Chapter 3, Problem 3.60HP

To determine

(a)

The Thevenin equivalent resistance seen by the load resistor $R_o$ .

Answer to Problem 3.60HP

The expression for the Thevenin resistance seen by the load resistor is $\left(\frac{R_1{R}_2}{R_1+R_2}\right)+\left(\frac{R_x{R}_3}{R_x+R_3}\right)$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

 

To calculate the Thevenin equivalent resistance, short circuit the voltage source and open circuit the load resistance, mark the values and redraw the circuit.

The required diagram is shown in Figure 2

 

From the above figure, the Thevenin equivalent resistance of the circuit is evaluated as,

  $\begin{array}{c}{R}_{Th}=\left(R_1\left|\right|R_2\right)+\left(R_x\left|\right|R_3\right)\\ =\left(\frac{R_1{R}_2}{R_1+R_2}\right)+\left(\frac{R_x{R}_3}{R_x+R_3}\right)\end{array}$   ....... (1)

Conclusion:

Therefore, the expression for the Thevenin resistance seen by the load resistor is $\left(\frac{R_1{R}_2}{R_1+R_2}\right)+\left(\frac{R_x{R}_3}{R_x+R_3}\right)$ .

To determine

(b)

The Thevenin equivalent network seen by $R_o$ and power dissipated.

Answer to Problem 3.60HP

The Thevenin equivalent circuit seen by the load is shown in Figure 4 and the power dissipated by the resistance $R_o$ is $0.032\times {10}^{-6}\text{W}$ .

Explanation of Solution

Calculation:

The conversion of $1\Omega$ into $\text{k}\Omega$ is given by,

  $1\Omega ={10}^{-3}\text{k}\Omega$

The conversion of $960\Omega$ into $\text{k}\Omega$ is given by,

  $960\Omega =960\times {10}^{-3}\text{k}\Omega$

Substitute $1\text{k}\Omega$ for $R_1$, $1\text{k}\Omega$ for $R_2$, $1\text{k}\Omega$ for $R_3$ and $960\Omega$ for $R_x$ in equation (1)

  $\begin{array}{c}{R}_0=\left(\frac{\left(1\text{k}\Omega \right)\left(1\text{k}\Omega \right)}{\left(1\text{k}\Omega \right)+\left(1\text{k}\Omega \right)}\right)+\left(\frac{\left(960\times {10}^{-3}\text{k}\Omega \right)\left(1\text{k}\Omega \right)}{\left(960\times {10}^{-3}\text{k}\Omega \right)+\left(1\text{k}\Omega \right)}\right)\\ =999\times {10}^{-3}\text{k}\Omega \end{array}$

The expression for the Thevenin equivalent resistance is given by,

  $R_{Th}=R_o$

Substitute $999\times {10}^{-3}\text{k}\Omega$ for $R_o$ in the above equation.

  $R_{Th}=999\times {10}^{-3}\text{k}\Omega$

The conversion of $\text{k}\Omega$ into $\Omega$ is given by,

  $1\text{k}\Omega ={10}^3\Omega$

The conversion of $999\times {10}^{-3}\text{k}\Omega$ into $\Omega$ is given by,

  $\begin{array}{c}999\times {10}^{-3}\text{k}\Omega =999\times {10}^{-3}\times {10}^3\Omega \\ =999\Omega \end{array}$

To calculate the Thevenin voltage open circuit the load resistance, mark the values and redraw the circuit.

The required diagram is shown in Figure 3

 

From the above figure the Thevenin equivalent voltage is calculated as,

  $\begin{array}{c}{V}_{Th}=\frac{\left(12\text{V}\right)\left(1\text{k}\Omega \right)}{\left(1\text{k}\Omega \right)+\left(1\text{k}\Omega \right)}-\frac{\left(12\text{V}\right)\left(996\times {10}^{-3}\text{k}\Omega \right)}{\left(1\text{k}\Omega \right)+\left(996\times {10}^{-3}\text{k}\Omega \right)}\\ =12\text{mV}\end{array}$

Mark the values and draw the Thevenin equivalent circuit.

The required diagram is shown in Figure 4

 

From Figure 4 the current that flows through the resistance of $R_o$ is calculated as,

  $\begin{array}{c}{i}_{R_o}=\frac{12\text{mV}}{999\Omega +500\Omega }\\ =8\times {10}^{-3}\text{mA}\end{array}$

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

  $1\text{mA}={10}^{-3}\text{A}$

The conversion of $8\times {10}^{-3}\text{mA}$ into $\text{A}$ is given by,

  $\begin{array}{c}8\times {10}^{-3}\text{mA}=8\times {10}^{-3}\times {10}^{-3}\text{A}\\ =\text{8}\times {10}^{-6}\text{A}\end{array}$

The expression for the power dissipated by the load is given by,

  $p_{R_o}=i_{R_o}^2\left(500\Omega \right)$

Substitute $\text{8}\times {10}^{-6}\text{A}$ for $i_{R_o}$ in the above equation.

