PracticeProblem3SolutionDetailed

# PracticeProblem3SolutionDetailed - Practice Problem 3...

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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

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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:
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## 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.

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PracticeProblem3SolutionDetailed - Practice Problem 3...

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