This is referring to a 50g
This is a discussion on Working with complex numbers and real variables - Hewlett Packard ; I'm wondering if it is possible to use variables in complex numbers(although the vars are real). I'm trying to do something like this: '(234*c,.0016)/(80,1000*c-.002) With complex numbers in the format (re,im). When I evaluate the above equation I get a ...
I'm wondering if it is possible to use variables in complex
numbers(although the vars are real).
I'm trying to do something like this:
'(234*c,.0016)/(80,1000*c-.002)
With complex numbers in the format (re,im). When I evaluate the above
equation I get a seemingly nonsensical answer. I've been scouring
google for hours and not having any luck trying to figure out how to
do what seemingly would be simply for the calculator. I've been doing
them by hand, but I get lost in the algebra.
This is referring to a 50g
On Jun 28, 7:57*pm, buz...@gmail.com wrote:
> This is referring to a 50g
I don't think you can use variables to perform the above operation
on complex numbers because the arguments must be real
numbers. Why don't you evaluate your variables and then use
the R->C (real to complex) command? In other words calculate
234 * c, leave the result in stack level 1, enter .0016 in stack
level 2, then use the R->C command. Again, enter 80 in stack
level 1, then 1000 * c - .002 in stack level 2, then use the R->C
command and multiply.
On Jun 28, 8:31 pm, mnhollin...@yahoo.com wrote:
> On Jun 28, 7:57 pm, buz...@gmail.com wrote:
>
> > This is referring to a 50g
>
> I don't think you can use variables to perform the above operation
> on complex numbers because the arguments must be real
> numbers. Why don't you evaluate your variables and then use
> the R->C (real to complex) command? In other words calculate
> 234 * c, leave the result in stack level 1, enter .0016 in stack
> level 2, then use the R->C command. Again, enter 80 in stack
> level 1, then 1000 * c - .002 in stack level 2, then use the R->C
> command and multiply.
The variable is unknown at the moment. The problem is a circuits
analysis and I'm attempting to find out, after this calculation
occurs, what value of c will give me a certain reult.
A bunch of elements are in a circuit and I'm attempting to find out
what value of 'c' will give me a certain phase angle at a certain
element:P The concept is easy, but the algebra is hideous!
Thanks
On Jun 28, 9:47*pm, buz...@gmail.com wrote:
> On Jun 28, 8:31 pm, mnhollin...@yahoo.com wrote:
>
> > On Jun 28, 7:57 pm, buz...@gmail.com wrote:
>
> > > This is referring to a 50g
>
> > I don't think you can use variables to perform the above operation
> > on complex numbers because the arguments must be real
> > numbers. Why don't you evaluate your variables and then use
> > the R->C (real to complex) command? In other words calculate
> > 234 * c, leave the result in stack level 1, enter .0016 in stack
> > level 2, then use the R->C command. Again, enter 80 in stack
> > level 1, then 1000 * c - .002 in stack level 2, then use the R->C
> > command and multiply.
>
> The variable is unknown at the moment. *The problem is a circuits
> analysis and I'm attempting to find out, after this calculation
> occurs, what value of c will give me a certain reult.
>
> A bunch of elements are in a circuit and I'm attempting to find out
> what value of 'c' will give me a certain phase angle at a certain
> element:P *The concept is easy, but the algebra is hideous!
>
> Thanks
I've never done any electrical engineering, could you elaborate a
bit more on the problem? If not, I believe John Meyers is an
electrical engineer and is also a regular contributor on this forum.
On Jun 28, 9:04 pm, mnhollin...@yahoo.com wrote:
> On Jun 28, 9:47 pm, buz...@gmail.com wrote:
>
>
>
> > On Jun 28, 8:31 pm, mnhollin...@yahoo.com wrote:
>
> > > On Jun 28, 7:57 pm, buz...@gmail.com wrote:
>
> > > > This is referring to a 50g
>
> > > I don't think you can use variables to perform the above operation
> > > on complex numbers because the arguments must be real
> > > numbers. Why don't you evaluate your variables and then use
> > > the R->C (real to complex) command? In other words calculate
> > > 234 * c, leave the result in stack level 1, enter .0016 in stack
> > > level 2, then use the R->C command. Again, enter 80 in stack
> > > level 1, then 1000 * c - .002 in stack level 2, then use the R->C
> > > command and multiply.
>
> > The variable is unknown at the moment. The problem is a circuits
> > analysis and I'm attempting to find out, after this calculation
> > occurs, what value of c will give me a certain reult.
