Chapter 28, Problem 31SP

For the circuit shown in Fig. 28-14, find (a) its equivalent resistance; (b) the current drawn from the power source; (c) the potential differences across ab, cd, and de; (d) the current in each resistor.

Fig. 28-14

To determine

The net resistance across the circuit provided in the figure 28-14 in the textbook.

Answer to Problem 31SP

Solution:

$15\text{ Ω}$

Explanation of Solution

Given data:

The resistances $10\text{ Ω}$ and $15\text{ Ω}$ across the $c\text{-}d$ circuit are in parallel combination.

The resistances $\text{9 Ω}$, $18\text{ Ω}$, and $30\text{ Ω}$ across the $d\text{-}e$ circuit are in parallel combination.

The resistance across circuit $a\text{-}b$ across the circuit is $4\text{ Ω}$.

The voltage across the terminal $a\text{ and }e$ is $300\text{V}$.

Formula used:

The expression for the equivalent resistance in series is written as

$R_{eq}=R_1+R_2+R_3+\mathrm{...}$

Here, $R_{eq}$ is the equivalent resistance of the resistances in series combination and $R_1$, $R_2$, and $R_3$ are resistances of the respective resistors.

The expression for the equivalent resistance in parallel is written as

$\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\mathrm{...}$

Here, $R_{eq}$ is the equivalent resistance of the resistances in parallel combination and $R_1$, $R_2$, and $R_3$ are resistances of the respective resistors.

Explanation:

Refer to the fig. 28-14from the textbook.

Draw the circuit diagram:

Since, the resistance $R_1$ and $R_2$ are in parallel combination.

Write the expression for the resistances across $c$-$d$, which are in parallel combination:

$\frac{1}{R_{cd}}=\frac{1}{R_1}+\frac{1}{R_2}$

Here, $R_{cd}$ is the net resistance across the $c$-$d$.

Substitute $10\text{ Ω}$ for $R_1$ and $15\text{ Ω}$ for $R_2$

$\begin{array}{c}\frac{1}{R_{cd}}=\frac{1}{10\text{ Ω}}+\frac{1}{15\text{ Ω}}\\ =\frac{15+10}{\left(10\right)\left(15\right)}\\ =\frac{1}{6}\end{array}$

Solve for $R_{cd}$

$R_{cd}=6\text{ Ω}$

Modify the above drawn circuit diagram by using $R_{cd}$ across the node $c$-$d$:

Understand that the resistors $R_3$, $R_4$, and $R_5$ are parallel combination to each other. Hence, the expression for the resistances across the node $d$-$e$ is written as

$\frac{1}{R_{de}}=\frac{1}{R_3}+\frac{1}{R_4}+\frac{1}{R_5}$

Here, $R_{de}$ is the equivalent resistance across the node $d$-$e$.

Substitute $9\text{ Ω}$ for $R_3$, $18\text{ Ω}$ for $R_4$, and $30\text{ Ω}$ for $R_5$

$\frac{1}{R_{de}}=\frac{1}{\text{9Ω}}+\frac{1}{\text{18Ω}}+\frac{1}{\text{30Ω}}$

Here, $R_{de}$ is the net resistance across the $d$-$e$.

$\begin{array}{c}{R}_{de}=\frac{\left(\text{18 Ω}\right)\left(\text{9 Ω}\right)\left(\text{30 Ω}\right)}{\left(\text{9 Ω}\right)\left(\text{30 Ω}\right)\text{+}\left(\text{30 Ω}\right)\left(\text{18 Ω}\right)+\left(\text{18 Ω}\right)\left(\text{9 Ω}\right)}\\ =\frac{\text{4860 Ω}}{\text{270 Ω+540 Ω}+\text{162 Ω}}\\ =5\text{ Ω}\end{array}$

Redraw the modified circuit diagram by using the calculated equivalent resistances across the nodes:

Since, all the resisters in the above diagram arein series combination.

Write the expression for the total resistance of the equivalent resistances across $a$-b, $c$-$d$, and $d$-$e$ across the circuit:

$R_{eq}=R_{ab}+R_{cd}+R_{de}$

Here, $R_{eq}$ is the equivalent resistance across the circuit,

Substitute $4\text{ Ω}$ for $R_{ab}$, $6\text{ Ω}$ for $R_{cd}$, and $5\text{ Ω}$ for $R_{de}$

$\begin{array}{c}{R}_{eq}=4\text{ Ω}+6\text{ Ω}+5\text{Ω}\\ =\text{15 }\Omega \end{array}$

Conclusion:

The net resistance across the point $a\text{-}e$ of the circuit is $15\text{ Ω}$.

To determine

The current drawn from the source if the voltage across the circuit is $300\text{ V}$.

