In the network in Fig. P5.10, find Io using superposition.Figure P5.10

In the circuit shown in Figure 1 , two \(6 \mathrm{k} \Omega\) resistors are connected in parallel. Calculate the equivalent resistance.

$$ \begin{aligned} R_{1} &=\frac{(6 \mathrm{k} \Omega)(6 \mathrm{k} \Omega)}{6 \mathrm{k} \Omega+6 \mathrm{k} \Omega} \\ &=\frac{36 \mathrm{k} \Omega}{12} \\ &=3 \mathrm{k} \Omega \end{aligned} $$

This \(3 \mathrm{k} \Omega\) resistor and \(6 \mathrm{k} \Omega\) resistors are in series. Calculate the equivalent resistance.

$$ \begin{aligned} R_{2} &=R_{1}+6 \mathrm{k} \Omega \\ &=3 \mathrm{k} \Omega+6 \mathrm{k} \Omega \\ &=9 \mathrm{k} \Omega \end{aligned} $$

The simplified circuit is shown in Figure 2.

Apply current division rule and calculate the current \(I_{o}^{\prime}\) through the \(6 \mathrm{k} \Omega\) resistor.

$$ \begin{aligned} I_{o}^{\prime} &=\left(\frac{9 \mathrm{k} \Omega}{9 \mathrm{k} \Omega+6 \mathrm{k} \Omega}\right)(-6 \mathrm{~mA}) \\ &=\left(\frac{9 \mathrm{k} \Omega}{15 \mathrm{k} \Omega}\right)(-6 \mathrm{~mA}) \\ &=-3.6 \mathrm{~mA} \end{aligned} $$

Hence, the current \(I_{o}^{\prime}\) through the \(6 \mathrm{k} \Omega\) resistor is \(-3.6 \mathrm{~mA}\).

Apply superposition principle to the circuit. Acting only voltage source, the current source is open-circuited as shown in Figure 3.

The \(6 \mathrm{k} \Omega\) resistor and \(6 \mathrm{k} \Omega\) resistor are in series. Calculate the equivalent resistance.

$$ \begin{aligned} R_{3} &=6 \mathrm{k} \Omega+6 \mathrm{k} \Omega \\ &=12 \mathrm{k} \Omega \end{aligned} $$

This \(12 \mathrm{k} \Omega\) resistor is in parallel with \(6 \mathrm{k} \Omega\) resistor. Calculate the equivalent resistance.

$$ \begin{aligned} R_{4} &=\frac{(12 \mathrm{k} \Omega)(6 \mathrm{k} \Omega)}{12 \mathrm{k} \Omega+6 \mathrm{k} \Omega} \\ &=\frac{72 \mathrm{k} \Omega}{18} \\ &=4 \mathrm{k} \Omega \end{aligned} $$

Apply ohms law and calculate the total current \(/\) through source.

$$ \begin{aligned} I &=\frac{12 \mathrm{~V}}{6 \mathrm{k} \Omega+4 \mathrm{k} \Omega} \\ &=\frac{12 \mathrm{~V}}{10 \mathrm{k} \Omega} \\ &=1.2 \mathrm{~mA} \end{aligned} $$

Apply current division rule and calculate the current \(I_{o}^{\prime \prime}\) through the \(6 \mathrm{k} \Omega\) resistor from Figure \(3 .\)

$$ \begin{aligned} I_{o}^{\prime \prime} &=\left(\frac{6 \mathrm{k} \Omega}{6 \mathrm{k} \Omega+6 \mathrm{k} \Omega+6 \mathrm{k} \Omega}\right) I \\ &=\left(\frac{6 \mathrm{k} \Omega}{18 \mathrm{k} \Omega}\right)(1.2 \mathrm{~mA}) \\ &=(0.333)(1.2 \mathrm{~mA}) \\ &=0.4 \mathrm{~mA} \end{aligned} $$

Hence, the current \(I_{o}^{\prime \prime}\) through the \(6 \mathrm{k} \Omega\) resistor is \(0.4 \mathrm{~mA}\).

Write the expression for the current \(I_{o}\) through the \(6 \mathrm{k} \Omega\) resistor.

$$ I_{o}=I_{o}^{\prime}+I_{o}^{\prime \prime} $$

Substitute \(-3.6 \mathrm{~mA}\) for \(I_{o}^{\prime}\) and \(0.4 \mathrm{~mA}\) for \(I_{o}^{\prime \prime}\).

$$ \begin{aligned} I_{o} &=-3.6 \mathrm{~mA}+0.4 \mathrm{~mA} \\ &=-3.2 \mathrm{~mA} \end{aligned} $$

Therefore, the current \(I_{o}\) through the \(6 \mathrm{k} \Omega\) resistor is \(-3.2 \mathrm{~mA}\).