  $\begin{array}{c}{p}_{R_o}={\left(\text{8}\times {10}^{-6}\text{A}\right)}^2\left(500\Omega \right)\\ =0.032\times {10}^{-6}\text{W}\end{array}$

Conclusion:

Therefore, the Thevenin equivalent circuit seen by the load is shown in Figure 4 and the power dissipated by the resistance $R_o$ is $0.032\times {10}^{-6}\text{W}$ .

To determine

(c)

The power dissipated by the Thevenin equivalent resistance seen by $R_o$ .

Answer to Problem 3.60HP

The power dissipated by the resistance $R_{Th}$ is $0.064\times {10}^{-6}\text{W}$ .

Explanation of Solution

Calculation:

The expression to calculate the power dissipated by the Thevenin resistance of the circuit is given by,

  $p_{R_{Th}}=i_{R_{Th}}^2{R}_{Th}$

Substitute $\text{8}\times {10}^{-6}\text{A}$ for $i_{R_o}$ and $999\Omega$ for $R_{Th}$ in the above equation.

  $\begin{array}{c}{p}_{R_{Th}}={\left(\text{8}\times {10}^{-6}\text{A}\right)}^2\left(999\Omega \right)\\ =0.064\times {10}^{-6}\text{W}\end{array}$

Conclusion:

Therefore, the power dissipated by the resistance $R_{Th}$ is $0.064\times {10}^{-6}\text{W}$ .

To determine

(d)

The power dissipated by the entire bridge when $R_o$ is open circuited.

Answer to Problem 3.60HP

The total power dissipated in the bridge is $\text{144}\text{.13}\text{mW}$ .

Explanation of Solution

Calculation:

Redraw the circuit for the resistance $R_o$ removed.

The required diagram is shown in Figure 5

 

The current through the resistance $R_1$ is given by,

  $\begin{array}{c}{i}_{R_1}=\frac{12\text{V}}{1\text{k}\Omega +1\text{k}\Omega }\\ =6\text{mA}\end{array}$

The expression for the power dissipated by the resistance is given by,

  $p_{R_1}={\left(i_{R_1}\right)}^2\left(1000\Omega \right)$

Substitute $6\text{mA}$ for $i_{R_1}$ in the above equation.

  $\begin{array}{c}{p}_{R_1}={\left(6\text{mA}\right)}^2\left(1000\Omega \right)\\ =36\text{mW}\end{array}$

The current through the resistance $R_2$ is given by,

  $\begin{array}{c}{i}_{R_2}=\frac{12\text{V}}{1\text{k}\Omega +1\text{k}\Omega }\\ =6\text{mA}\end{array}$

The expression for the power dissipated by the resistance $R_2$ is given by,

  $p_{R_2}={\left(i_{R_2}\right)}^2\left(1000\Omega \right)$

Substitute $6\text{mA}$ for $i_{R_2}$ in the above equation.

  $\begin{array}{c}{p}_{R_2}={\left(6\text{mA}\right)}^2\left(1000\Omega \right)\\ =36\text{mW}\end{array}$

The current through the resistance $i_{R_3}$ is given by,

  $\begin{array}{c}{i}_{R_3}=\frac{12\text{V}}{1\text{k}\Omega +996\times {10}^{-3}\text{k}\Omega }\\ =6.012\text{mA}\end{array}$

The expression for the power dissipated by the resistance $i_{R_3}$ is given by,

  $p_{R_3}={\left(i_{R_3}\right)}^2\left(1000\Omega \right)$

Substitute $6.012\text{mA}$ for $i_{R_3}$ in the above equation.

  $\begin{array}{c}{p}_{R_3}={\left(6.012\text{mA}\right)}^2\left(1000\Omega \right)\\ =36.14\text{mW}\end{array}$

The expression for the power dissipated by the resistance $i_{R_3}$ is given by,

  $p_{R_x}={\left(i_{R_x}\right)}^2\left(996\times {10}^{-3}\text{k}\Omega \right)$

Substitute $6.012\text{mA}$ for $i_{R_x}$ in the above equation.

  $\begin{array}{c}{i}_{R_x}={\left(6.012\text{mA}\right)}^2\left(996\times {10}^{-3}\text{k}\Omega \right)\\ =35.99\text{mW}\end{array}$

The expression for the total power dissipated by the bridge is given by,

  $p_T=p_{R_x}+p_{R_1}+p_{R_2}+p_{R_3}$

Substitute $35.99\text{mW}$ for $p_{R_x}$, $36.14\text{mW}$ for $p_{R_3}$, $36\text{mW}$ for $p_{R_2}$ and $36\text{mW}$ for $p_{R_1}$ in the above equation.

  $\begin{array}{c}{p}_T=36\text{mW}+36\text{mW}+36.14\text{mW}+35.99\text{mW}\\ =\text{144}\text{.13}\text{mW}\end{array}$

Conclusion:

Therefore, the total power dissipated in the bridge is $\text{144}\text{.13}\text{mW}$ .