document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=93127>Question 93127</A>: <I><SPAN STYLE=\"word-break: break-all;\"> if z1=2+j, z2=3+j and z3=j3\r<BR>\n" );
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document.write( "what is the equivalent impedance z if \r<BR>\n" );
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document.write( "1/z= 1/z1 + 1/z2 + 1/z3  ??  this was a question given to us for homework and it was not in any text book </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #67798 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=67798\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> You are given three impedances:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z%5B1%5D+=+2+%2B+j ALIGN=MIDDLE  ALT=\"z%5B1%5D+=+2+%2B+j\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z%5B2%5D+=+3+%2B+j ALIGN=MIDDLE  ALT=\"z%5B2%5D+=+3+%2B+j\"   DISABLED_style= \"max-width:90%; height:auto;\" > and<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z%5B3%5D+=+0+%2B+j3 ALIGN=MIDDLE  ALT=\"z%5B3%5D+=+0+%2B+j3\"   DISABLED_style= \"max-width:90%; height:auto;\" > <== notice that this presumes that the impedance is so inductive that the resistance of the device is negligible.<BR>\n" );
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document.write( "The real part of these impedance is resistance and the imaginary part (called the reactance<BR>\n" );
document.write( "and identified as the j terms) are inductances. (You know they are inductances because the <BR>\n" );
document.write( "sign of the j terms is plus. If the sign of a j term is negative, then it is a capacitance.)<BR>\n" );
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document.write( "For the first problem you are asked to find <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z%5Bt%5D+=+z%5B1%5D%2Bz%5B2%5D%2Bz%5B3%5D ALIGN=MIDDLE  ALT=\"z%5Bt%5D+=+z%5B1%5D%2Bz%5B2%5D%2Bz%5B3%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" >. This is a series connection<BR>\n" );
document.write( "of the three impedances and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z%5Bt%5D ALIGN=MIDDLE  ALT=\"z%5Bt%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" > is the equivalent impedance. You could take out the<BR>\n" );
document.write( "three impedances in a circuit and replace them by the equivalent impedance without effect.<BR>\n" );
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document.write( "In finding <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z%5Bt%5D ALIGN=MIDDLE  ALT=\"z%5Bt%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" > all you need to do is add the three impedances. You do this by<BR>\n" );
document.write( "adding their real parts and then adding their imaginary parts. Where I listed the three<BR>\n" );
document.write( "impedances above, you can just add them in their columns. Adding down the column of real<BR>\n" );
document.write( "parts you get 2 + 3 + 0 = 5. And adding down the column of imaginary parts you get <BR>\n" );
document.write( "j + j + j3 = +5j. (Note j is the same as j1. So the imaginary parts are j1 + j1 + j3 and<BR>\n" );
document.write( "the addition is j(1+1+3) = +j5).<BR>\n" );
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document.write( "So the answer to the first problem is the combination of these two parts, the real part <BR>\n" );
document.write( "which is 5 and the imaginary or reactive part which is +j5.  Therefore, we say:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z%5Bt%5D+=+5+%2B+j5 ALIGN=MIDDLE  ALT=\"z%5Bt%5D+=+5+%2B+j5\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "You convert this to polar form just as you would in coordinate geometry.  You start at<BR>\n" );
document.write( "the origin of an x-y coordinate system and you go +5 units along the x-axis. (That <BR>\n" );
document.write( "represents the real part of your answer.) Then from that point (+5 on the x-axis)<BR>\n" );
document.write( "you go vertically up (the +j direction) +5 units which represents the imaginary or reactive<BR>\n" );
document.write( "part of the answer. [On a graph you are now at the point (5,5).] Now draw a line connecting the<BR>\n" );
document.write( "origin with this point (5,5). What you have now is an Argand diagram of the answer, and the<BR>\n" );
document.write( "Argand diagram consists of the 5 units along the x-axis, the 5 vertical units from that<BR>\n" );
document.write( "point to the point 5,5 and the line connecting the origin to the point (5,5). Just remember<BR>\n" );
document.write( "that the real part of the answer (the resistance) is along the x-axis, and the imaginary part<BR>\n" );
document.write( "(the vertical) represents the j component of the answer.<BR>\n" );
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document.write( "To find the polar form of the answer you need to look at the Argand diagram and find the<BR>\n" );
