PracticeProblem3SolutionDetailed

PracticeProblem3SolutionDetailed - Practice Problem 3...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Practice Problem 3 Solution in Detail John McRaven Figure 1: 3) For the circuit in Figure 1, calculate the equivalent resistance for the circuit. Calculate the current, voltage, and power for each resistor and the battery. Solution: There are 3 resistors in this circuit. The 4Ω and 12Ω resistors are in parallel. The 5Ω resistor is in series with the combination of the 4Ω and 12Ω resistor. So, we must first combine the 4Ω and 12Ω resistor using the parallel resistor equation: 1 R P = 1 R 1 + 1 R 2 = 1 12 + 1 4 = 4 12 = 1 3 R P = 3Ω R P is now in series with the 5Ω resistor. We use the series equation to combine them. R S = R 1 + R 2 = 5 + R P = 5 + 3 = 8Ω 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
We have now reduced the circuit to one resistor, so 8Ω is the equivalent resistance. R eq = 5 + 1 1 4 + 1 12 = 8 We can now calculate the total current in the circuit. Whenever the circuit has been reduced to one battery and one resistor, the current is: I = E R eq = 16 8 = 2 A This makes the power coming out of the battery:
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/05/2011 for the course PHY 7B PHY7B taught by Professor Cebra during the Spring '10 term at UC Davis.

Page1 / 3

PracticeProblem3SolutionDetailed - Practice Problem 3...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online