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7BSp2009Q2Rubric

# 7BSp2009Q2Rubric - Physics 7B Quiz 2 Rubric a(30 of quiz...

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Physics 7B Quiz 2 Rubric a) (30% of quiz) Solution: Label the resistors as follows: R 1 = 5 Ω , R 2 = 2 Ω , R 3 = 4 Ω , R 4 = 7 Ω . First we see that since both resistors R 2 and R 3 connect points b and d, they both must have the same voltage drops across them, meaning that they are in parallel. We can find the equivalent resistance of this parallel combination that we will call R 5 . Adding these in parallel gives: so that we get R 5 = 1.33 Ω . Now we have an equivalent circuit where R 1 , R 5 and R 4 are all in series. Using the rule for adding resistors in series, we get the final equivalent diagram which only has one resistor, which we will call R 6 , and the battery. R 6 = R 1 + R 5 + R 4 = 13.33 Ω . Q 10.0 Completely correct answer. R 9.4 Minor math error or wrote correct formula but didn’t plug in final numbers. S 8.5 Had the correct idea, but implemented the formula for parallel equivalent resistance incorrectly. For example, used . T 6.0 Made some progress toward the answer but did not identify correctly which resistors were in series and which were in parallel. V 5.0 Added all the resistors as if they were in series. W 4.0 Major conceptual error such as adding three of the resistors as if they were in series but leaving the 4 Ω resistor out, saying that it is not important because no current will flow through it. b) (25% of quiz) Solution: From part a), we know that there is an equivalent circuit with one 13.33 Ω resistor and a battery with a voltage of 12V. The section of the circuit from e f is part of this equivalent circuit so it just carries the current in the equivalent circuit.

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