Chapter 18, Problem 83QAP

To determine

(a)

The equivalent resistance of given figure.

Answer to Problem 83QAP

The equivalent resistance of the circuit is $16\Omega$.

Explanation of Solution

Given:

The given circuit is shown below.

  

Figure 1
Resistors are as,
  $R_1=20\Omega$

  $R_2=12\Omega$

  $R_3=8\Omega$

Formula used:

For series combination, equivalent resistance can be written as,
  $R_{eq}=R_1+R_2+R_3+\mathrm{......}$

For parallel combination of resistors, equivalent resistance can be written as,
  $\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}+\mathrm{......}$

Calculation:

$R_1$ and $R_2$ are in parallel combination, so equivalent resistance of $R_1$ and $R_2$ will be,
  $\begin{array}{l}\frac{1}{R_{12}}=\frac{1}{R_1}+\frac{1}{R_2}\\ \frac{1}{R_{12}}=\frac{1}{20\Omega }+\frac{1}{12\Omega }=\frac{3+5}{60\Omega }=\frac{8}{60\Omega }=\frac{2}{15\Omega }\\ R_{12}=\frac{15}{2}\Omega =7.5\Omega \end{array}$

Now, $R_{12}$ and $R_3$ are in series, so, equivalent resistance can be written as,
  $\begin{array}{l}{R}_{eq}=R_{12}+R_3\\ R_{eq}=7.5\Omega +8\Omega =15.5\Omega \approx 16\Omega \end{array}$

Conclusion:

Thus, equivalent resistance of the given circuit is $16\Omega$.

To determine

(b)

The current through each resistor

Answer to Problem 83QAP

Current through $R_1=20\Omega$ is $0.12\text{A}.$

Current through $R_2=12\Omega$ is $0.20\text{A}.$

Current through $R_3=8\Omega$ is $0.32A.$

Explanation of Solution

Given:

Potential difference between a and b = $V_{ab}=5\text{V}$

  $R_1=20\Omega$

  $R_2=12\Omega$

  $R_3=8\Omega$

Formula used:

By ohm's law,
  $V=iR$

V = potential difference between two points
  $i$ = electric current
R = resistance
Calculation:

Total current = $i_{total}=\frac{V_{ab}}{R_{eq}}=\frac{5V}{15.5\Omega }=0.32\text{A}$

From the figure 1, total current through the circuit = current through $R_3=8\Omega$

So, $i_{total}=i_3=0.32\text{A}$

  

Again, from the figure
  $\begin{array}{l}{V}_{ab}=V_{ac}+V_{cb}\\ V_{ab}=V_{ac}+i_3{R}_3\\ V_{ac}=V_{ab}-i_3{R}_3\\ V_{ac}=5\text{V}-\left(0.32\text{A}\times 8\Omega \right)\\ V_{ac}=5\text{V}-2.56\text{V}\\ V_{ac}=2.44\text{V}\end{array}$

Now, by ohm's law,
  $\begin{array}{l}{i}_1=\frac{V_{ab}}{R_1}=\frac{2.44\text{V}}{20\Omega }=0.122\text{A}=0.12\text{A}\\ i_2=\frac{V_{ab}}{R_2}=\frac{2.44\text{V}}{12\Omega }=0.20333\text{}\text{A}=0.20\text{A}\end{array}$

Conclusion:

Thus, the current through $R_1=20\Omega$, $R_2=12\Omega$ and $R_3=8\Omega$ are $0.12\text{A},0.20\text{A}$ and $\text{0}\text{.32}\text{A}$ respectively.

To determine

(c)

The power dissipated in each resistor.

Answer to Problem 83QAP

Power dissipated in $R_1=20\Omega$ is $0.29\text{W}$

Power dissipated in $R_2=12\Omega$ is $0.49\text{W}$

Power dissipated in $R_3=8\Omega$ is $0.84\text{W}$

Explanation of Solution

Given:

From part b,
Current through $R_1=20\Omega$ is $0.12\text{A}.$

Current through $R_2=12\Omega$ is $0.20\text{A}.$

Current through $R_3=8\Omega$ is $0.32A.$

Formula used:

Power dissipated across resistance,
  $\begin{array}{l}P=i^{2}R\\ i=\text{ electric current}\\ R\text{ = electric resistance}\end{array}$

Calculation:

Power dissipated across $R_1=20\Omega$,
  $P_1=i_1{}^2{R}_1={(0.12\text{A})}^2\times 20\Omega =0.29\text{W}$

Power dissipated across $R_2=12\Omega$,
  $P_2=i_2{}^2{R}_2={(0.2033\text{A})}^2\times 12\Omega =0.49\text{W}$

Power dissipated across $R_3=8\Omega$,
  $P_1=i_1{}^2{R}_1={(0.324\text{A})}^2\times 8\Omega =0.84\text{W}$

Conclusion:

Power dissipated in $R_1=20\Omega$ is $0.29\text{W}$

Power dissipated in $R_2=12\Omega$ is $0.49\text{W}$

Power dissipated in $R_3=8\Omega$ is $0.84\text{W}$