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Unformatted text preview: EE 40: Introduction to Microelectronic Circuits Spring 2008: HW 6 Solution Venkat Anantharam March 14, 2006 Referenced problems from Hambley, 4th edition. 1. P5.56 Note that we have not been asked to find the voltage at the node between the 5 resistor and the 15 j inductor. We may therefore replace the series connection by a single impedance of the value 5 + 15 j before starting. KCL at node 1 gives 1 10 V 1 + 1 5 + j 15 ( V 1 V 2 ) = 1 KCL at node 2 gives 1 j 10 V 2 + 1 5 + j 15 ( V 2 V 1 ) = 1 e j/ 6 Rewriting, we see that ( 1 10 + 1 5 + 15 j ) V 1 ( 1 5 + 15 j ) V 2 = 1 (1) ( 1 5 + 15 j ) V 1 + ( 1 10 + 1 5 + 15 j ) V 2 = e j/ 6 (2) We choose to manipulate these two equations so that when we add them together the V 1 will cancel, giving a single equation in V 1 . Equation 1 can be multiplied by 2 * (5+ j 15) to eliminate the denominators, yielding 3(1 + j ) V 1 2 V 2 = 10 + j 30 . Multiplying this by 1 j (to get a real coefficient for V 1 ), we get 6 V 1 2(1 j ) V 2 = 40 + 20 j (3) 1 . Working with eq. 2 now, we first multiply through by 2 * (5+ j 15) to clear the denominators and obtain 2 V 1 + ( j 1) V 2 = 10(1 + j 3)( 3 3 + j 2 ) = 5( 3 3 + (3 3 + 1) j ) . (4) We can multiply this by 3, in order to add it to eq. 3. The sum of 3 * eq. 4 + eq. 3 is 5( j 1) V 2 = (15 3 5) + (45 3 + 35) j, or, dividing through by 5, ( j 1) V 2 = (3 3 1) + (9 3 + 7) j. (5) We can multiply each side by j + 1 to obtain 2 V 2 = (6 3 + 8) (12 3 + 6) j, or V 2 = (3 3 + 4) (6 3 + 3) j . It remains to find V 2 . We plug eq.5 into eq. 4, getting 2 V 1 + (3 3 1) + (9 3 + 7) j = (5 3 15) + (15 3 + 5) j, which we can solve for V 1 = (7 3) + (1 3 3) j. Therefore our final answer is V 2 = (3 3 + 4) (6 3 + 3) j = 9 . 1962 13 . 3923 i = 16 . 2457 6 55 . 5253 (6) V 1 = (7 3) + (1 3 3) j = 5 . 2679 4 . 1962 i = 55 . 5253 6 38 . 5394 (7) 2. P5.59 The KVL equations are 10 I 1 + j 20( I 1 I 2 ) = 0 j 20( I 2 I 1 ) + 10 + 15( I 2 I 3 ) = 0 15( I 3 I 2 ) + ( j 5) I 3 = 0 We simplify each equation individually to write: 2 (1 + j 2) I 1 = j 2 I 2 (8) j 20 I 1 + ( j 20 + 15) I 2 15 I 3 = 10 (9) 10 I 3 = (9 + j 3) I 2 (10) Plugging eq. 8 and eq. 10 into eq. 9, we have 4 j (2 j + 4) I 2 + ( j 20 + 15) I 2 3 2 (9 + 3 j ) I 2 = 10 which simplifies to I 2 = 10 9 . 5 . 5 j = 1 . 0497 . 0552 i = 1 . 0512 6 3 . 0102 ....
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