This document presents the solution to a circuit analysis problem. It is given that the supply voltage is 240V, frequency is 60Hz, active power is 1.5kW, and current is 9.615A.
The document calculates the power factor, resistive and inductive components, and determines that the load is inductive. It then calculates the value of inductance as 50.31365mH.
Finally, it calculates the required capacitance of 80.761uF to compensate the power factor to unity.
In epicyclic gear train ,the axis of at least one of the gears moves relative to the frame. A gear train having a relative motion of axes is called a planetary or an epicyclic gear train.
In epicyclic gear train ,the axis of at least one of the gears moves relative to the frame. A gear train having a relative motion of axes is called a planetary or an epicyclic gear train.
Charge Quantization and Magnetic MonopolesArpan Saha
This talk, given as a part of the Annual Seminar Weekend 2011, IIT Bombay, dealt with a homotopy-based
variant of the argument Dirac provided to show that the existence of a single magnetic monopole in the Universe
is a sufficient condition for the quantization of electric charge.
A numerical problem wherein the total inductance of an electromechanical energy conversion device is calculated furthermore the effect of changing the airgap length on the static force is also observed
SolutionsPlease see answer in bold letters.Note pi = 3.14.docxrafbolet0
Solution
s:
Please see answer in bold letters.
Note pi = 3.1415….
1. The voltage across a 15Ω is as indicated. Find the sinusoidal expression for the current. In addition, sketch the v and i waveform on the same axis.
Note: For the graph of a and b please see attached jpg photo with filename 1ab.jpg and for c and d please see attached photo with filename 1cd.jpg.
a. 15sin20t
v= 15sin20t
By ohms law,
i = v/r
i = 15sin20t / 15
i = sin20t A
Computation of period for graphing:
v= 15sin20t
i = sin20t
w = 20 = 2pi*f
f = 3.183 Hz
Period =1/f = 0.314 seconds
b. 300sin (377t+20)
v = 300sin (377t+20)
i = 300sin (377t+20) /15
i = 20 sin (377t+20) A
Computation of period for graphing:
v = 300sin (377t+20)
i = 20 sin (377t+20)
w = 377 = 2pi*f
f = 60 Hz
Period = 1/60 = 0.017 seconds
shift to the left by:
2pi/0.017 = (20/180*pi)/x
x = 9.44x10-4 seconds
c. 60cos (wt+10)
v = 60cos (wt+10)
i = 60cos (wt+10)/15
i = 4cos (wt+10) A
Computation of period for graphing:
let’s denote the period as w sifted to the left by:
10/180*pi = pi/18
d. -45sin (wt+45)
v = -45sin (wt+45)
i = -45sin (wt+45) / 15
i = -3 sin (wt+45) A
Computation of period for graphing:
let’s denote the period as w sifted to the left by:
45/180 * pi = 1/4*pi
2. Determine the inductive reactance (in ohms) of a 5mH coil for
a. dc
Note at dc, frequency (f) = 0
Formula: XL = 2*pi*fL
XL = 2*pi* (0) (5m)
XL = 0 Ω
b. 60 Hz
Formula: XL = 2*pi*fL
XL = 2 (60) (5m)
XL = 1.885 Ω
c. 4kHz
Formula: XL = 2*pi*fL
XL = = 2*pi* (4k)(5m)
XL = 125.664 Ω
d. 1.2 MHz
Formula: XL = 2*pi*fL
XL = 2*pi* (1.2 M) (5m)
XL = 37.7 kΩ
3. Determine the frequency at which a 10 mH inductance has the following inductive reactance.
a. XL = 10 Ω
Formula: XL = 2*pi*fL
Express in terms in f:
f = XL/2 pi*L
f = 10 / (2pi*10m)
f = 159.155 Hz
b. XL = 4 kΩ
f = XL/2pi*L
f = 4k / (2pi*10m)
f = 63.662 kHz
c. XL = 12 kΩ
f = XL/2piL
f = 12k / (2pi*10m)
f = 190.99 kHz
d. XL = 0.5 kΩ
f = XL/2piL
f = 0.5k / (2pi*10m)
f = 7.958 kHz
4. Determine the frequency at which a 1.3uF capacitor has the following capacitive reactance.
a. 10 Ω
Formula: XC = 1/ (2pifC)
Expressing in terms of f:
f = 1/ (2pi*XC*C)
f = 1/ (2pi*10*1.3u)
f = 12.243 kΩ
b. 1.2 kΩ
f = 1/ (2pi*XC*C)
f = 1/ (2pi*1.2k*1.3u)
f = 102.022 Ω
c. 0.1 Ω
f = 1/ (2pi*XC*C)
f = 1/ (2pi*0.1*1.3u)
f = 1.224 MΩ
d. 2000 Ω
f = 1/ (2pi*XC*C)
f = 1/ (2pi*2000*1.3u)
f = 61.213 Ω
5. For the following pairs of voltage and current, indicate whether the element is a capacitor, an inductor and a capacitor, an inductor, or a resistor and find the value of C, L, or R if insufficient data are given.
a. v = 55 sin (377t + 50)
i = 11 sin (377t -40)
Element is inductor
In this case voltage leads current (ELI) by exactly 90 degrees so that means the circuit is inductive and the element is inductor.
