Chapter 2, Problem 2.31HP

To determine

(a)

The total power dissipated by the ideal source.

Answer to Problem 2.31HP

The power delivered for the current of $0\text{mA}$ is $0\text{mW}$, for $10\text{mA}$ is $\text{150}\text{mW}$, for $20\text{mA}$ is $300\text{mW}$, for $\text{30}\text{mA}$ is $450\text{mW}$, for $80\text{mA}$ is $1200\text{mW}$ and for $100\text{mA}$ is $1500\text{mW}$ .

Explanation of Solution

Calculation:

The formula for the total power supplied or delivered by the ideal voltage source is given by,

  $p_{v_s}=v_s{i}_T$ ...... (1)

Substitute $0\text{mA}$ for ${\text{i}}_T$ and $15\text{V}$ for $v_s$ in the above expression.

  $\begin{array}{c}{p}_{v_s}=\left(15\text{V}\right)\left(0\text{mA}\right)\\ =0\text{W}\end{array}$

Substitute $10\text{mA}$ for ${\text{i}}_T$ and $15\text{V}$ for $v_s$ in equation (1).

  $\begin{array}{c}{p}_{v_s}=\left(15\text{V}\right)\left(10\text{mA}\right)\\ =150\text{mW}\end{array}$

Substitute $20\text{mA}$ for ${\text{i}}_T$ and $15\text{V}$ for $v_s$ in equation (1).

  $\begin{array}{c}{p}_{v_s}=\left(15\text{V}\right)\left(20\text{mA}\right)\\ =300\text{mW}\end{array}$

Substitute $30\text{mA}$ for ${\text{i}}_T$ and $15\text{V}$ for $v_s$ in equation (1).

  $\begin{array}{c}{p}_{v_s}=\left(15\text{V}\right)\left(30\text{mA}\right)\\ =450\text{mW}\end{array}$

Substitute $10\text{mA}$ for ${\text{i}}_T$ and $15\text{V}$ for $v_s$ in equation (1).

  $\begin{array}{c}{p}_{v_s}=\left(15\text{V}\right)\left(10\text{mA}\right)\\ =150\text{mW}\end{array}$

Substitute $80\text{mA}$ for ${\text{i}}_T$ and $15\text{V}$ for $v_s$ in equation (1).

  $\begin{array}{c}{p}_{v_s}=\left(15\text{V}\right)\left(80\text{mA}\right)\\ =1200\text{mW}\end{array}$

Substitute $100\text{mA}$ for ${\text{i}}_T$ and $15\text{V}$ for $v_s$ in equation (1).

  $\begin{array}{c}{p}_{v_s}=\left(15\text{V}\right)\left(100\text{mA}\right)\\ =1500\text{mW}\end{array}$

Conclusion:

Therefore, the power delivered for the current of $0\text{mA}$ is $0\text{mW}$, for $10\text{mA}$ is $\text{150}\text{mW}$, for $20\text{mA}$ is $300\text{mW}$, for $\text{30}\text{mA}$ is $450\text{mW}$, for $80\text{mA}$ is $1200\text{mW}$ and for $100\text{mA}$ is $1500\text{mW}$ .

To determine

(b)

The power dissipated within the non ideal source.

Answer to Problem 2.31HP

The power dissipated by the resistance of $100\Omega$ for $0\text{mA}$ current source is $0\text{mW}$, for $10\text{mA}$ current source is $\text{10}\text{mW}$, for $20\text{mA}$ current source is $40\text{mW}$, for current source $\text{30}\text{mA}$ is $90\text{mW}$, for current source $80\text{mA}$ is $640\text{mW}$ and for $100\text{mA}$ current source is $1000\text{mW}$ .

Explanation of Solution

Calculation:

The formula for the power delivered by the non ideal power source is given by,

  $p_{R_S}=i_T^2{R}_S$ ...... (2)

Substitute $0\text{mA}$ for ${\text{i}}_T$ and $100\Omega$ for $R_s$ the above equation.

  $\begin{array}{c}{p}_{R_S}={\left(0\text{mA}\right)}^2\left(100\Omega \right)\\ =0\text{W}\end{array}$

Substitute $10\text{mA}$ for ${\text{i}}_T$ and $100\Omega$ for $R_s$ in equation (2).

