Question 3.4: Use current division to find the current io and use voltage ...

We can use Eq. 3.32  i_{j}=\frac{v}{R_{j}} =\frac{R_{eq}}{R_{j}} i . if we can find the equivalent resistance of the four parallel branches containing resistors. Symbolically,

R_{eq}=\left(36+44\right)\parallel 10\parallel \left(40+10+30\right)\parallel 24=80\parallel 10\parallel 80\parallel 24=\frac{1}{\frac{1}{80}+ \frac{1}{10} + \frac{1}{80}+ \frac{1}{24}} =6\Omega .

Applying Eq. 3.32,

i_{o}=\frac{6}{24} \left(8 A\right) =2 A.

We can use Ohm’s law to find the voltage drop across the 24 Ω resistor:

v=\left(24\right) \left(2\right)=48 V.

This is also the voltage drop across the branch containing the 40Ω, the 10Ω and the 30Ω resistors in series. We can then use voltage division to determine the voltage dropv_{o} across the 30Ω resistor given that we know the voltage drop across the series-connected resistors, using Eq. 3.30 v_{j}=iR_{j}=\frac{R_{j}}{R_{eq}}v . To do this, we recognize that the equivalent resistance of the series-connected resistors is 40+10+30=80\Omega:

v_{o}=\frac{30}{80} \left(48 V\right)=18 V .