ee40sp08hwk6-sol

ee40sp08hwk6-sol - EE 40 Introduction to Microelectronic...

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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 5Ω resistor and the 15 j Ω inductor. We may therefore replace the series connection by a single impedance of the value 5 + 15 j Ω 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 jπ/ 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 πj/ 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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ee40sp08hwk6-sol - EE 40 Introduction to Microelectronic...

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