document.write( "hypotenuse of the right triangle.  The hypotenuse will be the magnitude of the answer and<BR>\n" );
document.write( "the angle between the x-axis and the hypotenuse will be the polar angle you need. <BR>\n" );
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document.write( "From geometry you may recall that a right triangle that has two legs that are equal in<BR>\n" );
document.write( "length is a 45-45-90 degree triangle.  Therefore, you know that the angle between the<BR>\n" );
document.write( "x-axis and the hypotenuse is 45 degrees.  You can use the Pythagorean theorem to find <BR>\n" );
document.write( "the length of the hypotenuse (or you could use trig). The Pythagorean theorem tells you to<BR>\n" );
document.write( "square each of the legs and add these squares to find the square of the hypotenuse.<BR>\n" );
document.write( "So for this problem we have:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=5%5E2+%2B+5%5E2+=+h%5E2 ALIGN=MIDDLE  ALT=\"5%5E2+%2B+5%5E2+=+h%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "and since <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=5%5E2+=+25 ALIGN=MIDDLE  ALT=\"5%5E2+=+25\"   DISABLED_style= \"max-width:90%; height:auto;\" > the problem becomes:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=25+%2B+25+=+50+=+h%5E2 ALIGN=MIDDLE  ALT=\"25+%2B+25+=+50+=+h%5E2\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "To find h just find the square root of 50. You can do it as follows:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=h+=+sqrt%2850%29+=+sqrt%2825%2A2%29+=+sqrt%2825%29%2Asqrt%282%29+=+5sqrt%282%29=7.0711 ALIGN=MIDDLE  ALT=\"h+=+sqrt%2850%29+=+sqrt%2825%2A2%29+=+sqrt%2825%29%2Asqrt%282%29+=+5sqrt%282%29=7.0711\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Now you know that the magnitude and angle (the polar form) is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=5sqrt%282%29 ALIGN=MIDDLE  ALT=\"5sqrt%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" > or 7.0711<BR>\n" );
document.write( "at an angle of 45 degrees which is often written as <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=5sqrt%282%29 ALIGN=MIDDLE  ALT=\"5sqrt%282%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >/45 or 7.0711/45 where the <BR>\n" );
document.write( "45 is underlined and is followed by the little \"o\" degrees symbol. (Hard to type the exact <BR>\n" );
document.write( "way it is written.)<BR>\n" );
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document.write( "Now for the second part of the problem. A little tougher because of the math manipulations,<BR>\n" );
document.write( "but pretty straightforward.<BR>\n" );
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document.write( "The given formula is:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2Fz%5Bt%5D+=+1%2Fz%5B1%5D%2B1%2Fz%5B2%5D%2B1%2Fz%5B3%5D ALIGN=MIDDLE  ALT=\"1%2Fz%5Bt%5D+=+1%2Fz%5B1%5D%2B1%2Fz%5B2%5D%2B1%2Fz%5B3%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "This is the formula for finding the equivalent impedance of three impedances connected in parallel.\r<BR>\n" );
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document.write( "There are several ways this could be done, but let's crank through a straightforward<BR>\n" );
document.write( "method.  Begin by substituting into the formula the values for the three impedances<BR>\n" );
document.write( "to get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2Fz%5Bt%5D+=+1%2F%282%2Bj%29%2B1%2F%283%2Bj%29%2B1%2F%28j3%29 ALIGN=MIDDLE  ALT=\"1%2Fz%5Bt%5D+=+1%2F%282%2Bj%29%2B1%2F%283%2Bj%29%2B1%2F%28j3%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Next recognize that the common denominator on the right side of this equation is:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=CD+=+%282%2Bj%29%2A%283%2Bj%29%2A%28j3%29 ALIGN=MIDDLE  ALT=\"CD+=+%282%2Bj%29%2A%283%2Bj%29%2A%28j3%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Let’s multiply this out combine terms. Remember in multiplying that j times j is j squared<BR>\n" );
document.write( "and by definition j squared is -1.<BR>\n" );
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document.write( "First let’s begin by multiplying <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%282%2Bj%29%2A%283+%2B+j%29+=+6+%2B+j2+%2B+j3+%2Bj%5E2+=+6+%2B+j2+%2B+j3+-1+=+5+%2B+j5 ALIGN=MIDDLE  ALT=\"%282%2Bj%29%2A%283+%2B+j%29+=+6+%2B+j2+%2B+j3+%2Bj%5E2+=+6+%2B+j2+%2B+j3+-1+=+5+%2B+j5\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( ". <BR>\n" );
document.write( "Next multiply that answer times j3:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=CD+=+%285%2Bj5%29%2A%28j3%29+=+j15+%2B+j%5E2+%2815%29+=+j15+%96+15 ALIGN=MIDDLE  ALT=\"CD+=+%285%2Bj5%29%2A%28j3%29+=+j15+%2B+j%5E2+%2815%29+=+j15+%96+15\"   DISABLED_style= \"max-width:90%; height:auto;\" > which in more standard form becomes:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=CD+=+-15+%2B+j15 ALIGN=MIDDLE  ALT=\"CD+=+-15+%2B+j15\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "If we put all three terms over the common denominator, each term's numerator consists of<BR>\n" );
