# - Catalyst

```Problem 7.3
The voltage and current of an electric load can be expressed by the following equations
v  340 sin (628.318 t  0.5236)
i  100 sin (628.318 t  0.87266)
Calculate the following:
a.
b.
c.
d.
e.
The rms voltage
The frequency of the current
The phase shift between current and voltage in degrees
The average voltage
Solution:
340
 240.41 V
2
a.
V
b.
  2 f  628.318 rad/sec
f
c.
628.318
 100 Hz
2
Phase shift
   0.5236  0.87266  0.3491 rad
  0.3491
180

 20 o
d.
Average voltage is zero (the voltage waveform is sinusoidal)
e.
Z




V 340 / 2

  20o  3.4   20o 
I 100 / 2
Problem 7.5
An ac source is feeding a load consists of a resistance and inductive reactance connected in series. The voltage and
current of the source are
v = 400 sin (377 t + 0.5236)
i = 100 sin (377 t + 0.87266)
Calculate the following:
V
A
a.
b.
c.
d.
e.
f.
The resistance and inductive reactance of the load impedance
The rms voltage across the resistance
The frequency of the supply voltage
Power factor at the source side, state leading or lagging
The-real power delivered to the load
The-reactive power delivered to the load
Solution:
0.5236 *180

Z
 4  20 o 
0.87266 *180
100

100
4 cos 20 o   265.78 V
VR  I R 
2
 377
f 

 60Hz
2 2
pf  cos(0.5236  0.87266)  0.939 leading
400 *100
P  V I cos  
0.939  18.78 kW
2
400
a.
b.
c.
d.
e.
Q
f.
V I 
2
2
 400 * 100 
2
P  
  18780  6.88 kVAr Capacitive
2


2
Problem 7.7
A load impedance consists of 25 resistance in series with 38 inductive reactance. The load is connected across a
240 V source. Compute the real and reactive powers consumed by the load.
Solution:
Z  252  382  45.5 ohm
I 
V 240

 5.28 A
Z 45.5
pf  cos( ) 
25
R

 0.55
45.5
Z
P  V I cos( )  240  5.28  0.55  696.96 W
sin( ) 
38
XL

 0.8352
45.5
Z
Q  V I sin( )  240  5.28  0.8352  1.058 kVAr
Problem 7.12
An electric load consists of 5 resistance, 20 capacitive reactance and 10 inductive reactance, all connected in
parallel. A voltage source of 100V is applied across the load. Compute the following:
a.
b.
c.
The real power of the load
The reactive power of the load
Solution:
a.
100 2
 2 kW
5
b.
100 2 100 2

 500 VAr
10
20
c.
S  2000 2  500 2  2.062 kVA
I 
2062
 20.62 A
100
```
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