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Unformatted text preview: Physics 21 Fall, 2009 Solution to HW9 2624 In the circuit shown in the figure above, find (a) the current in each branch and (b) the potential difference V ab of point a relative to point b . (a) Using the loop and junction equations, we can obtain a set of three equations to solve for the three currents. Junction: I 1 = I 2 + I 3 Upper loop: 10 − 2 I 1 − I 2 − 5 − 4 I 2 − 3 I 1 = 0 Lower loop: 5 + I 2 − 10 I 3 + 4 I 2 = 0 . Simplifying the two loop equations gives 5 I 1 + 5 I 2 = 5 − 5 I 2 + 10 I 3 = 5 or I 1 + I 2 = 1 − I 2 + 2 I 3 = 1 Now use the loop equation I 1 = I 2 + I 3 to eliminate I 1 from the first equation above: 2 I 1 + I 3 = 1 − I 2 + 2 I 3 = 1 Solving two equations in two unknowns, we get the final answer: I 1 = 0 . 800 A, I 2 = 0 . 200 A, I 3 = 0 . 600 A (b) If we start at b and go around loop 1 to a , we will get V a − V b . We can go in either direction. Here is the result for going counterclockwise: V a − V b = − 4 I 2 − 3 I 1 = − (0 . 2 A)(4Ω) − (0 . 8 A)(3 Ω) = − 3 . 20 V 2641 In the circuit shown in the figure both capacitors are initially charged to 45.0 V. (a) How long after closing the switch S will the potential across each capacitor be reduced to 15.0 V? (b) What will be the current at that time?...
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This note was uploaded on 11/06/2009 for the course PHYS 21 taught by Professor Hickman during the Spring '07 term at Lehigh University .
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 Hickman
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