- A circuit in parallel has Z1 = 2 - 3j and Z2 = 4 + j. Find the total impedance, phase angle and magnitude
- A circuit in parallel has Z1 = 5 - j and Z2 = 4 + 3j. Find the total impedance in polar form
- A circuit in series has Z1 = 2 - 3j and Z2 = 4 + j. Find the total impedance, phase angle and magnitude
- A circuit in series has Z1 = 5 - j and Z2 = 4 + 3j. Find the total impedance in polar form
- A car is being pulled with 2 forces. One force is 5 -3j and the other is 6 + j. What is the resultant force and show this is correct graphically.
Thursday, March 5, 2009
complex numbers questions
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Z1=2-3J.......1
ReplyDeleteZ2=4+J........2
=6+2J
Z1*Z2/Z1+Z2 ------formula
2-3J 4+J= 6+2J-12J-3J^2 -------multiply
J^2=-1............NB
9+10J
(9+10J/6+2J) (6-2J/6-2J)
(9+10J) (6-2J) =54-18J+60J-20J^2
=74-42J
A+B A-B
=a^2-B^2
(6+2J)(6-2J)
=6^2-(2J)^2
=36+4
=40
Zt=[(74/40) (42j/40)]
PHASE ANGLE
TAN x=imag/real
x=tan inverse 42/74
x=1.9 degrees
magnitude
2-3j+ 4+j
=6-2j
tel me if ah right nah people
sorry i made ah mistake for the phase angle
ReplyDeletetanX=imag/real
x=tan inverse 42/74
=tan inverse .5675
x=25.57 degrees
for question 2 since resistors are in parallel the total resistance can be found by dividing the product of the two resistors by the sum of both resistors. Thus Zt = Z1*Z2/Z1+Z2
ReplyDelete=(5-j)(4+3j) / (5-j)+(4+3j) to divide complex numbers the term is multiplied by the conjugate
working the numerator first the porduct of (5-j)*(4+3j) = (23+11j)
working the denominator = (5-j)+(4+3j)
=9-2j
using the conjugate = 9+2j the terms can be found.
=(23+llj/9-2j) * (9+2j/9+2j)
=working the numerator= 229+145j
working the denominator the rule (a+b)*(a-b)=a^2-b^2
=85
=1/85 * 229+145
continuing question 2. since the question asks to solve in polar form we use graphs to assist in visualising the question.
ReplyDelete1/85*229+144j=
2.69 + 1.7j
we use imaginary numbers on the y-axis and real numbers on the x-axis
2.69 is the opposite side and 1.7j is adjacent since these lengths create a triangle the hypotenuse will be the resultant thus the total impedance.
the angle produced by the resultant can be calculated using the tan function.
this will give us the angle=(tan inverse)(2.69/1.7) giving us a angle of 56.82 degrees.
to find the magnitude resultant line we use the cosine rule. the hypotenuse will be the resultant and the adjacent will be 1.7
resultant=1.7/cos56.82
=3.1ohms
for no. 5
ReplyDeletewell there 2 forces actin on the car
F1=5-3j F2=6+j
to find the resultant force........F1+F2
(5-3j)+(6+j)=11-2j
well if you plot this on a graph the real no.s being in your x-axis and imaginery bein your y-axis 11-2j will be the resultant force this can be oproved by plotting the other two forces onto the same graph and using the parallelogram of forces like in vectors to show this
5 ]since the resultant force is always produced when a force is applied to 2 ropes for example the force in which the direction the force is pulling away from the ropes .therefore the forces should be added
ReplyDelete5-3j
6+j
=11-2j
the graph must be plotted with tghe real values on the x axis and the imaginary on the y
just to show the relation to it in real life situation
for number 2 explain what is d polar form for me please people
ReplyDeletewel diesel i did some research to help you understand complex numbers easily, plz read this carefully and im sure you will find the ans to your question, wel it says the modulus is the length of the vector from the origin to the position of the complex number in the Argand diagram. We can find the modulus using Pythagoras' theorem. For the complex number 3+4i, for example, the modulus is 5, since 32+42=25.