document.write( "the product of the other two terms that were not its denominator. In other words the equation<BR>\n" );
document.write( "becomes:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2Fz%5Bt%5D+=+%28%28%283%2Bj%29%28j3%29%29%2FCD%29+%2B+%28%28%282%2Bj%29%28j3%29%29%2FCD%29+%2B%28%28%282%2Bj%29%283%2Bj%29%29%2FCD%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Now let’s multiply out each of the numerators in the three terms.<BR>\n" );
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document.write( "First numerator: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%283%2Bj%29%28j3%29+=+j9+%2B+j%5E%282%293+=+j9+-3+=+-3%2Bj9 ALIGN=MIDDLE  ALT=\"%283%2Bj%29%28j3%29+=+j9+%2B+j%5E%282%293+=+j9+-3+=+-3%2Bj9\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Second numerator: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%282%2Bj%29%28j3%29+=+j6%2Bj%5E%282%293+=+j6+-3+=+-3+%2Bj6 ALIGN=MIDDLE  ALT=\"%282%2Bj%29%28j3%29+=+j6%2Bj%5E%282%293+=+j6+-3+=+-3+%2Bj6\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Third numerator: <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%282%2Bj%29%283%2Bj%29+=+6+%2Bj2+%2Bj3+%2B+j%5E2+=+6%2Bj2%2Bj3-1+=+5%2Bj5 ALIGN=MIDDLE  ALT=\"%282%2Bj%29%283%2Bj%29+=+6+%2Bj2+%2Bj3+%2B+j%5E2+=+6%2Bj2%2Bj3-1+=+5%2Bj5\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Now add all the numerators over the common denominator:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2Fz%5Bt%5D+=+%28-9%2Bj9-3%2Bj6%2B5%2Bj5%29%2F%28-15%2Bj15%29+=+%28-1%2Bj20%29%2F%28-15%2Bj15%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Now we can get the answer for <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z%5Bt%5D ALIGN=MIDDLE  ALT=\"z%5Bt%5D\"   DISABLED_style= \"max-width:90%; height:auto;\" > by just inverting both sides of this equation to get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z%5Bt%5D+=+%28-15%2Bj15%29%2F%28-1%2Bj20%29 ALIGN=MIDDLE  ALT=\"z%5Bt%5D+=+%28-15%2Bj15%29%2F%28-1%2Bj20%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Just to make things a little more conventional, let’s factor -1 from both the numerator and <BR>\n" );
document.write( "denominator of this answer so that the real portions of answer are positive. This factoring<BR>\n" );
document.write( "results in:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z%5Bt%5D+=+%28%28-1%29%2815-j15%29%29%2F%28%28-1%29%281-j20%29%29+=+%28cross%28-1%29%2815+-+j15%29%29%2F%28cross%28-1%29%281-j20%29%29=+%2815-j15%29%2F%281-j20%29 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "This is not in a useful form yet. We can convert the denominator of this answer by multiplying<BR>\n" );
document.write( "it by its conjugate. The conjugate has the same real part and imaginary part, but the imaginary <BR>\n" );
document.write( "part has the opposite sign. In this case the denominator is <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1-j20 ALIGN=MIDDLE  ALT=\"1-j20\"   DISABLED_style= \"max-width:90%; height:auto;\" > so its conjugate is <BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1+%2B+j20 ALIGN=MIDDLE  ALT=\"1+%2B+j20\"   DISABLED_style= \"max-width:90%; height:auto;\" >. When we multiply the denominator by <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2Bj20 ALIGN=MIDDLE  ALT=\"1%2Bj20\"   DISABLED_style= \"max-width:90%; height:auto;\" >, we must also multiply<BR>\n" );
document.write( "the numerator by <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=1%2Bj20 ALIGN=MIDDLE  ALT=\"1%2Bj20\"   DISABLED_style= \"max-width:90%; height:auto;\" >. So we get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z%5Bt%5D+=+%2815-j15%29%2F%281-j20%29+%2A+%281%2Bj20%29%2F%281%2Bj20%29 ALIGN=MIDDLE  ALT=\"z%5Bt%5D+=+%2815-j15%29%2F%281-j20%29+%2A+%281%2Bj20%29%2F%281%2Bj20%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Multiplying the denominator by its conjugate results in:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%281-j20%29%2A%281%2Bj20%29+=+1+%2Bj20-j20+-j%5E%282%29%2820%29%5E2+=+1+-+%28-1%29%2820%5E2%29+=+1%2B400=+401 ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Then multiplying the numerator by the conjugate of the denominator:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%2815-j15%29%281%2Bj20%29+=+%2815%2A1%29+%2B+j%2815%2A20%29+%96j%2815%2A1%29+%96j%5E%282%29%2815%2A20%29%29+ ALIGN=MIDDLE   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "This further multiplies out to:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=15+%2B+j300+%96+j15+-+%28-1%29%28300%29 ALIGN=MIDDLE  ALT=\"15+%2B+j300+%96+j15+-+%28-1%29%28300%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Subtracting the two j terms shortens this to:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=15+%2Bj285+-+%28-1%29%28300%29 ALIGN=MIDDLE  