>
> > A bunch of elements are in a circuit and I'm attempting to find out
> > what value of 'c' will give me a certain phase angle at a certain
> > element:P The concept is easy, but the algebra is hideous!
>
> > Thanks
>
> I've never done any electrical engineering, could you elaborate a
> bit more on the problem? If not, I believe John Meyers is an
> electrical engineer and is also a regular contributor on this forum.
Here is a simple version of a problem, I'll try and word it out as
best I can.
There is a voltage source of 250cos(1000t)V. This is represented in a
phasor form as 250Vangle(0). Attached to the (+) side is and inductor
with an inductance of 5H, which equates to an impedance of 0+i(1000*5)
attached to this inductor is a capacitor and resistor parallel. The
resistor has a value of 12.5k, and the value of the capacitor is not
stated. Combining the impedance of the inductor and the resistor/
capacitor section comes up with the total impedance of:
((200m)(c)-i(16n))/((80u)+i(1000(c)-200u))
If my math that far was correct.
c:capacitance
m:milli's
u:micro's
Simplifying the above equation(getting the i out of the denominator)
gives the impedance of the entire circuit. To get the current through
the voltage source to be in phase with the voltage, I need to find
when the imaginary portion(reactance) is zero. It's just algebra for
the most part, but I'm always missing a sign or thinking a milli is a
micro or nano and just get lost in the math.
On Jun 29, 4:53*am, buz...@gmail.com wrote:
> On Jun 28, 9:04 pm, mnhollin...@yahoo.com wrote:
>
>
>
> > On Jun 28, 9:47 pm, buz...@gmail.com wrote:
>
> > > On Jun 28, 8:31 pm, mnhollin...@yahoo.com wrote:
>
> > > > On Jun 28, 7:57 pm, buz...@gmail.com wrote:
>
> > > > > This is referring to a 50g
>
> > > > I don't think you can use variables to perform the above operation
> > > > on complex numbers because the arguments must be real
> > > > numbers. Why don't you evaluate your variables and then use
> > > > the R->C (real to complex) command? In other words calculate
> > > > 234 * c, leave the result in stack level 1, enter .0016 in stack
> > > > level 2, then use the R->C command. Again, enter 80 in stack
> > > > level 1, then 1000 * c - .002 in stack level 2, then use the R->C
> > > > command and multiply.
>
> > > The variable is unknown at the moment. *The problem is a circuits
> > > analysis and I'm attempting to find out, after this calculation
> > > occurs, what value of c will give me a certain reult.
>
> > > A bunch of elements are in a circuit and I'm attempting to find out
> > > what value of 'c' will give me a certain phase angle at a certain
> > > element:P *The concept is easy, but the algebra is hideous!
>
> > > Thanks
>
> > I've never done any electrical engineering, could you elaborate a
> > bit more on the problem? If not, I believe John Meyers is an
> > electrical engineer and is also a regular contributor on this forum.
>
> Here is a simple version of a problem, I'll try and word it out as
> best I can.
>
> There is a voltage source of 250cos(1000t)V. *This is represented in a
> phasor form as 250Vangle(0). *Attached to the (+) side is and inductor
> with an inductance of 5H, which equates to an impedance of 0+i(1000*5)
> attached to this inductor is a capacitor and resistor parallel. *The
> resistor has a value of 12.5k, and the value of the capacitor is not
> stated. *Combining the impedance of the inductor and the resistor/
> capacitor section comes up with the total impedance of:
>
> *((200m)(c)-i(16n))/((80u)+i(1000(c)-200u))
> If my math that far was correct.
> c:capacitance
> m:milli's
> u:micro's
>
> Simplifying the above equation(getting the i out of the denominator)
> gives the impedance of the entire circuit. *To get the current through
> the voltage source to be in phase with the voltage, I need to find
> when the imaginary portion(reactance) is zero. *It's just algebra for
> the most part, but I'm always missing a sign or thinking a milli is a
> micro or nano and just get lost in the math.
So you want to choose c (real) such, that the imaginary part of the
above is =0?
If I enter the expression on my 50g and eval it, I get a fraction of
complex numbers. (Is that your "nonsensical answer"?).
Now have it get just the imaginary part (IM) and solve for c ('c'
SOLVE) (this implicitly assumes =0; for a different right hand side
enter= beforehand). That gives {c=7.406E-4 c=-7.386E-4}..
