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Question: 

Determine

  1. the equivalent resistance of the circuit shown in Fig. 19–48,
  2. the voltage across each resistor, and
  3. the current through each resistor.
Problem 16.

Figure 19-48.

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer: 
  1. $1350 \; \Omega$
  2. $V_3 = 8.8 \textrm{ V, } V_1 = V_2 = 3.2 \textrm{ V}$
  3. $I_3 = 0.0089 \textrm{ A, } I_2 = 0.0047 \textrm{ A, } I_1 = 0.0042 \textrm{ A}$

Giancoli 7th Edition, Chapter 19, Problem 16

(2:16)

Comments

I don't understand why those two are parallel and one is a series?

Hi Agp2196, $R_1$ and $R_2$ can be recognized as in parallel since they have both ends connected by wires. The bottom ends are connected by a wire, and the top ends are also connected by a wire. This means current at the bottom, coming from the positive end of the battery, has a choice of $R_1$ or $R_2$, and when you see current being given this choice, you know the resistors are in parallel. Resistor labelled $R_3$ is in series since there are no other options but to go through $R_3$ in order to continue through the circuit.

Hope that helps,
Mr. Dychko