Chapter 5, Problem 1P

Find $I_o$ in the network in Fig. P5.1 using linearity and the assumption that $I_o=\text{1 mA}\text{.}$

To determine

The value of current $I_o$ using linearity property.

Answer to Problem 1P

The value of $I_o\text{is}0.5\text{mA}$.

Explanation of Solution

Given:

A circuit is given with an assumption that $I_o$ = $1\text{mA}$.

Calculations:

Mark the node with voltage V and current in all branches.

As given $I_o$ = $1\text{mA}$ passing through the $\text{6kΩ}$ resistor, the voltage across it will be 6V (product of current and resistance).

The value of $I_3$ -

  $I_3=\frac{6}{2}=3\text{mA}$

From circuit, $I_2$ is sum of $I_o\text{and}{I}_3$

  $I_2$ =1+3=4 $\text{mA}$

Applying KVL,

  $\begin{array}{l}V=3I_2+6\\ =3\times 4+6\\ V=18\text{V}\end{array}$

Therefore, $I_1$ -

  $\begin{array}{l}{I}_1=\frac{V}{6}\\ =\frac{18}{6}\\ I_1=3\text{mA}\end{array}$

Where

  $\begin{array}{l}{I}_S=I_1+I_2\\ =3+4\\ I_S=7\text{mA}\end{array}$

Let the voltage source be denoted by $V_S$, then apply the KVL in the circuit-

  $\begin{array}{l}{V}_S=2I_S+V\\ =2\times 7+18\\ V_S=32\text{V}\end{array}$

Therefore if

  $\begin{array}{l}{V}_S=32\text{V}\to I_o=1\text{mA}\\ \text{Then}\\ V_S=64\text{V}\to I_o=2\text{mA}\end{array}$