I haven't heard of negative capacitances yet, so the first one should
be your answer.
good luck,
M.
And one more addition: the 50g is very capable in handling units. If
you always enter the units along with the values it will take care of
nanos, micros, millis etc. + will fail if untis are incompatible
(here's a glitch: it can't automatically convert ¡ÆC to K or ¡ÆF, but
you're probably not gonna run into that all too often in el.eng.).
I'm too lazy for actual conversion, so I just add e.g. 0_V if I know
my result should be a voltage. That forces the calc to evaluate sth.
like mA*§Ù into a meaningful unit. UBASE evaluates all units into
m,kg,s,A.
greetings, M.
On Jun 29, 2:31 pm, Matthias Rampke
wrote:
> On Jun 29, 4:53 am, buz...@gmail.com wrote:
>
>
>
> > On Jun 28, 9:04 pm, mnhollin...@yahoo.com wrote:
>
> > > On Jun 28, 9:47 pm, buz...@gmail.com wrote:
>
> > > > On Jun 28, 8:31 pm, mnhollin...@yahoo.com wrote:
>
> > > > > On Jun 28, 7:57 pm, buz...@gmail.com wrote:
>
> > > > > > This is referring to a 50g
>
> > > > > I don't think you can use variables to perform the above operation
> > > > > on complex numbers because the arguments must be real
> > > > > numbers. Why don't you evaluate your variables and then use
> > > > > the R->C (real to complex) command? In other words calculate
> > > > > 234 * c, leave the result in stack level 1, enter .0016 in stack
> > > > > level 2, then use the R->C command. Again, enter 80 in stack
> > > > > level 1, then 1000 * c - .002 in stack level 2, then use the R->C
> > > > > command and multiply.
>
> > > > The variable is unknown at the moment. The problem is a circuits
> > > > analysis and I'm attempting to find out, after this calculation
> > > > occurs, what value of c will give me a certain reult.
>
> > > > A bunch of elements are in a circuit and I'm attempting to find out
> > > > what value of 'c' will give me a certain phase angle at a certain
> > > > element:P The concept is easy, but the algebra is hideous!
>
> > > > Thanks
>
> > > I've never done any electrical engineering, could you elaborate a
> > > bit more on the problem? If not, I believe John Meyers is an
> > > electrical engineer and is also a regular contributor on this forum.
>
> > Here is a simple version of a problem, I'll try and word it out as
> > best I can.
>
> > There is a voltage source of 250cos(1000t)V. This is represented in a
> > phasor form as 250Vangle(0). Attached to the (+) side is and inductor
> > with an inductance of 5H, which equates to an impedance of 0+i(1000*5)
> > attached to this inductor is a capacitor and resistor parallel. The
> > resistor has a value of 12.5k, and the value of the capacitor is not
> > stated. Combining the impedance of the inductor and the resistor/
> > capacitor section comes up with the total impedance of:
>
> > ((200m)(c)-i(16n))/((80u)+i(1000(c)-200u))
> > If my math that far was correct.
> > c:capacitance
> > m:milli's
> > u:micro's
>
> > Simplifying the above equation(getting the i out of the denominator)
> > gives the impedance of the entire circuit. To get the current through
> > the voltage source to be in phase with the voltage, I need to find
> > when the imaginary portion(reactance) is zero. It's just algebra for
> > the most part, but I'm always missing a sign or thinking a milli is a
> > micro or nano and just get lost in the math.
>
> So you want to choose c (real) such, that the imaginary part of the
> above is =0?
>
> If I enter the expression on my 50g and eval it, I get a fraction of
> complex numbers. (Is that your "nonsensical answer"?).
>
> Now have it get just the imaginary part (IM) and solve for c ('c'
> SOLVE) (this implicitly assumes =0; for a different right hand side
> enter= beforehand). That gives {c=7.406E-4 c=-7.386E-4}.
> I haven't heard of negative capacitances yet, so the first one should
> be your answer.
>
> good luck,
> M.
On Jun 29, 7:40 am, Matthias Rampke
wrote:
> And one more addition: the 50g is very capable in handling units. If
> you always enter the units along with the values it will take care of
> nanos, micros, millis etc. + will fail if untis are incompatible
> (here's a glitch: it can't automatically convert ¡ÆC to K or ¡ÆF,but
> you're probably not gonna run into that all too often in el.eng.).
>
> I'm too lazy for actual conversion, so I just add e.g. 0_V if I know
> my result should be a voltage. That forces the calc to evaluate sth.