  $\begin{array}{c}{p}_{R_S}={\left(10\text{mA}\right)}^2\left(100\Omega \right)\\ =10\text{mW}\end{array}$

Substitute $20\text{mA}$ for ${\text{i}}_T$ and $100\Omega$ for $R_s$ in equation (2).

  $\begin{array}{c}{p}_{R_S}={\left(20\text{mA}\right)}^2\left(100\Omega \right)\\ =40\text{mW}\end{array}$

Substitute $30\text{mA}$ for ${\text{i}}_T$ and $100\Omega$ for $R_s$ in equation (2).

  $\begin{array}{c}{p}_{R_S}={\left(30\text{mA}\right)}^2\left(100\Omega \right)\\ =90\text{mW}\end{array}$

Substitute $80\text{mA}$ for ${\text{i}}_T$ and $100\Omega$ for $R_s$ in equation (2).

  $\begin{array}{c}{p}_{R_S}={\left(80\text{mA}\right)}^2\left(100\Omega \right)\\ =640\text{mW}\end{array}$

Substitute $100\text{mA}$ for ${\text{i}}_T$ and $100\Omega$ for $R_s$ in equation (2).

  $\begin{array}{c}{p}_{R_S}={\left(100\text{mA}\right)}^2\left(100\Omega \right)\\ =1000\text{mW}\end{array}$

Conclusion:

Therefore, the power dissipated by the resistance of $100\Omega$ for $0\text{mA}$ current source is $0\text{mW}$, for $10\text{mA}$ current source is $\text{10}\text{mW}$, for $20\text{mA}$ current source is $40\text{mW}$, for current source $\text{30}\text{mA}$ is $90\text{mW}$, for current source $80\text{mA}$ is $640\text{mW}$ and for $100\text{mA}$ current source is $1000\text{mW}$ .

To determine

(c)

The amount of power supplied by the load resistor.

Answer to Problem 2.31HP

The power supplied to the load resistance of $100\Omega$ for $0\text{mA}$ current source is $0\text{mW}$, for $10\text{mA}$ current source is $\text{140}\text{mW}$, for $20\text{mA}$ current source is $260\text{mW}$, for current source $\text{30}\text{mA}$ is $360\text{mW}$, for current source $80\text{mA}$ is $560\text{mW}$ and for $100\text{mA}$ current source is $500\text{mW}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

Apply KCL to the above circuit.

  $\begin{array}{l}{v}_T-v_S+i_T{R}_S=0\\ v_T=v_S-i_T{R}_S\end{array}$ ...... (3)

The formula for the load power is given by,

  $p_{R_0}=v_T{i}_T$ ...... (4)

The conversion of $\text{mA}$ into $\text{A}$ is given by,

  $1\text{mA}={10}^{-3}\text{A}$

The conversion of $\text{10}\text{mA}$ into $\text{A}$ is given by,

  $10\text{mA}=10\times {10}^{-3}\text{A}$

The conversion of $\text{20}\text{mA}$ into $\text{A}$ is given by,

  $20\text{mA}=20\times {10}^{-3}\text{A}$

The conversion of $\text{30}\text{mA}$ into $\text{A}$ is given by,

  $30\text{mA}=30\times {10}^{-3}\text{A}$

The conversion of $\text{80}\text{mA}$ into $\text{A}$ is given by,

  $80\text{mA}=80\times {10}^{-3}\text{A}$

The conversion of $\text{100}\text{mA}$ into $\text{A}$ is given by,

  $100\text{mA}=100\times {10}^{-3}\text{A}$

Substitute $15\text{V}$ for $v_s$, $0\text{mA}$ for ${\text{i}}_T$ and $100\Omega$ for $R_s$ in equation (3)

  $\begin{array}{c}{v}_T=15\text{V}-{\left(0\times {10}^{-3}\text{A}\right)}^2\left(100\Omega \right)\\ =15\text{V}\end{array}$

Substitute $15\text{V}$ for $v_T$ and $0\text{mA}$ for ${\text{i}}_T$ in equation (4).

  $\begin{array}{c}{p}_{R_0}=\left(15\text{V}\right)\left(0\text{mA}\right)\\ =0\text{W}\end{array}$

Substitute $15\text{V}$ for $v_s$, $10\text{mA}$ for ${\text{i}}_T$ and $100\Omega$ for $R_s$ in equation (3).

