Chapter 18, Problem 79QAP

To determine

The power dissipated in each resistor.

Answer to Problem 79QAP

  $P_1=45\text{W}$

  $P_2=90\text{W}$

  $P_3=28.125\text{W}$

  $P_4=56.25\text{W}$

Explanation of Solution

Given:

Resistors,

  $\begin{array}{l}{R}_1=5\Omega \\ R_2=10\Omega \\ R_3=8\Omega \\ R_4=16\Omega \end{array}$

Voltage, $V=45V$

Formula:

According to Ohm's law the potential difference is given by,

  $\begin{array}{l}V=IR\\ \text{Where},\\ V=\text{voltage}\\ I=\text{current}\\ R=\text{resistance}\end{array}$

Net resistance in series is given by,

  $\begin{array}{l}{R}_s=R_1+R_2\\ \text{Where},\\ R_s=\text{net resistance}\\ R_1\text{ and }{R}_2=\text{resistors connected in series}\end{array}$

Net resistance in parallel is given by,

  $\begin{array}{l}\frac{1}{R_p}=\frac{1}{R_1}+\frac{1}{R_2}\\ \text{Where},\\ R_p=\text{net resistance}\\ R_1\text{ and }{R}_2=\text{resistors connected in parallel}\end{array}$

Charge in the conductor is given by,

  $\begin{array}{l}Q=It\\ \text{Where},\\ Q=\text{charge}\\ I=\text{current}\\ t=\text{time}\end{array}$

Calculation:

Resistor $R_1\text{ and }{R}_2$ are connected in series. So, the net resistance in the series is

  $\begin{array}{l}{R}_{s_1}=R_1+R_2\\ R_{s_1}=5+10\\ R_{s_1}=15\Omega \end{array}$

Resistor $R_3\text{ and }{R}_4$ are connected in series. So, the net resistance in the series is

  $\begin{array}{l}{R}_{s_2}=R_3+R_4\\ R_{s_2}=8+16\\ R_{s_2}=24\Omega \end{array}$

Resistor $R_{s_1}\text{and}{R}_{s_2}$ are connected in parallel. So, the net resistance in the parallel is

  $\begin{array}{l}\frac{1}{R_p}=\frac{1}{R_{s_1}}+\frac{1}{R_{s_2}}\\ \frac{1}{R_p}=\frac{1}{15}+\frac{1}{24}\\ R_p=\frac{15\times 24}{15+24}\\ R_p=9.23\Omega \end{array}$

Thus, net current in the circuit is,

  $\begin{array}{l}I=\frac{V}{R}\\ I=\frac{45}{9.23}\\ I=4.875\text{A}\end{array}$

Now, current from $R_{s_1}$ is,

  $i_1=\frac{V}{R_{s_1}}=\frac{45}{15}=3A$

So, power dissipated in $R_1$ is,

  $\begin{array}{l}{P}_1=i_1^2\times R_1\\ P_1=9\times 5\\ P_1=45\text{W}\end{array}$

Power dissipated in $R_2$ is,

  $\begin{array}{l}{P}_2=i_1^2\times R_2\\ P_2=9\times 10\\ P_2=90\text{W}\end{array}$

Now, current from $R_{s_2}$ is,

  $\begin{array}{l}{i}_2=I-i_2\\ i_2=4.875-3\\ i_2=1.875\text{A}\end{array}$

Power dissipated in $R_3$ is,

  $\begin{array}{l}{P}_3=i_2^2\times R_2\\ P_3={\left(1.875\right)}^2\times 8\\ P_3=28.125\text{W}\end{array}$

Power dissipated in $R_4$ is,

  $\begin{array}{l}{P}_4=i_2^2\times R_2\\ P_4={\left(1.875\right)}^2\times 16\\ P_4=56.25\text{W}\end{array}$