XL = 55/11 = 5 Ω
we know the w=2pif so
w= 377=2pif
f= 60 Hz
To compute for th.
1. 0
0
1
1
2
2
3
3
4
4
5
5
6
6
7
7
8
8
9
9
10
10
11
11
12
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14
14
A A
B B
C C
D D
E E
F F
G G
H H
I I
J J
V1
240 Vrms
60 Hz
0°
R1
16.224
U1
9.615 A
+ -
L1
50.31365mH
XWM1
V I
U3
9.615 A
+
-
V2
240 Vrms
60 Hz
0°
R2
16.224
U4
6.25 A
+ -
L2
50.31365mH
XWM2
V I
C2
80.761 F
U5
7.307 A
+
-
U6
9.615 A
+
-
Author: Samir Ahmadov
P = 1.5 kW
cosphi = 0.65
P = 1.5 kW
cosphi = 1
Finding of active R and reactive L parameters of single phase circuit if known active power, current, supply voltage and frequency
Compensation of power factor to make cosphi equal 1
Task: To find value of active and reactive resistance R1 and XL1, then to find a value of C in
order to compencate power factor to 1.
Given: U=240, f=60Hz, active power P=1.5kW, I=9.615A; Find: cosphi, R1, L1 and C (for compensation)
Solution: First we have to find real power factor of system, we need to do following:
S=U x I=240 x 9.615 = 2308 VA; cosphi = P/S = 1500/2308 = 0.65
Important: Due to such power factor and common industrial case we define that load has inductive nature,
then Z must be R1+jXL1 with "+" and also inductive load current lags behind from voltage.
Now we have to find full resistance Z (impedance or complex Z) of circuit, we need to do following by
using COMPLEX NUMBER method:
we have to define complex current complexI. As we know cosphi=0.65 then phi = arccos(0.65)=49.458.
Now we can be sure that complexI=9.615e^-j49.458 (we put phi with "-" as soon as we know that on
inductive load current lags behind from voltage), then reversed complexI*=9.615e^j49.458
Apparent power S= complexU x complexI*(reversed) = 240 x 9.615e^j49.458 = 2308e^j49.458
S= complexU x complexI*(reversed) = complexI x Z x complexI*(reversed) (see TOE lecture)
complexZ= S/(complexI x Z x complexI*(reversed)) = 2308e^j49.458/(9.615e^-j49.458 x 9.615e^j49.458) =
= 2308e^j49.458/(9.615 x 9.615) = 2308e^j49.458/92.448225 = 24.96e^j49.458, so now we know
that complexZ=24.965e^j49.458 , then by using Euler formula:
a x e^jphi = a x cosphi + ja x sinphi = k+jy=sqr(k^2+y^2)e^jarctg(y/k)
a x e^-jphi = a x cosphi - ja x sinphi = k-jy=sqr(k^2+y^2)e^-jarctg(y/k)
then we can find:
Z= 24.96 x cos 49.458 + j24.96 x sin 49.458 = 16.224 + j18.9678; which is means that
R1= 16.224 Om; XL1=18.9678, then we know that XL1 = w x L=2 x п x f x L; L=XL1/(2 x п x f)
L = 50.3mH
Now we need to compencate the power factor in order to make cosphi=1; for this we have to compencate Reactive power. To find reactive power we have to use formula:
Reactive power of our circuit is Q = sqr (S^2 - P^2) = sqr (2308^2 - 1500^2) = sqr (5326864 - 2250000) = 1754 VAR = 1.754 kVAR; Now we need to find a value of C which will compencate power
factor to 1. For this case reactive power of compensation unit should be defined by formula: Q=U x I (usually Q = U x I x sinphi, but when we are talking about power factor compencation unit where
we have to find a value of C to compencate power factor, then we should not use sinphi in this formula, it is because compensation unit consist of only capacitor and as we know in branch with
only capacitor absolute value of phi=90 and then sin90 =1, for better understanding using complex valued functions we can see that reactance of capacitor bank Rc=-jXc, or let's show like 0-jXc, then
Rc=sqr(0^2+Xc^2)e^-jarctg(Xc/0), as we know any number deviding by 0 is infinite and arctg of infinite is pi/2 or 90 degree, then sin(90)=1. Note that if we are using complex valued functions then
we have to show minus "-" before j90 for capacitance and plus "+" for inductance).
So Q = U^2/Xc1, then Xc1 = U^2/Q = 240^2/1754 = 32.8 Om, then Xc1 = 1/(w x C), then C=1/(w x Xc1)=1/(2 x 3.14 x 60 x 32.8) = 80.7uF (mikrofarad)