal to the outgoing current, current at node B= current at node C: I₂=I₅=9.614 mA

Using the voltage E₅ and Power P_5, we calculate resistance R₅ and current I₅

$\begin{array}{l}{E}_5=\frac{P_5}{I_5}=\frac{0.0203}{9.614\text{x}{10}^{-3}}=2.11\text{V}\\ \\ R_5=\frac{{\left(E_5\right)}^2}{P_5}=\frac{{\left(2.11\right)}^2}{0.0203}=220\Omega \end{array}$

The total voltage drop across R7 and R8 i.e. E₇ and E₈ will be

$E_2+E_5+E_6=4.8+2.11+1.726=8.63\text{ V}$

The total power consumed in the branch containing resistors R7 and R8 will be 0.0518+0.0726 = 0.1244 W

Using the total power consumed and the total voltage, we can compute the total current in the network of R7 and R8.

$I_7=I_8=\frac{P_8}{E_8}=\frac{0.1244}{8.63}=14.41\text{ mA}$

Using the voltage and Power, we calculate R₇, R₈ and current I₇, I₈

$\begin{array}{l}{E}_7=\frac{P_7}{I_7}=\frac{0.0518}{14.41\text{x}{10}^{-3}}=3.6\text{V}\\ \\ R_7=\frac{{\left(E_7\right)}^2}{P_7}=\frac{{\left(2.11\right)}^2}{0.0518}=86\Omega \end{array}$

$\begin{array}{l}{E}_8=\frac{P_8}{I_8}=\frac{0.0726}{14.41\text{x}{10}^{-3}}=5.04\text{V}\\ \\ R_8=\frac{{\left(E_8\right)}^2}{P_8}=\frac{{\left(5.04\right)}^2}{0.0726}=350\Omega \end{array}$

At node A,

$\begin{array}{l}{I}_T=I_1=I_2+I_7\\ =9.614+14.41\\ =24.02\text{ mA}\end{array}$

Since the incoming current in a network is equal to the outgoing current, current at node A=current at node D: I₁ = I₉ =I_T=24.02 mA

Using the current I₁ and Power P_1, we calculate resistance R₁ and voltage E₁

$\begin{array}{l}{E}_T=E_1+E_7+E_8+E_9\\ =3.4+3.6+5.04+12\\ =24.04\text{ V}\end{array}$

Using the current I₉ and Power P_9, we calculate resistance R₉ and voltage E₉

$\begin{array}{l}{E}_9=\frac{P_9}{I_9}=\frac{0.288}{24.02\text{x}{10}^{-3}}=12\text{ V}\\ \\ R_9=\frac{{\left(E_9\right)}^2}{P_9}=\frac{{\left(12\right)}^2}{0.288}=500\Omega \end{array}$

Total voltage applied across the network will be,

$\begin{array}{l}{E}_T=E_1+E_7+E_8+E_9\\ =3.4+3.6+5.04+12\\ =24.04\text{ V}\end{array}$

$\begin{array}{l}{R}_T=\left(R_1+R_9\right)+\left\{\left\{\left(\left(R_3+R_4\right)//R_6\right)+R_2+R_5\right\}//\left(R_7+R_8\right)\right\}\\ \text{ =}\left(R_1+R_9\right)+\frac{1}{\frac{1}{\left\{\left(\frac{1}{\frac{1}{R_3+R_4}+\frac{1}{R_6}}\right)+\left(R_2+R_5\right)\right\}}+\frac{1}{\left(R_7+R_8\right)}}\\ =\left(144+500\right)+\frac{1}{\frac{1}{\left\{\left(\frac{1}{\frac{1}{124+324}+\frac{1}{300}}\right)+\left(499+220\right)\right\}}+\frac{1}{\left(86+360\right)}}\\ =\left(644\right)+\frac{1}{\frac{1}{\left\{\left(\frac{1}{\frac{1}{448}+\frac{1}{300}}\right)+\left(719\right)\right\}}+\frac{1}{\left(446\right)}}\\ =\left(644\right)+\frac{1}{\frac{1}{\left\{180+\left(719\right)\right\}}+\frac{1}{\left(446\right)}}\\ =\left(644\right)+\frac{1}{\frac{1}{\left\{899\right\}}+\frac{1}{\left(446\right)}}\\ =\left(644\right)+298\\ =942\Omega \end{array}$