> like mA*§Ù into a meaningful unit. UBASE evaluates all units into
> m,kg,s,A.
>
> greetings, M.
>
> On Jun 29, 2:31 pm, Matthias Rampke
> wrote:
>
> > On Jun 29, 4:53 am, buz...@gmail.com wrote:
>
> > > On Jun 28, 9:04 pm, mnhollin...@yahoo.com wrote:
>
> > > > On Jun 28, 9:47 pm, buz...@gmail.com wrote:
>
> > > > > On Jun 28, 8:31 pm, mnhollin...@yahoo.com wrote:
>
> > > > > > On Jun 28, 7:57 pm, buz...@gmail.com wrote:
>
> > > > > > > This is referring to a 50g
>
> > > > > > I don't think you can use variables to perform the above operation
> > > > > > on complex numbers because the arguments must be real
> > > > > > numbers. Why don't you evaluate your variables and then use
> > > > > > the R->C (real to complex) command? In other words calculate
> > > > > > 234 * c, leave the result in stack level 1, enter .0016 in stack
> > > > > > level 2, then use the R->C command. Again, enter 80 in stack
> > > > > > level 1, then 1000 * c - .002 in stack level 2, then use the R->C
> > > > > > command and multiply.
>
> > > > > The variable is unknown at the moment. The problem is a circuits
> > > > > analysis and I'm attempting to find out, after this calculation
> > > > > occurs, what value of c will give me a certain reult.
>
> > > > > A bunch of elements are in a circuit and I'm attempting to find out
> > > > > what value of 'c' will give me a certain phase angle at a certain
> > > > > element:P The concept is easy, but the algebra is hideous!
>
> > > > > Thanks
>
> > > > I've never done any electrical engineering, could you elaborate a
> > > > bit more on the problem? If not, I believe John Meyers is an
> > > > electrical engineer and is also a regular contributor on this forum..
>
> > > Here is a simple version of a problem, I'll try and word it out as
> > > best I can.
>
> > > There is a voltage source of 250cos(1000t)V. This is represented in a
> > > phasor form as 250Vangle(0). Attached to the (+) side is and inductor
> > > with an inductance of 5H, which equates to an impedance of 0+i(1000*5)
> > > attached to this inductor is a capacitor and resistor parallel. The
> > > resistor has a value of 12.5k, and the value of the capacitor is not
> > > stated. Combining the impedance of the inductor and the resistor/
> > > capacitor section comes up with the total impedance of:
>
> > > ((200m)(c)-i(16n))/((80u)+i(1000(c)-200u))
> > > If my math that far was correct.
> > > c:capacitance
> > > m:milli's
> > > u:micro's
>
> > > Simplifying the above equation(getting the i out of the denominator)
> > > gives the impedance of the entire circuit. To get the current through
> > > the voltage source to be in phase with the voltage, I need to find
> > > when the imaginary portion(reactance) is zero. It's just algebra for
> > > the most part, but I'm always missing a sign or thinking a milli is a
> > > micro or nano and just get lost in the math.
>
> > So you want to choose c (real) such, that the imaginary part of the
> > above is =0?
>
> > If I enter the expression on my 50g and eval it, I get a fraction of
> > complex numbers. (Is that your "nonsensical answer"?).
>
> > Now have it get just the imaginary part (IM) and solve for c ('c'
> > SOLVE) (this implicitly assumes =0; for a different right hand side
> > enter= beforehand). That gives {c=7.406E-4 c=-7.386E-4}.
> > I haven't heard of negative capacitances yet, so the first one should
> > be your answer.
>
> > good luck,
> > M.
Thanks Matthias that worked for one that I've done by hand(so I was
able to verify my answer), but I can't seen to get it to work for the
described circuit. I do get the same answers when I do it by hand, so
perhaps I'm messing some math up but it seems fairly straight
forward... I'm getting two positive values for c. In this straight
RLC circuit, that shouldn't be the case. I'll have to figure out what
I'm doing wrong. I appreciate the help though, I was about to go buy
a TI and see if that could do it!
On Jun 29, 7:40 am, Matthias Rampke
wrote:
> And one more addition: the 50g is very capable in handling units. If
> you always enter the units along with the values it will take care of
> nanos, micros, millis etc. + will fail if untis are incompatible
> (here's a glitch: it can't automatically convert ¡ÆC to K or ¡ÆF, but
> you're probably not gonna run into that all too often in el.eng.).
