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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: Instantaneous, Power, Average, Current, Division, Inductor, Capacitor, Complex, Conservation
Typology: Exercises
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Chapter 11, Problem 1.
If v ( t ) = 160 cos 50 t V and i ( t ) = –20 sin(50 t – 30°) A, calculate the instantaneous power
and the average power.
Chapter 11, Solution 1.
v( t)= 160 cos( 50 t )
i( t)=- 20 sin( 50 t− 30 °)= 2 cos( 50 t− 30 °+ 180 °− 90 ° )
i( t)= 20 cos( 50 t+ 60 ° )
p( t)=v(t)i(t)=( 160 )( 20 )cos( 50 t)cos( 50 t+ 60 ° )
p( t) = 800 + 1600 cos( 100 t + 60 ° )W
( 160 )( 20 )cos( 60 ) 2
V I cos( ) 2
P = m m θv−θi = °
Chapter 11, Problem 2.
Given the circuit in Fig. 11.35, find the average power supplied or absorbed by each
element.
Figure 11.
For Prob. 11.2.
Chapter 11, Solution 2.
Using current division,
j 1 Ω I 2 I (^1)
Vo
-j4 Ω 2 ∠ 0
o A 5 Ω
1
j j j I j j j
2
j j j
.
For the inductor and capacitor, the average power is zero. For the resistor,
2 2 1
Vo = 5 I 1 = −2.6471 − j 4.
(^1) * 1 ( 2.6471 4.4118) 2 2.6471 4. 2 2
S = V I o = − − j x = − − j
Hence the average power supplied by the current source is 2.647 W.
Figure 11.
For Prob. 11.4.
o o o o o
V j j j
For the 5-Ω resistor,
1
o Vo o I
The average power dissipated by the resistor is
2 2 1 1 1
P = I R = x x =
For the 8-Ω resistor,
Vo o I j
The average power dissipated by the resistor is
2 2 2 2 2
P = I R = x x =
The complex power supplied is
1
o o S = V Is = < < = + j
Adding P 1 and P 2 gives the real part of S, showing the conservation of power.
Chapter 11, Problem 6.
For the circuit in Fig. 11.38, i (^) s t
3 = 6 cos 10 A. Find the average power absorbed by the
50- Ω resistor.
Figure 11.
For Prob. 11.6.
Chapter 11, Solution 6.
3 3
− ⎯⎯→ = =
j 25 j 10 x 40 x 10
j C
3 6
ω
μ → −
We apply nodal analysis to the circuit below.
Vo 20I (^) x
I (^) x
j
50
6 ∠ 0
o
–j25 10
50 j 25
10 j 20
−
But 50 j 25
o x −
=. Substituting this and solving for Vo leads
( 0. 02 j 0. 04 0. 012802 j 0. 009598 0. 016 j 0. 008 )V 6
50 j 25
( 50 j 25 )
( 10 j 20 )
10 j 20
o
o
o
(0.0232 – j0.0224)Vo = 6 or Vo = 6/(0.03225∠–43.99˚ = 186.05∠43.99˚
For power, all we need is the magnitude of the rms value of I (^) x.
|Ix| = 186.05/55.9 = 3.328 and |Ix| (^) rms = 3.328/1.4142 = 2.
We can now calculate the average power absorbed by the 50-Ω resistor.
Pavg = (2.353)
2 x50 = 276.8 W.
Chapter 11, Problem 8.
In the circuit of Fig. 11.40, determine the average power absorbed by the 40- Ω resistor.
Figure 11.
For Prob. 11.8.
