chap10

# chap10 - Chapter 10 Solution 1 =1 10 cos t 45 10 45 5 sin t...

This preview shows pages 1–6. Sign up to view the full content.

Chapter 10, Solution 1. ω 1 = ° → ° 45 - 10 ) 45 t cos( 10 ° ° + 60 - 5 ) 30 t sin( 5 j L j H 1 = ω j - C j 1 F 1 = ω The circuit becomes as shown below. V o 3 + 10 -45 ° V + 2 I o j 5 -60 ° V Applying nodal analysis, j - j ) 60 - 5 ( 3 ) 45 - 10 ( o o o V V V = ° + ° o j 60 - 15 45 - 10 j V = ° + ° ° = ° + ° = 9 . 247 15.73 150 - 15 45 - 10 o V Therefore, = ) t ( v o 15.73 cos(t + 247.9 ° ) V Chapter 10, Solution 2. ω 10 = ° π 45 - 4 ) 4 t 10 cos( 4 ° π + 150 - 20 ) 3 t 10 sin( 20 10 j L j H 1 = ω 5 j - 2 . 0 j 1 C j 1 F 02 . 0 = = ω

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
The circuit becomes that shown below. I o V o 10 j10 -j5 20 -150 ° V + 4 -45 ° A Applying nodal analysis, 5 j - 10 j 45 - 4 10 ) 150 - 20 ( o o o V V V + = ° + ° o ) j 1 ( 1 . 0 45 - 4 150 - 20 V + = ° + ° ° = + ° + ° = = 98 . 150 816 . 2 ) j 1 ( j 45 - 4 150 - 2 10 j o o V I Therefore, = ) t ( i o 2.816 cos(10t + 150.98 ° ) A Chapter 10, Solution 3. ω 4 = ° → 0 2 ) t 4 cos( 2 -j16 90 - 16 ) t 4 sin( 16 = ° 8 j L j H 2 = ω 3 j - ) 12 1 )( 4 ( j 1 C j 1 F 12 1 = = ω The circuit is shown below. V o + 2 0 ° A -j16 V 4 -j3 6 1 j8
Applying nodal analysis, 8 j 6 1 2 3 j 4 16 j - o o o + + = + V V V o 8 j 6 1 3 j 4 1 1 2 3 j 4 16 j - V + + + = + ° = ° ° = + = 02 . 35 - 835 . 3 88 . 1 2207 . 1 15 . 33 - 682 . 4 04 . 0 j 22 . 1 56 . 2 j 92 . 3 o V Therefore, = ) t ( v o 3.835 cos(4t – 35.02 ° ) V Chapter 10, Solution 4. 16 4 , 10 - 16 ) 10 t 4 sin( = ω ° → ° 4 j L j H 1 = ω j - ) 4 1 )( 4 ( j 1 C j 1 F 25 . 0 = = ω j4 1 I x -j 16 -10 ° V + V 1 0.5 I x + V o j 1 2 1 4 j ) 10 - 16 ( 1 x 1 = + ° V I V But 4 j ) 10 - 16 ( 1 x V I ° = So, j 1 8 j ) ) 10 - 16 (( 3 1 1 = ° V V

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
4 j 1 - 10 - 48 1 + ° = V Using voltage division, ° = + ° = = 04 . 69 - 232 . 8 ) 4 j 1 j)(- - (1 10 - 48 j 1 1 1 o V V Therefore, = ) t ( v o 8.232 sin(4t – 69.04 ° ) V Chapter 10, Solution 5. Let the voltage across the capacitor and the inductor be V x and we get: 0 3 j V 2 j V 4 30 10 I 5 . 0 V x x x x = + + ° x x x x x V 5 . 0 j 2 j V I but 30 30 I 5 . 1 V ) 4 j 6 j 3 ( = = ° = + Combining these equations we get: A 38 . 97 615 . 4 25 . 1 j 3 30 30 5 . 0 j I 25 . 1 j 3 30 30 V or 30 30 V ) 75 . 0 j 2 j 3 ( x x x ° = + ° = + ° = ° = + Chapter 10, Solution 6. Let V o be the voltage across the current source. Using nodal analysis we get: 0 10 j 20 V 3 20 V 4 V o x o = + + where o x V 10 j 20 20 V + = Combining these we get: 30 j 60 V ) 3 5 . 0 j 1 ( 0 10 j 20 V 3 10 j 20 V 4 20 V o o o o + = + = + + + = + = + + = 5 . 0 j 2 ) 3 ( 20 V or 5 . 0 j 2 30 j 60 V x o 29.11 –166 ˚ V .
Chapter 10, Solution 7. At the main node, + + + = + → + + = + 50 1 30 j 20 j 40 1 V 3 j 196 . 5 20 j 40 058 . 31 j 91 . 115 50 V 30 j V 30 6 20 j 40 V 15 120 o o V 154 08 . 124 0233 . 0 j 04 . 0 7805 . 4 j 1885 . 3 V o = + = Chapter 10, Solution 8. , 200 = ω 20 j 1 . 0 x 200 j L j mH 100 = = ω 100 j 10 x 50 x 200 j 1 C j 1 F 50 6 = = ω µ The frequency-domain version of the circuit is shown below.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 91

chap10 - Chapter 10 Solution 1 =1 10 cos t 45 10 45 5 sin t...

This preview shows document pages 1 - 6. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online