Chap8_Practice (1)

# Chap8_Practice (1) - Problem 8.3 Determine(a the average...

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Problem 8.3 Determine (a) the average and (b) rms values of the periodic current waveform shown in Fig. P8.3. 3 4 5 6 8 7 2 1 0 1 2 3 4 t (s) i (A) Figure P8.3: Waveform for Problem 8.3. Solution: (a) i ( t ) = braceleftBigg 2 A , for 0 t 2 s 4 A , for 2 t 4 s T = 4 s . I av = 1 4 bracketleftbigg integraldisplay 2 0 2 dt + integraldisplay 4 2 4 dt bracketrightbigg = 1 4 bracketleftbig 2 t | 2 0 + 4 t | 4 2 bracketrightbig = 3 A . (b) I rms = braceleftbigg 1 4 bracketleftbigg integraldisplay 2 0 4 dt + integraldisplay 4 2 16 dt bracketrightbiggbracerightbigg 1 / 2 = braceleftbigg 1 4 bracketleftbig 4 t | 2 0 + 16 t | 4 2 bracketrightbig bracerightbigg 1 / 2 = 3 . 16 A .

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Problem 8.11 In the circuit of Fig. P8.11, i s ( t ) = 0 . 2sin10 5 t A, R = 20 , L = 0 . 1 mH, and C = 2 μ F. Show that the sum of the complex powers for the three passive elements is equal to the complex power of the source. R L C i s ( t ) + _ Figure P8.11: Circuit for Problem 8.11. Solution: + _ 20 - j 5 j 10 V I R I s I L I C I s = 0 . 2 0 A Z L = j ϖ L = j 10 5 × 10 4 = j 10 Z C = j ϖ C = j 10 5 × 2 × 10 6 = j 5 . V 20 + V j 10 + V j 5 = I s = 0 . 2 V = 1 . 79 e j 63 . 4 V . I R = V 20 = 1 . 79 e j 63 . 4 20 A . S R = 1 2 VI R = ( 1 . 79 ) 2 2 × 20 = 0 . 08 VA . I L = V j 10 = 1 . 79 10 e j 153 . 4 S L = 1 2 VI L = 1 . 79 2 e j 63 . 4 × 1 . 79 10 e j 153 . 4 = 0 . 16 e j 90 = 0 + j 0 . 16 VA . I C = V j 5 = 1 . 79 5 e j 26 . 6 A . S C = 1 2 VI C = 1 2 1 . 79 e j 63 . 4 × 1 . 79 5 e j 26 . 6 = 0 . 32 e j 90 = 0 j 0 . 32 VA . S T = S R + S L + S C = 0 . 08 + j 0 . 16 j 0 . 32 = 0 . 08 j 0 . 16 VA . For the source, S s = 1 2 VI s = 1 2 1 . 79 e j 63 . 4 × 0 . 2 = 0 . 179 e j 63 . 4 = 0 . 179cos63 . 4 j 0 . 179sin63 . 4
= 0 . 08 j 0 . 16 VA . Hence S T = S s .

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Problem 8.16 Determine the power dissipated in R L of the circuit in Fig. P8.16. 3 R L = 4 6 - j 1 j 3 b a + _ + _ 8 V 45 o 2 I C I C Figure P8.16: Circuit for Problem 8.16. Solution: 3 R L = 4 6 - j 1 j 3 b a + _ + _ 8 V 45 o 2 I C I C I 1 I 2 I 3 8 e j 45 + 3 I 1 + j 3 I 1 j ( I 1 I 2 ) = 0 j ( I 2 I 1 ) 2 I C + 6 ( I 2 I 3 ) = 0 6 ( I 3 I 2 )+ 4 I 3 = 0 Also, I C = I 1 I 2 .
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