Chapter A1.2, Problem 2P

Draw a suitable tree and use general loop analysis to find i₁₀ in the circuit of (a) Fig. A1.13a by writing just one equation with i₁₀ as the variable; (b) Fig. A1.13b by writing just two equations with i₁₀ and i₃ as the variables.

To determine

The value of the current $i_{10}$ for the given circuit diagram.

Answer to Problem 2P

The value of the current $i_{10}$ is $-4.00\text{mA}$.

Explanation of Solution

Given data:

The given diagram is shown in Figure 1.

 

Calculation:

Draw the tree diagram of the given network.

The required diagram is shown in Figure 2.

 

The current $i_5$ through the $5\text{kΩ}$ resistor using KCL is given as,

$\begin{array}{c}{i}_5=0.4i_{10}-i_{10}\\ =-0.6i_{10}\end{array}$

The modified diagram is shown in Figure 3.

 

The conversion from $\text{k}\Omega$ to $\Omega$ is given by,

$\text{1}\text{k}\Omega =1000\Omega$

The conversion from $\text{10}\text{k}\Omega$ to $\Omega$ is given by,

$\text{10}\text{k}\Omega =10000\Omega$

The conversion from $\text{5}\text{k}\Omega$ to $\Omega$ is given by,

$\text{5}\text{k}\Omega =5000\Omega$

The conversion from $\text{20}\text{k}\Omega$ to $\Omega$ is given by,

$\text{20}\text{k}\Omega =20000\Omega$

The conversion from $\text{mA}$ to $\text{A}$ is given by,

$\text{1}\text{mA}={10}^{-3}\text{A}$

The conversion from $\text{5}\text{mA}$ to $\text{A}$ is given by,

$\text{5}\text{mA}=5\times {10}^{-3}\text{A}$

Write the KVL for the loop $bcdeb$.

$\begin{array}{l}10000\left(i_{10}\right)+20000\left(5\times {10}^{-3}\text{+0}\text{.6}{i}_{10}\right)+5000\left(0.6i_{10}\right)=0\\ 25000\left(i_{10}\right)=-100\\ i_{10}=-4.00\times {10}^{-3}\text{A}\end{array}$

The conversion from $\text{A}$ to $\text{mA}$ is given by,

$\text{1}\text{A}={10}^3\text{mA}$

The conversion from $-4.00\times {10}^{-3}\text{A}$ to $\text{mA}$ is given by,

$\begin{array}{c}-4.00\times {10}^{-3}\text{A}=-4.00\times {10}^{-3}\times {10}^3\text{mA}\\ =-\text{4}\text{.00}\text{mA}\end{array}$

Hence, the current $i_{10}$ is,

$i_{10}=-4.00\text{mA}$

Conclusion:

Therefore, the value of the current $i_{10}$ is $-4.00\text{mA}$.

To determine

The value of the current $i_{10}$ for the given circuit diagram.

Answer to Problem 2P

The value of the current $i_{10}$ is $4.69\text{A}$.

Explanation of Solution

Given data:

The given diagram is shown in Figure 4.

 

Calculation:

Draw tree diagram for the given network.

The required diagram is shown in Figure 5.

Write the KVL for the loop $abceda$.

$\begin{array}{l}20\left(2+i_3\right)+4\left(2+i_3+3i_3\right)+10i_{10}-100+6i_3=0\\ 42i_3+10i_{10}=52\end{array}$        (1)

Write the KVL for the loop $decd$.

$\begin{array}{l}100-10i_{10}+24\left(2+4i_3-i_{10}\right)=0\\ 96i_3-34i_{10}=-148\\ i_3=\frac{34i_{10}-148}{96}\end{array}$

Substitute $\frac{34i_{10}-148}{96}$ for $i_3$ in equation (1).

$\begin{array}{l}42\left(\frac{34i_{10}-148}{96}\right)+10i_{10}=52\\ \frac{1428i_{10}-6216+960i_{10}}{96}=52\\ 2388i_{10}=4992+6216\\ i_{10}=4.69\text{A}\end{array}$

Conclusion:

Therefore, the value of the current $i_{10}$ is $4.69\text{A}$.