Answer to Problem 31SP

Solution:

$20\text{ A}$

Explanation of Solution

Given data:

Voltage across the circuit is $300\text{ V}$.

Formula used:

The expression for theOhm`s law if net current $I$ is flowingacross circuit ofresistance $R$ having voltage $V$ is written as

$V=IR$

Explanation:

Redraw the equivalent circuit diagram:

In the above diagram, $I_{ae}$ is the current that flows through the circuit.

Rewrite the expression of the Ohm`s law for the net current across circuit $a\text{-}e$:

$V_{ae}=I_{ae}{R}_{ae}$

Here, $V_{ae}$ is the voltage across the node $a\text{-}e$.

Rearrange for $I_{ae}$

$I_{ae}=\frac{V_{ae}}{R_{eq}}$

Here, $I_{ae}$ is the net resistance across circuit $a\text{-}e$, $V_{ae}$ is the voltage across circuit $a\text{-}e$, and $R_{eq}$ is the equivalent resistance across circuit $a\text{-}e$.

Substitute $300\text{ V}$ for $V_{ae}$ and $15\text{ Ω}$ for $R_{eq}$

$\begin{array}{c}{I}_{ae}=\frac{300\text{ V}}{15\text{ Ω}}\\ =20\text{ A}\end{array}$

Conclusion:

Therefore, the current drawn from the source is $20\text{ A}$.

To determine

The voltage difference across circuits $a\text{-}b$, $c\text{-}d$, and $d\text{-}e$ if the net voltage across the circuit, resistance across each circuit, and current across the circuit are known.

Answer to Problem 31SP

Solution:

$V_{ab}=80\text{ V}$, $V_{cd}=120\text{ V}$, and $V_{de}=100\text{ V}$

Explanation of Solution

Given data:

The current across $a\text{-}e$ is $20\text{ A}$.

The resistances of the resistor are provided in the figure as $4\text{ Ω}$, $10\text{ Ω}$, $15\text{ Ω}$, $9\text{ Ω}$, $18\text{ Ω}$, and $30\text{ Ω}$ .

Formula used:

The expression for the net Voltage $V$ across the circuit of resistance $R$, having net current from the source $I$, is written as

$V=IR$

Explanation:

The current in the series combination of circuits $a\text{-}b$, $c\text{-}d$, and $d\text{-}e$ is same, which is equal to the net current across $a\text{-}e$.

Write the expression for the voltage across circuit $a\text{-}b$:

$V_{ab}=I_{ae}{R}_{ab}$

Here, $V_{ab}$ is the voltage across circuit $a\text{-}b$ and $R_{ab}$ is the resistance across circuit $a\text{-}b$.

Substitute $20\text{ A}$ for $I_{ae}$ and $4\text{ Ω}$ for $R_{ab}$

$\begin{array}{c}{V}_{ab}=\left(20\text{ A}\right)\left(4\text{ Ω}\right)\\ =\text{80 V}\end{array}$

Write the expression for the voltage across circuit $c\text{-}d$:

$V_{bc}=I_{ae}{R}_{bc}$

Here, $V_{bc}$ is the voltage across circuit $b\text{-}c$ and $R_{bc}$ is the resistance across circuit $b\text{-}c$.

Substitute $20\text{ A}$ for $I_{ae}$ and $6\text{ Ω}$ for $R_{bc}$

$\begin{array}{c}{V}_{bc}=\left(20\text{ A}\right)\left(6\text{ Ω}\right)\\ =120\text{ V}\end{array}$

Write the expression for the voltage across circuit $d\text{-}e$:

$V_{de}=I_{ae}{R}_{de}$

Here, $V_{de}$ is the voltage across circuit $d\text{-}e$ and $R_{de}$ is the resistance across circuit $d\text{-}e$.

Substitute $20\text{ A}$ for $I_{ae}$ and $5\text{ Ω}$ for $R_{ab}$

$\begin{array}{c}{V}_{de}=\left(20\text{ A}\right)\left(5\text{ Ω}\right)\\ =10\text{0 V}\end{array}$

Conclusion:

The voltages across circuits $a\text{-}b$ is $80\text{ V}$, across $c\text{-}d$ is $120\text{ V}$, andacross $d\text{-}e$ is $100\text{ V}$.

To determine

The current across each resistor if the voltage and current across each circuit are known.

Answer to Problem 31SP

Solution:

$I_4=20\text{A}$, $I_{10}=12\text{A}$, $I_{15}=8\text{A}$, $I_9=11.1\text{A}$, $I_{18}=5.6\text{A}$, and $I_{30}=3.3\text{A}$

Explanation of Solution

Given data:

The voltage across the circuit $a\text{-}b$ is $80\text{ V}$ .

