Chapter 3, Problem 3.48HP

To determine

The value of temperature $T$ .

Answer to Problem 3.48HP

The value of the temperature is $0.956°\text{C}$ .

Explanation of Solution

Calculation:

The given diagram is shown in Figure 1

  

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

  $1\text{k}\Omega ={10}^3\Omega$

The conversion of $12\text{k}\Omega$ into $\Omega$ is given by

  $12\text{k}\Omega =12\times {10}^3\Omega$

The conversion of $24\text{k}\Omega$ into $\Omega$ is given by

  $24\text{k}\Omega =24\times {10}^3\Omega$

The conversion of $3\text{k}\Omega$ into $\Omega$ is given by

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

The conversion of $10\text{k}\Omega$ into $\Omega$ is given by

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

Consider the source $V_{S1}$ only, mark the values and redraw the circuit.

The required diagram is shown in Figure 2

  

Apply KVL to the first loop.

  $\begin{array}{l}-24+\left(12\times {10}^3\right)\left(i_1-i_2\right)+\left(12\times {10}^3\right)\left(i_1-i_3\right)=0\\ \left(\text{24}\times {10}^3\right)i_1-\left(12\times {10}^3\right)i_2-\left(12\times {10}^3\right)i_3=24\\ i_1-0.5i_2-0.5i_3=0.01\end{array}$   ....... (1)

Apply KVL in the second loop.

  $\begin{array}{l}\left(3\times {10}^3\right)i_2+\left(10\times {10}^3\right)\left(i_2-i_3\right)+\left(12\times {10}^3\right)\left(i_2-i_1\right)=0\\ -12i_1+25i_2-10i_3=0\end{array}$   ....... (2)

Apply KVL in the third loop.

  $\begin{array}{l}\left(24\times {10}^3\right)i_3+\left(12\times {10}^3\right)\left(i_3-i_1\right)+\left(10\times {10}^3\right)\left(i_3-i_2\right)=0\\ -12i_1-10i_2+46i_3=0\end{array}$   ....... (3)

From equation (1), equation (2) and equation (3) the values of current are,

  $\begin{array}{l}{i}_1=2.1\times {10}^{-3}\text{A}\\ i_2=1.3\times {10}^{-3}\text{A}\\ i_3=0.8\times {10}^{-3}\text{A}\end{array}$

The expression to calculate the value of voltage across the resistance $R_3$ is given by

  $V_{R_1}=\left(i_2-i_3\right)10\times {10}^3\Omega$

Substitute $2.1\times {10}^{-3}\text{A}$ for $i_1$, $1.3\times {10}^{-3}\text{A}$ for $i_2$ and $0.8\times {10}^{-3}\text{A}$ for $i_3$ in the above equation.

  $\begin{array}{c}{V}_{R_1}=\left(1.3\times {10}^{-3}\text{A}-0.8\times {10}^{-3}\text{A}\right)\left(10\times {10}^3\Omega \right)\\ =5\text{V}\end{array}$

Short circuit all the sources expect the source $V_{S2}$ and redraw the circuit.

The required diagram is shown in Figure 3

  

Apply KVL to loop 1.

  $\begin{array}{l}\left(12\times {10}^3\right)\left(i_1-i_2\right)+\left(12\times {10}^3\right)\left(i_1-i_3\right)+V_{S2}=0\\ 24i_1-12i_2-12i_3=-0.01V_{S2}\end{array}$

Substitute $T$ for $V_{S2}$ in the above equation.

  $24i_1-12i_2-12i_3=-0.01T$   ....... (4)

Apply KVL to the loop $2$ .

  $\begin{array}{l}\left(3\times {10}^3\right)i_2+\left(10\times {10}^3\right)\left(i_2-i_3\right)+\left(12\times {10}^3\right)\left(i_2-i_1\right)=0\\ -12i_1+25i_2-10i_3=0\end{array}$   ....... (5)

Apply KVL in the loop $3$ .

  $\begin{array}{l}\left(24\times {10}^3\right)i_3+\left(12\times {10}^3\right)\left(i_3-i_1\right)+\left(10\times {10}^3\right)\left(i_3-i_2\right)-V_{S2}=0\\ -12i_1+10i_2+46i_3=-0.01V_{S2}\end{array}$

Substitute $T$ for $V_{S2}$ in the above equation.

  $-12i_1+10i_2+46i_3=-0.01T$   ....... (6)

From equation (4), equation (5) and equation (6) the value of currents is,

  $\begin{array}{l}{i}_1=-0.52\times {10}^{-3}T\text{A}\\ i_2=-0.23\times {10}^{-3}T\text{A}\\ i_3=0.029\times {10}^{-3}T\text{A}\end{array}$

The expression to calculate the voltage across the resistance $R_3$ is given by,

  $V_{R_3}=\left(i_2-i_3\right)\left(10\times {10}^3\Omega \right)$

Substitute $-0.23\times {10}^{-3}T\text{A}$ for $i_2$ and $0.029\times {10}^{-3}T\text{A}$ for $i_3$ in the above equation.

  $\begin{array}{c}{V}_{R_3}=\left(-0.23\times {10}^{-3}T\text{A}-0.029\times {10}^{-3}T\text{A}\right)\left(10\times {10}^3\Omega \right)\\ =-2.59T\end{array}$

The expression to calculate the voltage across the resistance $R_3$ by superposition theorem is given by,

  $V_{R_3}=-2.59T+5$

Substitute $-2.524\text{V}$ for $V_{R_3}$ in the above equation.

  $\begin{array}{l}-2.524\text{V}=-2.59T+5\\ T=\frac{2.476}{2.59}\\ T=0.956°\text{C}\end{array}$

Conclusion:

Therefore, the value of the temperature is $0.956°\text{C}$ .