ReplyDeleteThe argument is the angle that the vector makes with the horizontal x-axis. So the complex number 3+3i has argument 45o, since the line from the origin to the point (3,3) is at 45o to the horizontal axis.
These terms modulus and argument may be familiar to you from using polar coordinates. If you are not familiar with polar coordinates, you may find it useful to look in the MathHelp notebook on Coordinates and Graphs.
When using polar coordinates, instead of the Cartesian coordinates x and y, we define the position of a point in terms of its distance from the origin and its angle from the x-axis.
The position of the point A can be described as (2,2) in the Cartesian coordinates x and y. ALternatively its position can be described as (r,q), where r is the distance from A to the origin and q is the angle shown.
In this case we can work out that r must be about 2.8 (the square root of 22+22) and that q must be 45o.
We calculate the value of q by looking at the ratio of the y value and the x value. By its definition, tan(q) is equal to the length of the side opposite the angle q, divided by the length of the side adjacent to the angle q.
In other words, tan(q)=y/x. In this case y=2 and x=2, so y/x=1. That means q=45o, since tan(45o)=1.
To find out what x and y are, if we know r and q, we can use the definitions of sine and cosine to give the following formulae:
x=rcos(q)
y=rsin(q).
OK so that was a quick revision of polar coordinates.. What does it have to do with complex numbers?
As you may have guessed, we can write any complex number in terms of its modulus and argument, instead of its Cartesian coordinates (remember that its Cartesian coordinates are simply the real and imaginary parts of the complex number).
So for example, think about the complex number 3+3i.
We know that its modulus is the square root of 32+32, which is about 4.2 and that its argument is 45o, which is p/4 in radians.
Here's the crunch. We can write 3+3i as:
4.2cos(p/4) + 4.2 i sin(p/4).
More generally, we can write any complex number in two different ways, either
x+iy
OR
rcos(q)+irsin(q),
We can also write this second form as
r(cos(q)+isin(q)).
The form of writing a complex number that involves r and q rather than x and y is called the polar form of the complex number.
Question 3
ReplyDeleteThe total impedance in a series circuit is the addition of all the impedances in the circuit.
Therefore if Z1= 2-3j and Z2= 4+j
Then Ztotal = Z1+Z2
2-3j
4+j
6-2j
So Ztotal = 6-2j
The phase angle is the inverse tangent of the imaginary number divided by the real
So tanα= imag/real
= -2/6
Therefore α= tan-1 -2/6
= -18.435
Phase angle = 360-18.435
= 341.565o
Magnitude is the resultant of the real and imaginary
Therefore magnitude = √(-2)2 +62
=√40
no 1
ReplyDeleteR1 = 2 - 3j
R2 = 4 + j
RT = (R1 X R2)/(R1 + R2)
therefore RT = [(2-3j)(4+j)]/[(2-3j)+(4+j)]
RT = (8+2j-12j-3j^2)/(6-2j)
RT = (11-10j)/(6-2j)
= [(11-10j)/(6-2j)] x [(6+2j)/(6+2j)]
= (66+22j-60j-20j^2)/(36+4)
= (86-38j)/40
= (1/40)(86-38j)
RT = (2.15 - 0.95j)
phase angle-
tan x = opp/adj
tan x = -0.95/2.15
x = -23.8
since the phase angle is in the forth quadrant it will be:
x = 360-(-23.8)
= 383.8 degrees
The magnitude will be:
r = (-0.95^2 + 2.15^2)^1/2
= 2.4
no 2
ReplyDeleteR1 = (5-j)
R2 = (4+3j)
for paralell circuit:
RT = (R1 x R2)/(R1 + R2)
RT = [(5-j)(4+3j)]/[(5-j)+(4+3j)]
RT = (20 + 15j - 4j - 3j^2)/(9 + 2j)
RT = (23 + 11j)/(9+2j)
= [(23 + 11j)/(9+2j)]x[(9-2j)/(9-2j)]
= (229 + 53j)/(81+4)
= (229 + 53j)/ 85
RT = (2.69 + 0.62j)
magnitude, r = (2.69^2 + 0.62^2)^1/2
= 2.76
the phase angle is:
tan x = opp/adj
tan x = 0.62/2.69
x = 12.98 degrees
RT in polar form:
= 2.76 (cos 12.98 + j sin 12.98)
total impedance in the series circuit:
ReplyDeleteRT = (2-3j) + (4+j)
= (6-2j)
the phase is:
tan x = opp/adj
tan x = (-2/6)
x = -18.4
since the phase angle is in the forth quadrant, it will be:
= 360 - (-18.4)
= 378.4 degrees
the magnitude will be:
r = [6^2 + (-2^2)]^1/2
= 6.3
no 4
ReplyDeletetotal impedance in series circuit:
RT = (5-j) + (4+3j)
= (9 + 2j)
the phase angle is:
tan x = opp/adj
tan x = 2/9
x = 12.5 degrees
the magnitude is:
r = (2^2 + 9^2)^1/2
= 9.2
RT in polar form:
= 9.2(cos 12.5 + j sin 12.5)
This comment has been removed by the author.
