Download presentation

Presentation is loading. Please wait.

Published byJanice Benjamin Modified over 5 years ago

1
DC Circuits Quiz #1 Solution 1

2
Mr. McCall Find R T Add series resistors together. R 1,2 = R 1 + R 2 = 20.0 Ω + 14.0 Ω = 34.0 Ω R 4,5 = R 4 + R 5 = 16.0 Ω + 12.0 Ω = 28.0 Ω Add parallel resistors together. 1/R 3-5 = 1/R 3 + 1/R 4-5 = 1/18.0 Ω + 1/28.0 Ω = 11.0 Ω Add series resistors together. R T = R 1 + R 2-5 = 34.0 Ω + 11.0 Ω = 45.0 Ω R T = 45.0 Ω 2

3
V T = I T R T V T /R T = I T 45.0 V/45.0 Ω = I T = 1.0 A ` V 1,2 = I 1,2 R 1,2 V 1,2 = (1.0 A) (34.0 Ω) = 34.0 V 34.0 V V T = V 1 + V 2 45.0 V – 34.0 V = 11.0 V 11.0 V 1.0 A V 4,5 /R 4,5 = I 4,5 11.0 V/ 28 Ω = I 4 =.393 A.393 A V 4 = I 4 R 4 V 4 = (.393 A) (16.0 Ω) = V 4 = 6.29 V Mr. McCall 3

4
Find R T Add series resistors together. R 1,2 = R 1 + R 2 = 20.0 Ω + 12.0 Ω = 32.0 Ω R 4,5 = R 4 + R 5 = 14.0 Ω + 16.0 Ω = 30.0 Ω Add parallel resistors together. 1/R 3-5 = 1/R 3 + 1/R 4-5 = 1/18.0 Ω + 1/30.0 Ω = 11.3 Ω Add series resistors together. R T = R 1 + R 2-5 = 32.0 Ω + 11.3 Ω = 43.3 Ω R T = 43.3 Ω 4

5
Mr. McCall V T = I T R T V T /R T = I T 45.0 V/43.3 Ω = I T = 1.04 A V 1,2 = I 1,2 R 1,2 V 1,2 = (1.04 A) (32.0 Ω) = 33.3 V V T = V 1,2 + V 3 45.0 V – 33.3 V = 11.7 V V 3 /R = I 3 11.7 V/ 18 Ω = I 3 =.650 A 1.04 A 33.3 V 11.7 V.650 A V 3 = 11.7 V R T = 43.3 Ω 5

6
Find R T Add series resistors together. R 1,2 = R 1 + R 2 = 20.0 Ω + 12.0 Ω = 32.0 Ω R 4,5 = R 4 + R 5 = 14.0 Ω + 16.0 Ω = 30.0 Ω Add parallel resistors together. 1/R 3-5 = 1/R 3 + 1/R 4-5 = 1/18.0 Ω + 1/30.0 Ω = 11.7 Ω Add series resistors together. R T = R 1 + R 2-5 = 32.0 Ω + 11.7 Ω = 43.3 Ω R T = 43.3 Ω 6 Mr. McCall

7
7 R T = 43.3 Ω V T = I T R T V T /R T = I T 45.0 V/43.3 Ω = I T = 1.04 A I 2 = 1.04 A V 2 = I 2 R 2 V 2 = (1.04 A) (12.0 Ω) = 12.5 V V 2 = 12.5 V 1.04 A

8
Problem Solving Guidelines 1)Find the total resistance. 2)Find the total current through the battery. 3)Find the resistor in series or in parallel to the battery and determine what this resistor has in common with the battery. If the resistor is in series with the battery, then the resistor has the same current as the current through the battery. If the resistor is in parallel with the battery, then the resistor has the same voltage as the battery. 4)Use Ohm’s Law or Kirchhoff’s Loop rule or Junction rule to determine the needed value of current or voltage. 8

9
END 9

Similar presentations

© 2020 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google