The voltage across the circuit $c\text{-}d$ is $120\text{ V}$ .

The voltage across the circuit $d\text{-}e$ is $100\text{ V}$ .

The current across the circuit $a\text{-}b$, $c\text{-}d$, and $d\text{-}e$ is $20\text{ A}$.

Formula used:

The expression for the net voltage $V$ across the circuit of resistance $R$, having net current from the source $I$, is written as

$V=IR$

Explanation:

In this, wewill use the fact that voltage across each resistor in a parallel combination of resistors remains same.

According to the Ohm’s law, the current across the $4\text{ Ω}$ resistor is

$V_{ab}=I_4{R}_{ab}$

Rearrange for $I_4$

$I_4=\frac{V_{ab}}{R_{ab}}$

Here $I_4$ is the current across $4\text{ Ω}$ resistor.

Substitute $4\text{ Ω}$ for $R_{ab}$ and $80\text{ V}$ for $V_{ab}$

$\begin{array}{c}{I}_4=\frac{80\text{ V}}{\left(4\text{ Ω}\right)}\\ =20\text{ A}\end{array}$

The voltage across the parallel combination of resistance $10\text{ Ω}$ and $15\text{ Ω}$ is equal to $V_{cd}$.

Write the expression for the current across $10\text{Ω}$ resistor:

$V_{cd}=I_{10}{R}_1$

Rearrange for $I_{10}$

$I_{10}=\frac{V_{cd}}{R_1}$

Here, $I_{10}$ is the current across $10\text{ Ω}$ resistor.

Substitute $120\text{ V}$ for $V_{ab}$ and $10\text{ Ω}$ for $R_1$

$\begin{array}{c}{I}_{10}=\frac{120\text{ V}}{\left(\text{10 Ω}\right)}\\ =12\text{ A}\end{array}$

Write the expression for the current across $15\text{Ω}$ resistor:

$V_{cd}=I_{15}{R}_2$

Rearrange for $I_{15}$

$I_{15}=\frac{V_{cd}}{R_2}$

Here, $I_{15}$ is the current across $15\text{ Ω}$ resistor.

Substitute $120\text{ V}$ for $V_{ab}$ and $15\text{ Ω}$ for $R_2$

$\begin{array}{c}{I}_{15}=\frac{120\text{ V}}{\left(\text{15 Ω}\right)}\\ =8\text{ A}\end{array}$

The voltage across the parallel combination of resistors $30\text{ Ω}$, $9\text{ Ω}$, and $18\text{ Ω}$ is equal to $V_{de}$.

Write the expression for the current across $9\text{Ω}$ resistor:

$V_{de}=I_9{R}_3$

Rearrange for $I_9$

$I_9=\frac{V_{de}}{R_3}$

Here, $I_9$ is the current across $9\text{ Ω}$ resistor.

Substitute $100\text{ V}$ for $V_{de}$ and $9\text{ Ω}$ for $R_3$

$\begin{array}{c}{I}_9=\frac{100\text{ V}}{\left(\text{9 Ω}\right)}\\ =11.11\text{ A}\end{array}$

Write the expression for the current across $18\text{Ω}$ resistor:

$V_{de}=I_{18}{R}_4$

Rearrange for $I_{18}$

$I_{18}=\frac{V_{de}}{R_4}$

Here, $I_{18}$ is the current across $18\text{ Ω}$ resistor.

Substitute $100\text{ V}$ for $V_{de}$ substitute $R_4$ for $18\Omega$

$\begin{array}{c}{I}_{18}=\frac{100\text{ V}}{\left(\text{18 Ω}\right)}\\ =5.56\text{ A}\end{array}$

Write the expression for the current across $30\text{Ω}$ resistor:

$V_{de}=I_{30}{R}_5$

Rearrange for $I_{30}$

$I_{30}=\frac{V_{de}}{R_5}$

Here, $I_{30}$ is the current across $30\text{ Ω}$ resistor.

Substitute $100\text{ V}$ for $V_{de}$ Substitute $30\text{ Ω}$ for

$\begin{array}{c}{I}_{30}=\frac{100\text{ V}}{\left(\text{30 Ω}\right)}\\ =3.33\text{ A}\end{array}$

Conclusion:

The current across the $4\text{ Ω}$ resistance is $\text{20 A}$, across $10\text{ Ω}$ is $12\text{ A}$, across $15\text{ Ω}$ is $8\text{ A}$, across $9\text{ Ω}$ is $11.11\text{ A}$, across $18\text{ Ω}$ is $5.56\text{ A}$, and across $30\text{ Ω}$ is $3.33\text{ A}$.