>
> I'm too lazy for actual conversion, so I just add e.g. 0_V if I know
> my result should be a voltage. That forces the calc to evaluate sth.
> like mA*§Ù into a meaningful unit. UBASE evaluates all units into
> m,kg,s,A.
>
> greetings, M.
>
> On Jun 29, 2:31 pm, Matthias Rampke
> wrote:
>
> > On Jun 29, 4:53 am, buz...@gmail.com wrote:
>
> > > On Jun 28, 9:04 pm, mnhollin...@yahoo.com wrote:
>
> > > > On Jun 28, 9:47 pm, buz...@gmail.com wrote:
>
> > > > > On Jun 28, 8:31 pm, mnhollin...@yahoo.com wrote:
>
> > > > > > On Jun 28, 7:57 pm, buz...@gmail.com wrote:
>
> > > > > > > This is referring to a 50g
>
> > > > > > I don't think you can use variables to perform the above operation
> > > > > > on complex numbers because the arguments must be real
> > > > > > numbers. Why don't you evaluate your variables and then use
> > > > > > the R->C (real to complex) command? In other words calculate
> > > > > > 234 * c, leave the result in stack level 1, enter .0016 in stack
> > > > > > level 2, then use the R->C command. Again, enter 80 in stack
> > > > > > level 1, then 1000 * c - .002 in stack level 2, then use the R->C
> > > > > > command and multiply.
>
> > > > > The variable is unknown at the moment. The problem is a circuits
> > > > > analysis and I'm attempting to find out, after this calculation
> > > > > occurs, what value of c will give me a certain reult.
>
> > > > > A bunch of elements are in a circuit and I'm attempting to find out
> > > > > what value of 'c' will give me a certain phase angle at a certain
> > > > > element:P The concept is easy, but the algebra is hideous!
>
> > > > > Thanks
>
> > > > I've never done any electrical engineering, could you elaborate a
> > > > bit more on the problem? If not, I believe John Meyers is an
> > > > electrical engineer and is also a regular contributor on this forum.
>
> > > Here is a simple version of a problem, I'll try and word it out as
> > > best I can.
>
> > > There is a voltage source of 250cos(1000t)V. This is represented in a
> > > phasor form as 250Vangle(0). Attached to the (+) side is and inductor
> > > with an inductance of 5H, which equates to an impedance of 0+i(1000*5)
> > > attached to this inductor is a capacitor and resistor parallel. The
> > > resistor has a value of 12.5k, and the value of the capacitor is not
> > > stated. Combining the impedance of the inductor and the resistor/
> > > capacitor section comes up with the total impedance of:
>
> > > ((200m)(c)-i(16n))/((80u)+i(1000(c)-200u))
> > > If my math that far was correct.
> > > c:capacitance
> > > m:milli's
> > > u:micro's
>
> > > Simplifying the above equation(getting the i out of the denominator)
> > > gives the impedance of the entire circuit. To get the current through
> > > the voltage source to be in phase with the voltage, I need to find
> > > when the imaginary portion(reactance) is zero. It's just algebra for
> > > the most part, but I'm always missing a sign or thinking a milli is a
> > > micro or nano and just get lost in the math.
>
> > So you want to choose c (real) such, that the imaginary part of the
> > above is =0?
>
> > If I enter the expression on my 50g and eval it, I get a fraction of
> > complex numbers. (Is that your "nonsensical answer"?).
>
> > Now have it get just the imaginary part (IM) and solve for c ('c'
> > SOLVE) (this implicitly assumes =0; for a different right hand side
> > enter= beforehand). That gives {c=7.406E-4 c=-7.386E-4}.
> > I haven't heard of negative capacitances yet, so the first one should
> > be your answer.
>
> > good luck,
> > M.
Here is the problem if it helps at all,
http://www.kx9.net/prob.pdf
I keep getting 40nF and 160nF for the value of C to get I through the
whole circuit to be in phase with the source. And that doesn't make
sense.
On Sat, 28 Jun 2008 21:04:56 -0500:
> I believe John Meyers is an electrical engineer...
My Electrical Engineering career ended abruptly
after my Lionel train set broke
However, I do know that in the HP48/49/50,
while "complex number objects" may consist only
of two real-valued parts, complex-valued
algebraic *expressions* '(A,B)'
where A and B may also be any expressions
are perfectly legitimate,
and mean the same as 'A+B*i'
http://www.lionelcollectors.org/
[r->] [OFF]