ALT=\"15+%2Bj285+-+%28-1%29%28300%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "The last term involves minus a minus term which is positive:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=15+%2B+j285+%2B+300 ALIGN=MIDDLE  ALT=\"15+%2B+j285+%2B+300\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "The two real parts add and the numerator becomes:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=315+%2B+j285 ALIGN=MIDDLE  ALT=\"315+%2B+j285\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "This gets put over the real denominator which we found to be 401:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z%5Bt%5D+=+%28315+%2B+j285%29%2F401+=+315%2F401+%2B+j%28285%2F401%29 ALIGN=MIDDLE  ALT=\"z%5Bt%5D+=+%28315+%2B+j285%29%2F401+=+315%2F401+%2B+j%28285%2F401%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Dividing the two terms by 401 results in:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=z%5Bt%5D+=+0.7855+%2B+j%280.7107%29 ALIGN=MIDDLE  ALT=\"z%5Bt%5D+=+0.7855+%2B+j%280.7107%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "The Argand diagram will start at the origin and go to the right on the x-axis 0.7855 units. <BR>\n" );
document.write( "From that point it will go vertically upward 0.7107 units. You are now at the point <BR>\n" );
document.write( "(0.7855, 0.7107). Draw a line from the origin to that point. Again, as in the first <BR>\n" );
document.write( "problem, this is the Argand diagram for the equivalent impedance. <BR>\n" );
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document.write( "Next we will use the Argand diagram to find the polar form. The polar form requires that<BR>\n" );
document.write( "you have a magnitude and an angle. The magnitude is the length of the hypotenuse in the<BR>\n" );
document.write( "Argand diagram, and you can use either the Pythagorean theorem or trig functions to get<BR>\n" );
document.write( "the magnitude. This time let’s use trig.  We know that the tangent is defined by the side<BR>\n" );
document.write( "opposite divided by the adjacent side of the angle.  In this problem the side opposite the<BR>\n" );
document.write( "angle we are looking for is the vertical side and its length is 0.7107. The adjacent side<BR>\n" );
document.write( "is the horizontal side of the triangle and its length is 0.7855.  So the tangent of the angle<BR>\n" );
document.write( "we are looking for is:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=tanA+=+0.7107%2F0.7855+=+0.9048 ALIGN=MIDDLE  ALT=\"tanA+=+0.7107%2F0.7855+=+0.9048\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "That means that the angle has a tangent of 0.9048. Using the the tan^(-1) function key<BR>\n" );
document.write( "on a calculator we find that the angle with a tangent of 0.9048 is 42.1388 degrees. So<BR>\n" );
document.write( "we have half of what we need for the polar form.  We still need the hypotenuse and<BR>\n" );
document.write( "we can get that using either the sine or cosine function. Let’s use the sine. The <BR>\n" );
document.write( "definition of the sine is:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sinA+=+%28side+opposite%29%2F%28hypotenuse%29++ ALIGN=MIDDLE  ALT=\"sinA+=+%28side+opposite%29%2F%28hypotenuse%29++\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "If we multiply both sides of this equation by the hypotenuse we get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=%28hypotenuse%29%2AsinA+=+side+opposite ALIGN=MIDDLE  ALT=\"%28hypotenuse%29%2AsinA+=+side+opposite\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Then divide both sides by <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sinA+ ALIGN=MIDDLE  ALT=\"sinA+\"   DISABLED_style= \"max-width:90%; height:auto;\" > to get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=hypotenuse+=+%28side+opposite%29%2FsinA ALIGN=MIDDLE  ALT=\"hypotenuse+=+%28side+opposite%29%2FsinA\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Now we can plug in values. We know the side opposite is 0.7107 and <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=sintA ALIGN=MIDDLE  ALT=\"sintA\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
document.write( "is the sine of 42.1388 degrees. From a calculator  this is 0.6709.  So our equation<BR>\n" );
document.write( "with these values is:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=hypotenuse+=+0.7107%2F0.6709+=+1.0593 ALIGN=MIDDLE  ALT=\"hypotenuse+=+0.7107%2F0.6709+=+1.0593\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "So the polar form of this impedance is 1.0593/42.1388 which is read as 1.0593 ohms at 42.1388<BR>\n" );
document.write( "degrees.<BR>\n" );
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document.write( "Lots of work. Hope this all makes sense to you and you can figure out from this the basics of <BR>\n" );
document.write( "finding equivalent impedances for series and parallel circuits. <BR>\n" );
document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );