Chapter 28, Problem 39SP

Referring to the circuit in Fig. 28-22, determine (a) the equivalent resistance between terminals A and B. If a 15.0-V dc power supply were placed across A and B, (b) how much current would flow through the 1.0- $\Omega$ resistor? (c) Calculate the net power that would be dissipated by all the resistors.

Fig. 28-22

To determine

The equivalent resistance between terminals A and B of the circuit given in Fig. 28-22.

Answer to Problem 39SP

Solution:

$5.0\Omega$

Explanation of Solution

Given data:

Refer to the circuit given in Fig. 28-22.

The power supply between terminals A and B is $15\text{ V}$ dc.

Formula used:

The expression for equivalent resistance in a parallel connection is,

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

Here, $R_1$ and $R_2$ are the resistors connected in parallel and $R_{eq}$ is the equivalent resistance.

The expression for the equivalent resistance in a series connection is,

$R_{eq}=R_1+R_2$

Here, $R_1$ and $R_2$ are the resistors connected in series and $R_{eq}$ is the equivalent resistance.

The current always flows in a closed path and follows the minimum resistance path.

Explanation:

Figure 1 shows the direction of current flow in the circuit.

There are seven nodes a, b, c, d, e, f, and g as shown in Figure 1.

The nodes a and c are shorted, and thus there is no resistance between a and c.

Calculate the equivalent resistance between nodes a and b.

The $4\Omega$ and $4\Omega$ resistances are connected in parallel between nodes a and b.

Recall the expression for equivalent resistance in a parallel connection.

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

Here, $R_1$ and $R_2$ are the resistors connected in parallel and $R_{eq}$ is the equivalent resistance.

Substitute $4\Omega$ for $R_1$, $4\Omega$ for $R_2$, and $R_{ab}$ for $R_{eq}$

$\begin{array}{l}\frac{1}{R_{ab}}=\frac{1}{4\Omega }+\frac{1}{4\Omega }\\ \frac{1}{R_{ab}}=\frac{2}{4\Omega }\\ R_{ab}=2\Omega \end{array}$

The equivalent resistance between nodes b and d is zero because nodes b and d are shorted. The $4\Omega$ and $2\Omega$ resistors between nodes b and d are the ineffective resistors in the circuit, and there is no current flow in the circuit because nodes b and d are shorted.

The equivalent resistance between nodes b and d is not in the circuit, because terminal d is open-circuited.

The $2\Omega$ and $2\Omega$ resistors are connected in series between nodes d and e.

Calculate the equivalent resistance between nodes d and e.

Recall the expression for the equivalent resistance in a series connection.

$R_{eq}=R_1+R_2$

Here, $R_1$ and $R_2$ are the resistors connected in series and $R_{eq}$ is the equivalent resistance.

Substitute $2\Omega$ for $R_1$, $2\Omega$ for $R_2$, and $R_{de}$ for $R_{eq}$

$\begin{array}{c}{R}_{de}=2\Omega +2\Omega \\ =4\Omega \end{array}$

The equivalent resistance between nodes e and c is not in the circuit, because terminal c is open-circuited.

The $2.5\Omega$ and $2.5\Omega$ resistors are connected in series with $5\Omega$ resistance parallel between nodes e and f.

Calculate $R_S$, the series combination of the $2.5\Omega$ and $2.5\Omega$ resistors.

Recall the expression for the equivalent resistance in a series connection.

$R_{eq}=R_1+R_2$

Here, $R_1$ and $R_2$ are the resistors connected in series and $R_{eq}$ is the equivalent resistance.

Substitute $2.5\Omega$ for $R_1$, $2.5\Omega$ for $R_2$, and $R_S$ for $R_{eq}$

$\begin{array}{c}{R}_S=2.5\Omega +2.5\Omega \\ =5\Omega \end{array}$

Now the $R_S=5\Omega$ and $5\Omega$ resistances are connected in parallel.

Calculate the equivalent resistance between nodes e and f.

Recall the expression for equivalent resistance in a parallel connection.

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

Here, $R_1$ and $R_2$ are the resistors connected in parallel and $R_{eq}$ is the equivalent resistance.

Substitute $5\Omega$ for $R_1$, $5\Omega$ for $R_2$, and $R_{ef}$ for $R_{eq}$

$\begin{array}{c}\frac{1}{R_{ef}}=\frac{1}{5\Omega }+\frac{1}{5\Omega }\\ \frac{1}{R_{ef}}=\frac{2}{5\Omega }\\ R_{ef}=2.5\Omega \end{array}$

Now the $R_{ef}$ and $0.5\Omega$ resistors are connected in series with $6\Omega$ resistance parallel between nodes e and g.

Calculate $R_{S1}$, the series combination of the $R_{ef}$ and $0.5\Omega$ resistors.

Recall the expression for the equivalent resistance in a series connection.

$R_{eq}=R_1+R_2$

Here, $R_1$ and $R_2$ are the resistors connected in series and $R_{eq}$ is the equivalent resistance.

Substitute $R_{ef}$ for $R_1$, $0.5\Omega$ for $R_2$, and $R_{S1}$ for $R_{eq}$.

$R_{S1}=R_{ef}+0.5\Omega$

Substitute $2.5\Omega$ for $R_{ef}$.

$\begin{array}{c}{R}_{S1}=2.5\Omega +0.5\Omega \\ =3.0\Omega \end{array}$

Now the $R_{S1}=3\Omega$ and $6\Omega$ resistances are connected in parallel.

Calculate the equivalent resistance between nodes e and g.

Recall the expression for equivalent resistance in a parallel connection.

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

Here, $R_1$ and $R_2$ are the resistors connected in parallel and $R_{eq}$ is the equivalent resistance.

Substitute $3\Omega$ for $R_1$, $6\Omega$ for $R_2$, and $R_{eg}$ for $R_{eq}$

$\begin{array}{l}\frac{1}{R_{eg}}=\frac{1}{3\Omega }+\frac{1}{\text{6 }\Omega }\\ \frac{1}{R_{eg}}=\frac{3}{\text{6 }\Omega }\\ R_{eg}=2\Omega \end{array}$

Figure 2 shows the reduced form of Figure 1.

Calculate the equivalent resistance between nodes a and e.

Nodes b and d are shorted.

The resistor $R_{ab}$ and $R_{de}$ are connected in series, with $3\Omega$ resistance parallel between nodes a and e, shown in figure 2.

Calculate $R_{S2}$, the series combination of the resistors $R_{ab}$ and $R_{de}$

Recall the expression for the equivalent resistance in a series connection.

$R_{eq}=R_1+R_2$

Here, $R_1$ and $R_2$ are the resistors connected in series and $R_{eq}$ is the equivalent resistance.

Substitute $R_{ab}$ for $R_1$, $R_{de}$ for $R_2$, and $R_{S2}$ for $R_{eq}$

$R_{S2}=R_{ab}+R_{de}$

Again substitute $2\Omega$ for $R_{ab}$, $\text{4 }\Omega$ for $R_{de}$.

$\begin{array}{c}{R}_{S2}=2\Omega +\text{4 }\Omega \\ =6\Omega \end{array}$

Now, the $3\Omega$ and $R_{S2}=6\Omega$ resistances are connected in parallel.

Calculate the equivalent resistance between nodes a and e.

Recall the expression for equivalent resistance in a parallel connection.

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

Here, $R_1$ and $R_2$ are the resistors connected in parallel and $R_{eq}$ is the equivalent resistance.

Substitute $3\Omega$ for $R_1$, $6\Omega$ for $R_2$, and $R_{ae}$ for $R_{eq}$

$\begin{array}{l}\frac{1}{R_{ae}}=\frac{1}{3\Omega }+\frac{1}{\text{6 }\Omega }\\ \frac{1}{R_{ae}}=\frac{3}{\text{6 }\Omega }\\ R_{ae}=2\Omega \end{array}$

Figure 3 shows the reduced form of Figure 2.

Calculate the equivalent resistance between terminals A and B.

The resistances $R_{ae}$, $R_{eg}$, and $1\Omega$ are connected in series.

Recall the expression for the equivalent resistance in a series connection.

$R_{eq}=R_1+R_2+R_3$

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

Substitute $R_{ae}$ for $R_1$, $R_{eg}$ for $R_2$, $1.0\Omega$ for $R_3$, and $R_{\text{AB}}$ for $R_{eq}$

$R_{\text{AB}}=R_{ae}+R_{eg}+1.0\Omega$

Substitute $2.0\Omega$ for $R_{ae}$ and $2.0\Omega$ for $R_{eg}$.

$\begin{array}{c}{R}_{\text{AB}}=2.0\Omega +2.0\Omega +1.0\Omega \\ =5.0\Omega \end{array}$

Conclusion:

Therefore, the equivalent resistance between terminals A and B is $5.0\Omega$.

To determine

The amount of the current flow from the $1.0\Omega$ resistor in the circuit given in Fig. 28-22.

Answer to Problem 39SP

Solution:

$3.0\text{ A}$

Explanation of Solution

Given data:

Refer to Figure 3.

The power supply between terminals A and B is $15.0\text{ V}$ dc.

The equivalent resistance between terminals A and B is $5.0\Omega$.

Formula used:

Write the expression for current from Ohm’s law.

$I=\frac{V}{R}$

Here, $R$ is the resistance of the resistor, $V$ is the potential difference across the resistor, and $I$ is the current through the resistor.

Explanation:

Calculate the amount of the current flow through the $1.0\Omega$ resistor.

The current flow through the $1.0\Omega$ resistor is equal to the total current flow in the circuit shown in Figure 3.

Recall the expression for current from Ohm’s law.

$I=\frac{V}{R}$

Here, $R$ is the resistance of the resistor, $V$ is the potential difference across the resistor, and $I$ is the current through the resistor.

Substitute $15.0\text{ V}$ for $V$ and $5.0\Omega$ for $R$.

$\begin{array}{c}I=\frac{15.0\text{ V}}{5.0\Omega }\\ =3.0\text{ A}\end{array}$

Conclusion:

Therefore, the current flow through the $1.0\Omega$ resistor is $3.0\text{ A}$.

To determine

The net power dissipated by all resistors in the circuit given in Fig. 28-22.

Answer to Problem 39SP

Solution:

$45\text{ W}$

Explanation of Solution

Given data:

The power supply between terminals A and B is $15.0\text{ V}$ dc.

The total current flow between terminals A and B is $3.0\text{ A}$.

Formula used:

Write the expression for power.

$P=VI$

Here, $V$ is the supply voltage in the circuit, $P$ is the power, and $I$ is the current flow in the circuit.

The net power dissipated by all resistors in the given circuit is equal to the total power delivered by the power supply because the power supply does not have any internal resistance.

Explanation:

Calculate the total power delivered by the power supply.

Recall the expression for power.

$P=VI$

Here, $V$ is the supply voltage in the circuit, $P$ is the power, and $I$ is the current flow in the circuit.

Substitute $15.0\text{ V}$ for $V$ and $3.0\text{ A}$ for $I$

$\begin{array}{c}P=\left(15.0\text{ V}\right)\left(3.0\text{ A}\right)\\ =45\text{ W}\end{array}$

The total power delivered is equal to the power dissipated by all the resistors.

Conclusion:

Therefore, the power dissipated by all resistors in the given circuit is $45\text{ W}$.