Question 2.1: Find a single equivalent resistance for the network shown in...
Find a single equivalent resistance for the network shown in Figure 2.3(a).
First,we look for a combination of resistances that is in series or in parallel.
In Figure 2.3(a), R_3 and R_4 are in series (In fact, as it stands, no other two resistances in this network are either in series or in parallel.) Thus, our first step is to combine R_3 and R_4 , replacing them by their equivalent resistance. Recall that for a series combination, the equivalent resistance is the sum of the resistances in series:
R_{eq1} = R_3 + R_4 = 5 + 15 = 20 Ω
Figure 2.3(b) shows the network after replacing R_3 and R_4 by their equivalent resistance. Now we see that R_2 and R_{eq1} are in parallel. The equivalent resistance for this combination is
R_{eq2} = \frac {1}{1/R_{eq1}+1/R_2} = \frac{1}{1/20+1/20} = 10 Ω
Making this replacement gives the equivalent network shown in Figure 2.3(c). Finally, we see that R_1 and R_{eq2} are in series. Thus, the equivalent resistance for the entire network is
R_{eq} = R_1 +R_{eq2} = 10 + 10 = 20 Ω