Question 3.1: Find is,i1 and i2 in the circuit shown in Fig. 3.9.

We begin by noting that the 3\Omega  resistor is in series with the 6\Omega  resistor. We therefore replace this series combination with a  9\Omega resistor, reducing the circuit to the one shown in Fig. 3.10(a).We now can replace the parallel combination of the 9\Omega and 18\Omega resistors with a single resistance of \frac{\left(18\times9 \right)}{\left(18+9 \right) } or 6\Omega Figure 3.10(b) shows this further reduction of the circuit. The nodes x and y marked on all diagrams facilitate tracing through the reduction of the circuit. From Fig. 3.10(b) you can verify  that i_{s} equals \frac{120}{10}, or 12 A. Figure 3.11 shows the result at this point in the analysis. We added the voltage v_{1} to help clarify the subsequent discussion. Using Ohm’s law we compute the value of  v_{1} :

v_{1} = \left(12\right)\left(6\right) = 72 V. .

But  v_{1}. is the voltage drop from node x to node y, so we can return to the circuit shown in Fig. 3.10(a) and again use Ohm’s law to calculatei_{1} andi_{2} Thus,

i_{1} = \frac{v_{1}}{18} = \frac{72}{18}=4A ,

i_{2} = \frac{v_{1}}{9} = \frac{72}{9}=8A .

We have found the three specified currents by using series-parallel reductions in combination with Ohm’s law.