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Homework 06-solutions

# Homework 06-solutions - sanne(as42476 Homework 06 Yao(59110...

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sanne (as42476) – Homework 06 – Yao – (59110) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points Consider the combination of resistors shown in the figure. 4 . 5 Ω 1 Ω 4 . 9 Ω 9 Ω 7 . 8 Ω 2 . 2 Ω 3 . 7 Ω a b What is the resistance between point a and point b ? Correct answer: 7 . 92496 Ω. Explanation: Let’s redraw the figure i R 1 i R 2 i R 5 i R 4 i R 3 i R 6 i R 7 a b Let : R 1 = 4 . 5 Ω , R 2 = 1 Ω , R 3 = 4 . 9 Ω , R 4 = 9 Ω , R 5 = 7 . 8 Ω , R 6 = 2 . 2 Ω , and R 7 = 3 . 7 Ω . Basic Concepts: Equivalent resistance. Ohm’s Law. There are two rules for adding up resis- tances. If the resistances are in series, then R series = R 1 + R 2 + R 3 + · · · + R n . If the resistances are parallel, then 1 R parallel = 1 R 1 + 1 R 2 + 1 R 3 + · · · + 1 R n . Solution: The key to a complex arrange- ments of resistors like this is to split the prob- lem up into smaller parts where either all the resistors are in series, or all of them are in parallel. It is easier to visualize the problem if you redraw the circuit each time you add them. i R 1 i R 2 i R 5 i R 4 i R 367 a b Step 1: The three resistors on the right are all in series, so R 367 = R 3 + R 6 + R 7 = (4 . 9 Ω) + (2 . 2 Ω) + (3 . 7 Ω) = 10 . 8 Ω . i R 1 i R 2 i R 3675 i R 4 a b Step 2: R 5 and R 367 are connected paral- lel, so R 3675 = parenleftbigg 1 R 5 + 1 R 367 parenrightbigg 1 = R 5 R 367 R 5 + R 367 = (7 . 8 Ω) (10 . 8 Ω) 18 . 6 Ω = 4 . 52903 Ω .

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sanne (as42476) – Homework 06 – Yao – (59110) 2 i R 1 i R 36752 i R 4 a b Step 3: R 2 and R 3675 are in series, so R 23675 = R 2 + R 3675 = (1 Ω) + (4 . 52903 Ω) = 5 . 52903 Ω . Step 4: R 23675 and R 4 are parallel, so R 236754 = parenleftbigg 1 R 4 + 1 R 23675 parenrightbigg 1 = R 4 R 23675 R 4 + R 23675 = (9 Ω) (5 . 52903 Ω) 14 . 529 Ω = 3 . 42496 Ω . i R 1 i R 367524 a b Step 5: Finally, R 1 and R 236754 are in se- ries, so the equivalent resistance of the circuit is R eq = R 1 + R 236754 = (4 . 5 Ω) + (3 . 42496 Ω) = 7 . 92496 Ω . 002 (part 2 of 2) 10.0 points If the current in the right-hand 2 . 2 Ω resis- tor is 6 A , what is the potential difference between points a and b ? Correct answer: 183 . 046 V. Explanation: Let : R 6 = 6 A . Current, resistance, and voltage drop are related by V = I R . Since all the resis- tances and the current through one resistor are known, all the voltages and currents can be calculated. Then the same current through R 6 must go through R 3 and R 7 since they are in series. This is the same as saying there is 6 A going through R 367 (see previous part). V 8 = I 8 R 367 = (6 A) (10 . 8 Ω) = 64 . 8 V . Now since R 367 and R 5 share the same end- points (they are connected in parallel), the potential difference across each must be the same. So, V 5 = 64 . 8 V , and the current through R 5 is I 5 = V 5 R 5 = 64 . 8 V 7 . 8 Ω = 8 . 30769 A . A total of 14 . 3077 A goes through R 2 and is split, with 8 . 30769 A going through R 5 and 6 A going through R 367 . From Step 3 in the previous part, we know that the R 2 , R 5 , R 367 combination is equiva- lent to R 23675 . This means that the potential difference across R 23675 is V 10 = R 23675 I 10 = (14 . 3077 A) (5 . 52903 Ω) = 79 . 1077 V , which is also the voltage drop across R 4 .
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