ReplyDeleteno 5
ReplyDeletethe total impedance is:
= (5-3j) + (6+j)
= (11 - 2j)
sorry miss dont know how to draw the graph on the blog, but can do it on paper very well!
can anyone do the graph on the blog for qes 5?
ReplyDeletethe resultant force for Q5 is the two sets of numbers added together. but how to state polar forms?
ReplyDeleteThis comment has been removed by the author.
ReplyDeletein polar forms how do you kno wen to use sin or cos pr tan or is there a formula?
ReplyDeleteif a graph coda go up on this it woda rel good..lol
ReplyDeleteQuestion:
ReplyDeleteGiven that the phase angle of a current is 30۫ and total impedence in the circuit is given by R+5j, find the resistance R, and hence find the total impedence. Express the total impendence in polar form.
just thought of it. nice stuff to get comments next to your name.
i don't think u have 2 use sin and cos to put it into polar form. i remember miss told my class that putting something into polar form means plotting it on a graph. in the case of complex numbers y= imaginery and x=real
ReplyDeletei'm not sure but i think that from the graph you can figure out if to use sin,cos or tan by drawing aright angle triangle.
ReplyDeletefor dj halftime qu:
ReplyDeletesin30=5/r
then r=5/sin 30
r=10
1)a) Z1 = 2 - 3j , Z2 = 4 + j
ReplyDeleteTotalImpedance= (z1 x z2)/(z1 + z2)
= (2 - 3j x 4 + j)/(2 - 3j + 4 + j)
(2 - 3j x 4 + j)=(8 + 2j - 12j - 3j^2)
=(8 + 2j - 12j - 3(-1))
=(11 - 10j)
(2 - 3j + 4 + j)=(6 - 2j)
(11 - 10j)/(6 - 2j) , multiply by the conjugate of the denominator.
(11 - 10j)(6 - 2j) =(66 - 22j - 60j + 20j^2)
=(46 - 82j)
(6 - 2j)(6 - 2j) is a difference in squares.
= (36 + 4j^2)
= 32
so T.I. =(46 - 82j)/32
The real values go on the x-axis and imaginary values go on the y-axis...for question 2 since resistors are in parallel the total resistance can be found by dividing the product of the two resistors by the sum of the both resistors. Using the conjugate the terms can be found.
