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This is solution to problems related Electrical Circuit Analysis course. It was given by Prof. Gurnam Kanth at Punjab Engineering College. Its main points are: Determinants, Phasor, Form, Voltages, Series, Sinusoids, Linear, Network, Associated, Impedance
Typology: Exercises
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Evaluate these determinants:
(a) j
j j
− − +
(b) ∠ ° ∠ °
(c)
j j
j j
j j
Chapter 9, Solution 15.
(a)
10 j 6 2 j 3
= -10 – j6 + j10 – 6 + 10 – j
= –6 – j
(b) ∠ ° ∠ °
= 57.96 + j15.529 + 63.03 – j11.
= 120.99 + j4.
(c)
j 1 j
1 j j 0
1 j 1 j
j 1 j
1 j j 0
= 1 1 0 1 0 j ( 1 j) j ( 1 j)
2 2
= 1 − 1 ( 1 −j+ 1 +j)
= 1 – 2 = –
Transform the following sinusoids to phasors:
(a) -10 cos (4 t + 75
o )
(b) 5 sin(20 t - 10
o )
(c) 4 cos2 t + 3 sin 2 t
Chapter 9, Solution 16.
(a) -10 cos(4t + 75°) = 10 cos(4t + 75° − 180 °)
= 10 cos(4t − 105 °)
The phasor form is 10 ∠ -105 °
(b) 5 sin(20t – 10°) = 5 cos(20t – 10° – 90°)
= 5 cos(20t – 100°)
The phasor form is 5 ∠ -100 °
(c) 4 cos(2t) + 3 sin(2t) = 4 cos(2t) + 3 cos(2t – 90°)
The phasor form is 4∠ 0 ° + 3∠-90° = 4 – j3 = 5 ∠ -36.87 °
Chapter 9, Problem 17.
Two voltages v 1 and v 2 appear in series so that their sum is v = v 1 + v 2. If v 1 = 10
cos(50t - 3
π (^) )V and v 2 = 12cos(50 t^ + 30^
o ) V, find v.
Chapter 9, Solution 17.
o o o V = V + V = < − + < = − j + + j = < −
15.62 cos(50 9.805 ) V
o v = t − = 15.62cos(50t–9.8˚) V
Using phasors, find:
(a) 3cos(20 t + 10º) – 5 cos(20 t - 30º)
(b) 40 sin 50 t + 30 cos(50 t - 45º)
(c) 20 sin 400 t + 10 cos(400 t + 60º) -5 sin(400 t - 20º)
Chapter 9, Solution 19.
(a) 3 ∠ 10 ° − 5 ∠-30° = 2.954 + j0.5209 – 4.33 + j2.
= -1.376 + j3.
= 3.32∠114.49°
Therefore, 3 cos(20t + 10°) – 5 cos(20t – 30°) = 3.32 cos(20t + 114.49 ° )
(b) 40 ∠-90° + 30∠-45° = -j40 + 21.21 – j21.
= 21.21 – j61.
= 64.78∠-70.89°
Therefore, 40 sin(50t) + 30 cos(50t – 45°) = 64.78 cos(50t – 70.89 ° )
(c) Using sinα = cos(α − 90 °),
20 ∠-90° + 10∠ 60 ° − 5 ∠-110° = -j20 + 5 + j8.66 + 1.7101 + j4. = 6.7101 – j6.
= 9.44∠-44.7°
Therefore, 20 sin(400t) + 10 cos(400t + 60°) – 5 sin(400t – 20°)
= 9.44 cos(400t – 44.7 ° )
Chapter 9, Problem 20.
A linear network has a current input 4cos ( ω t + 20º)A and a voltage output 10
cos( ω t +110º) V. Determine the associated impedance.
Chapter 9, Solution 20.
o o I = < V = <
10 110 2.5 90 2. 4 20
o o o
Z j I
Simplify the following:
(a) f ( t ) = 5 cos(2 t + 15(º) – 4sin(2 t -30º)
(b) g ( t ) = 8 sin t + 4 cos( t + 50º)
t t t dt 0
( 10 cos 40 50 sin 40 )
Chapter 9, Solution 21.
(a)
o o o o F = 5 ∠ 15 − 4 ∠− 30 − 90 = 6. 8296 + j 4. 758 = 8. 3236 ∠ 34. 86
( ) 8. 324 cos( 30 34. 86 )
o f t = t +
(b)
o o o G = 8 ∠− 90 + 4 ∠ 50 = 2. 571 − j 4. 9358 = 5. 565 ∠− 62. 49
( ) 5. 565 cos( 62. 49 )
o g t = t −
j
o o ∠ + ∠− ω= ω
i.e.
o o o H = 0. 25 ∠− 90 + 1. 25 ∠− 180 =−j 0. 25 − 1. 25 = 1. 2748 ∠− 168. 69
h(t) = 1.2748cos(40t – 168.69°)
Chapter 9, Problem 22.
An alternating voltage is given by v ( t ) = 20 cos(5 t - 30
o ) V. Use phasors to find
−∞
t
vtdt dt
dv 10 v ( t ) 4 2 ()
Assume that the value of the integral is zero at t = - ∞.
Chapter 9, Solution 22.
−∞
t
vtdt dt
dv 10 v ( t ) 4 2 ()
o V j
F V j V , 5 , 20 30
= 10 + 4 − ω= = ∠− ω
ω
o F = 10 V+j 20 V−j 0. 4 V=( 10 +j 20. 4 )( 17. 32 −j 10 )= 454. 4 ∠ 33. 89
f (t) 454. 4 cos( 5 t 33. 89 )
o = +
Using phasors, determine i ( t ) in the following equations:
(a) 2 3 () 4 cos( 2 45 )
o it t dt
di
(b) 10 (^) ∫ + + 6 ()= 5 cos( 5 + 22 )
o it t dt
di idt
Chapter 9, Solution 25.
(a)
2j ω I + 3 I = 4 ∠- 45 °, ω= 2
I ( 3 + j 4 )= 4 ∠- 45 °
3 j
Therefore, i(t) = 0.8 cos(2t – 98.13 ° )
(b)
j 6 5 22 , 5 j
10 + ω + = ∠ ° ω= ω
(- j 2 + j 5 + 6 ) I = 5 ∠ 22 °
6 j 3
Therefore, i(t) = 0.745 cos(5t – 4.56 ° )
Chapter 9, Problem 26.
The loop equation for a series RLC circuit gives
t i idt t dt
di 2 cos 2
Assuming that the value of the integral at t = - ∞ is zero, find i ( t ) using the phasor
method.
Chapter 9, Solution 26.
j
j 2 = ∠ ° ω= ω
ω + +
j 2
j 2 2 ⎟= ⎠
2 j 1. 5
Therefore, i(t) = 0.4 cos(2t – 36.87 ° )
A parallel RLC circuit has the node equation
50 100 110 cos(377 10 )
dv (^) o v v dt t dt
= + (^) ∫ = −
Determine v ( t ) using the phasor method. You may assume that the value of the integral at
t = - ∞ is zero.
Chapter 9, Solution 27.
j
j 50 100 = ∠ ° ω= ω
ω + +
j 100 V j 377 50
Therefore, v(t) = 0.289 cos(377t – 92.45 ° ).
Chapter 9, Problem 28.
Determine the current that flows through an 8- Ω resistor connected to a voltage source
vs = 110 cos 377 t V.
Chapter 9, Solution 28.
110 cos( 377 t)
R
v(t) i( t)
s 13.75 cos(377t) A.