  $\begin{array}{c}{v}_T=15\text{V}-{\left(10\times {10}^{-3}\text{A}\right)}^2\left(100\Omega \right)\\ =14\text{V}\end{array}$

Substitute $14\text{V}$ for $v_T$ and $10\text{mA}$ for ${\text{i}}_T$ in equation (4).

  $\begin{array}{c}{p}_{R_0}=\left(14\text{V}\right)\left(10\text{mA}\right)\\ =140\text{mW}\end{array}$

Substitute $15\text{V}$ for $v_s$, $20\text{mA}$ for ${\text{i}}_T$ and $100\Omega$ for $R_s$ in equation (3).

  $\begin{array}{c}{v}_T=15\text{V}-{\left(20\times {10}^{-3}\text{A}\right)}^2\left(100\Omega \right)\\ =13\text{V}\end{array}$

Substitute $13\text{V}$ for $v_T$ and $20\text{mA}$ for ${\text{i}}_T$ in equation (4).

  $\begin{array}{c}{p}_{R_0}=\left(13\text{V}\right)\left(20\text{mA}\right)\\ =260\text{mW}\end{array}$

Substitute $15\text{V}$ for $v_s$, $30\text{mA}$ for ${\text{i}}_T$ and $100\Omega$ for $R_s$ in equation (3).

  $\begin{array}{c}{v}_T=15\text{V}-{\left(30\times {10}^{-3}\text{A}\right)}^2\left(100\Omega \right)\\ =12\text{V}\end{array}$

Substitute $12\text{V}$ for $v_T$ and $30\text{mA}$ for ${\text{i}}_T$ in equation (4).

  $\begin{array}{c}{p}_{R_0}=\left(12\text{V}\right)\left(30\text{mA}\right)\\ =360\text{mW}\end{array}$

Substitute $15\text{V}$ for $v_s$, $80\text{mA}$ for ${\text{i}}_T$ and $100\Omega$ for $R_s$ in equation (3).

  $\begin{array}{c}{v}_T=15\text{V}-{\left(80\times {10}^{-3}\text{A}\right)}^2\left(100\Omega \right)\\ =7\text{V}\end{array}$

Substitute $7\text{V}$ for $v_T$ and $80\text{mA}$ for ${\text{i}}_T$ in equation (4).

  $\begin{array}{c}{p}_{R_0}=\left(9\text{V}\right)\left(80\text{mA}\right)\\ =560\text{mW}\end{array}$

Substitute $15\text{V}$ for $v_s$, $100\text{mA}$ for ${\text{i}}_T$ and $100\Omega$ for $R_s$ in equation (3).

  $\begin{array}{c}{v}_T=15\text{V}-{\left(100\times {10}^{-3}\text{A}\right)}^2\left(100\Omega \right)\\ =5\text{V}\end{array}$

Substitute $5\text{V}$ for $v_T$ and $100\text{mA}$ for ${\text{i}}_T$ in equation (4).

  $\begin{array}{c}{p}_{R_0}=\left(5\text{V}\right)\left(100\text{mA}\right)\\ =500\text{mW}\end{array}$

Conclusion:

Therefore, the power supplied to the load resistance of $100\Omega$ for $0\text{mA}$ current source is $0\text{mW}$, for $10\text{mA}$ current source is $\text{140}\text{mW}$, for $20\text{mA}$ current source is $260\text{mW}$, for current source $\text{30}\text{mA}$ is $360\text{mW}$, for current source $80\text{mA}$ is $560\text{mW}$ and for $100\text{mA}$ current source is $500\text{mW}$ .

To determine

(d)

The terminal voltage $v_T$ and the power supplied to the load resistor as a function of terminal current $i_T$ .

Answer to Problem 2.31HP

The plot between the terminal voltage and the load current is shown in Figure 2 and plot for the load current and power delivered to the load is shown in Figure 3

Explanation of Solution

Calculation:

The plot for the terminal voltage and total current is shown in Figure 2.

  

The plot between for the power delivered to the load versus the total current is shown in Figure 3

  

Conclusion:

Therefore, the plot between the terminal voltage and the load current is shown in Figure 2 and plot for the load current and power delivered to the load is shown in Figure 3