Giancoli's Physics: Principles with Applications, 7th Edition
19
DC Circuits
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19-1: Emf and Terminal Voltage
19-2: Resistors in Series and Parallel
19-3: Kirchhoff's Rules
19-4: Emfs Combined, Battery Charging
19-5: Capacitors in Series and Parallel
19-6: RC Circuits
19-8: Ammeters and Voltmeters

Question by Giancoli, Douglas C., Physics: Principles with Applications, 7th Ed., ©2014, Reprinted by permission of Pearson Education Inc., New York.
Problem 16
Q
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.
Figure 19-48.
A
1. $1350 \; \Omega$
2. $V_3 = 8.8 \textrm{ V, } V_1 = V_2 = 3.2 \textrm{ V}$
3. $I_3 = 0.0089 \textrm{ A, } I_2 = 0.0047 \textrm{ A, } I_1 = 0.0042 \textrm{ A}$
COMMENTS
By Agp2196 on Sat, 3/21/2015 - 8:47 PM

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

By Mr. Dychko on Sun, 3/22/2015 - 4:54 AM

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

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