TRANSCRIBE THE FOLLOWING TEXT IN DIGITAL FORMAT

Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
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TRANSCRIBE THE FOLLOWING TEXT IN DIGITAL FORMAT

(8)
Current through the load
Power
ab Lorbed
S =
S
Power factor of
=
4₁₂ = £₂-fR
by
load
load
*
10 Lo- 4L35
7-104 L-18-84
Avg power delivered to load
Apparent power delivered
or
A
supplied by
VL CL
80L35 x (7-1042-18-84 ) *
80 L35 x 7-104 L18.84
568-32 [53-84 VA =
=
335 32
to load
Power factor = Coso
P-f=
=
Source to load
Cos 53.84
0.59
568-32 L53-89
335 32 +458-86 1
=
568.32
VA
Transcribed Image Text:(8) Current through the load Power ab Lorbed S = S Power factor of = 4₁₂ = £₂-fR by load load * 10 Lo- 4L35 7-104 L-18-84 Avg power delivered to load Apparent power delivered or A supplied by VL CL 80L35 x (7-1042-18-84 ) * 80 L35 x 7-104 L18.84 568-32 [53-84 VA = = 335 32 to load Power factor = Coso P-f= = Source to load Cos 53.84 0.59 568-32 L53-89 335 32 +458-86 1 = 568.32 VA
Given
(d)
C = 4 L35
have
that
We
Power supplied
by each
€
each component
To find above parameter we
202
which
Power
to calculate
or absorb
V
the
resistor
to
(a) Power Supplied by source
Circuit
5 =
/
the
£x 20
*
S= V₁₂ £5²
800 (35
(a)
Avg power supplied by source
(d) Apparent power supplied by source
(b)
10 Lo
Power supplied to load
Amp
4L35 x 20
the
will first find the voltage acralt
is
given by
202 resistor
PR = 111² R
90 L30 x (1010)
=
2012
*
80 L35
Lood
80 L35 v
655.32 +458.86 i
655.32 W
= 800 VA
PR= 320 W
(es. Apparent power across the 2012 resistor will also be 320 VA
the resistance has a unit power factor
because
S = 320 VA
4²x 20
= 800 L35 VA
320 W
To find power supplied to load we will find the current
flowing
across
load I
voltage
across it.
VL = VR
As they are in parallel }
Transcribed Image Text:Given (d) C = 4 L35 have that We Power supplied by each € each component To find above parameter we 202 which Power to calculate or absorb V the resistor to (a) Power Supplied by source Circuit 5 = / the £x 20 * S= V₁₂ £5² 800 (35 (a) Avg power supplied by source (d) Apparent power supplied by source (b) 10 Lo Power supplied to load Amp 4L35 x 20 the will first find the voltage acralt is given by 202 resistor PR = 111² R 90 L30 x (1010) = 2012 * 80 L35 Lood 80 L35 v 655.32 +458.86 i 655.32 W = 800 VA PR= 320 W (es. Apparent power across the 2012 resistor will also be 320 VA the resistance has a unit power factor because S = 320 VA 4²x 20 = 800 L35 VA 320 W To find power supplied to load we will find the current flowing across load I voltage across it. VL = VR As they are in parallel }
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