Ee40sp08hwk2-sol - EE40 Spring 2008 Homework 2 Solutions 1 a The circuit can be redrawn as And then the portion can be replaced by a(2 4/6 = 6/6 =

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EE40 Spring 2008 Homework 2 Solutions 1) a) The circuit can be redrawn as And then the portion: can be replaced by a (2+4)//6 = 6//6 = 3 resistor as shown in the drawing. Then we get, by similar reasoning (3+3)//3 = 6//3 = 2 for the larger parallel block. Therefore the total resistance is R eq = 2+2+8+2+2 = 16 . b) Shorting c to d means no voltage drop between those two points, so no current on the 8 or 2 resistors, and we can just ignore them. Hence the total resistance is R eq = 2+2+0+2 = 6 . 2) Hambley P 2.17: Combining the 20 resistor with the 30 gives 20*30/50 = 60/5=12 , then we note as in the drawing above that 60 // (8+12) = 60 // 20 = 120/8=15, for a total resistance of 7+15=22 .
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3) Hambley P 2.24 Label the blacked nodes in the text by d,a,b,c going counterclockwise from top left. The total current flowing into and out of the parallel combination of the 6 and 12 resistors on the right is 2A. By the current divider formula, i 2 = 2*(1/12) /( 1/12 + 1/6 ) = 2/3A. The total current flowing into and out of the parallel combination of the 12 and 24 resistors on the left is 2A. By the current divider formula, A i 3 4 2 24 / 1 12 / 1 12 / 1 1 = + = . Thus v a -v b = 12*i 2 = 8V v b v c = 8*2 = 16V v c v d = 12*i 1 = 16V By KVL, v = v a - v d = 8+16+16 = 40V 4) Hambley P 2.32 With the switch open, by the voltage division formula v 2 = 10*R 2 / (R 2 + 6) V. Since
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This note was uploaded on 04/07/2008 for the course EE 40 taught by Professor Chang-hasnain during the Spring '07 term at University of California, Berkeley.

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Ee40sp08hwk2-sol - EE40 Spring 2008 Homework 2 Solutions 1 a The circuit can be redrawn as And then the portion can be replaced by a(2 4/6 = 6/6 =

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