ReplyDeleteQUESTION 1
ReplyDeleteWHAT DO WE HAVE R1 = 2 - 3j,R2 = 4 + j
remember formular
R total
Rt = (R1 X R2)/(R1 + R2)
= [(2-3j)(4+j)]/[(2-3j)+(4+j)]
= (8+2j-12j-3j^2)/(6-2j)
= (11-10j)/(6-2j)
= [(11-10j)/(6-2j)] x [(6+2j)/(6+2j)]
= (66+22j-60j-20j^2)/(36+4)
= (86-38j)/40
= (1/40)(86-38j)
THEREFORE THE TOTAL RESISTANCER
= (2.15 - 0.95j)
FOR THE PHASE ANGLE
tan x = opp/adj
tan x = -0.95/2.15
x = -23.8
In this case the angle is in the 4th quadrant
x = 360-(-23.8)
= 383.8
MAGNITUDE
R = (-0.95^2 + 2.15^2)^1/2
= 2.4
QUESTION 2
R1 = (5-j),R2 = (4+3j)
We KNOW THAT IN PARALLEL CIRCUITS
Rtotal = (R1 x R2)/(R1 + R2)
= [(5-j)(4+3j)]/[(5-j)+(4+3j)]
= (20 + 15j - 4j - 3j^2)/(9 + 2j)
= (23 + 11j)/(9+2j)
=[(23 + 11j)/(9+2j)]x[(9-2j)/(9-2j)]
= (229 + 53j)/(81+4)
= (229 + 53j)/ 85
RTotal = (2.69 + 0.62j)
THE MAGNITUDE
R = (2.69^2 + 0.62^2)^1/2
= 2.76
PNASE ANGLE
tan x = opp/adj
tan x = 0.62/2.69
x = 12.98
RTotal in polar form:
= 2.76 (cos 12.98 + j sin 12.98)
Tell me if am correct but the most important thing to remeber in these question is knowing the symbols for a resistor,inductor and capacitor since the formula will be given.
ReplyDeleteQuestion 1
ReplyDeleteA circuit in parallel has Z1 = 2 - 3j and Z2 = 4 + j. Find the total impedance, phase angle and magnitude
T1 = 2 - 3j
T2 = 4 + j
T = (T1 X T2)/(T1 + T2)
therefore
T = [(2-3j)(4+j)]/[(2-3j)+(4+j)]
T = (8+2j-12j-3j^2)/(6-2j)
T = (11-10j)/(6-2j)
= [(11-10j)/(6-2j)] x [(6+2j)/(6+2j)]
= (66+22j-60j-20j^2)/(36+4)
= (86-38j)/40
= (1/40)(86-38j)
T = (2.15 - 0.95j)
The phase angle
tan x = opp/adj
tan x = -0.95/2.15
x = -23.8
However the phase angle is in the forth quadrant so it will be:
x = 360-(-23.8)
= 383.8 degrees
The magnitude will therefore be:
r = (-0.95^2 + 2.15^2) x 1/2
it is multiplied by 1/2 since it is the squared root
= 2.4
just to clear things up with Boy-b, inorder to figure out how to write your answer in polar form you must know the formula which is... impedance in polar form - r ( cos@ + jsin@ )i dont think 'tan' is used here, in the formula 'r' is the length of the hypotenuse, which u can find using pythagorus theorem and then the angle for @ must be found, at this point u can use sin, cos or tan to find @ depending on the triangle and the length of the sides given...once 'r' is calculated and @ is found jus plug in the values into the formula and that is the answer written in polar form.
ReplyDeleteany1 have any idea how u can put up a graph on this thing? it would be pretty helpful lol
ReplyDeletesome important things to remember when doing questions like these :
ReplyDelete- 'R' is the resistor & represents the 'real' voltage
- 'VL' is the inductor
- 'VC' is the capacitor
- VL - VC represents the imaginary quantity
- in a parallel circuit total impedence is
ZT = Z1 Z2 / Z1 + Z2....product / sum
- the phase angle can be found easily by sketching the graph & the triangle and finding '@' using sin, cos or tan
oh and it can be helpful jus to note the symbols of the resistor, capacitor and inductor, just incase a diagram is given, lol
ReplyDeleteQuestion 4 :
ReplyDeletetotal impedance in series circuit can be found by adding both quantities of Z1 and Z2 :
ZT = (5-j) + (4+3j)
= (9 + 2j)
the resultant r is: sq.rt( i^2 + j^2)
r = (2^2 + 9^2)^1/2
= 9.2
the phase angle @ is found by
tan @ = opp/adj
tan @ = 2/9
@ = 12.5 degrees
Therefore, plug in the values
Total Impedence in polar form using the formula
r ( cos@ + jsin@)
= 9.2( cos 12.5 + j sin 12.5 )