CK-12-Physics-Concepts - Intermediate

19.3. Combined Series-Parallel Circuits http://www.ck12.org

Solution:We start by simplifying the parallel resistorsR 2 andR 3.

1

R 23

=

1

180 Ω

+

1

220 Ω

=

1

99 Ω

R 23 = 99 Ω

We then simplifyR 1 andR 23 which are series resistors.

RT=R 1 +R 23 = 110 Ω+ 99 Ω= 209 Ω

We can then find the total current,IT=VRTT = 20924 VΩ= 0. 11 A

All the current must pass throughR 1 , soI 1 = 0. 11 A.

The voltage drop throughR 1 is( 110 Ω)( 0. 11 A) = 12 .6 volts.

Therefore, the voltage drop throughR 2 andR 3 is 11.4 volts.

I 2 =VR 22 =^11180.^4 ΩV= 0. 063 AandI 3 =VR^33 =^11220.^4 ΩV= 0. 052 A

Summary

• Combined circuit problems should be solved in steps.

Practice

Questions

Video teaching the process of simplifying a circuit that contains both series and parallel parts.

http://www.youtube.com/watch?v=In3NF8f-mzg

MEDIA

Click image to the left for use the URL below.

URL: http://www.ck12.org/flx/render/embeddedobject/63608

Follow up questions:

1. In a circuit that contains both series and parallel parts, which parts of the circuit are simplified first?