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quiz_solutions05

# Fundamentals of Physics, (Chapters 21- 44) (Volume 2)

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PY212 Solutions Quiz 5 First we identify all the current and assign them an arbitrary direction. There are three different currents: i 1 moving through ε 1 , i 2 moving through ε 2 , and i 3 moving through ε 3 . We shall assume for now that they are all going upwards (this is obviously a wrong assumption, but the sign of each current will indicate the correct direction at the end). The junction rule now tells us that 0 3 2 1 out in = + + = i i i i i Now consider the loops. There are two unique loops and we can walk through them in a clockwise manner. Starting from point b , we obtain 0 5 6 2 0 2 1 2 2 2 1 1 1 1 1 = + = + + i i R i R i R i ε for the left loop, and 0 6 5 2 0 3 2 1 3 3 1 3 2 2 2 = + = + + + i i R i R i R i for the right loop. We now have three equations and three unknowns (each of the currents). Hence, in principle, we can solve this system of equations. There are many different ways to do this. Your calculator might have a special method; check your manual how to solve a set of equations with several unknowns. Here we’ll show how one can do it without a calculator. Let us first remove
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Unformatted text preview: i 3 from the equations by writing 2 1 3 i i i − − = . Plugging this into the other two equations we obtain 2 1 2 1 2 2 1 11 6 ) ( 6 5 2 5 6 2 i i i i i i i − − = − − + − = + − = We now have two equations and two unknowns, so we’re heading in the right direction. Now rewrite the upper equation to 1 2 6 2 5 i i + = and substitute this into the other: A. 33 . 3 1 2 . 19 4 . 6 2 . 13 4 . 4 6 5 6 2 11 6 2 1 1 1 1 1 − = − = − = ⇒ − − − = + − − = i i i i i (a) and (b) We see that i 1 = 0.33 A and it’s directed down . (c) and (d) The current i 2 is actually zero: ( ) ( ) 5 / ) 3 / 1 ( 6 2 5 / 6 2 1 2 = − + = + = i i . (e) and (f) The current i 3 is given by A 33 . 2 1 3 + = − − = i i i , so it’s directed up . (g) To find V a – V b we walk from point b to a : V. . 5 5 5 2 2 2 2 = − = − i R i...
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