Hello all,

I am new to the ClassPad. I can get it to do pretty much what I want except 1 thing.

It will not let me enter an Inverse Polar entry. I get an Invalid Dimmension error.

I can convert polar to Rectangular and vice versa but if I try to divide either into one, it tells me to get lost.

ToPol([1,1]) works

ToRect([10, <(45)]) The "greater than" sign here being the angle sign on the CP. Works.

1/([10,<(45)]) gives dimmension error.

1/([1,1]) same error

Anyone know if this is possible in a CP?

My classmates can do them on thier TIs so I would THINK a CP can do it as well

?

Inverse polar coordanates are used in figuring Voltage, Current...etc in electronic AC circuits so it should be capable of doing it since that is not an uncommon use of a calculator.

Any help would be appreciated.

Regards,

Charles

# Inverse Polar

Started by
Chuckster7
, Nov 21 2007 07:42 PM

4 replies to this topic

### #1

Posted 21 November 2007 - 07:42 PM

### #2

Posted 22 November 2007 - 04:59 PM

The mistake you made is that you cannot divide a number with a vector!

### #3 Guest_Guest_Chuckster7_*_*

Posted 26 November 2007 - 03:45 PM

The mistake you made is that you cannot divide a number with a vector!

Well.....TI will do it. If a TI can do it, you'd think a classpad could as well.

My classmates take their TI-89s and do it no problem.

This is supposed to be some whiz-bang engenerring calculator and I cant even use it to find impeadence in an AC circuit.

To find impeadence in a parallel AC circuit:

1/(1/Z1+1/Z2+1/Zx)=Ztotal

Z=Impeadence

Impeadence can be written in 2 ways, Polar or Rectangular.

100 @0deg is 100volts at 0 deg phase shift or.....pure resistive

100 @-90deg is 100 volts at a -90 shift or........... pure capacative

100 @+90deg ...same thing but pure inductive

I need to be able to take either polar or rectangular expressions, aka impeaedence, and apply it to the above formula.

any ideas?

### #4 Guest_Guest_Chuckster7_*_*

Posted 26 November 2007 - 05:02 PM

There are two strategies for calculating total current and total impedance. First, we could calculate total impedance from all the individual impedances in parallel (ZTotal = 1/(1/ZR + 1/ZL + 1/ZC), and then calculate total current by dividing source voltage by total impedance (I=E/Z). However, working through the parallel impedance equation with complex numbers is no easy task, with all the reciprocations (1/Z). This is especially true if you're unfortunate enough not to have a calculator that handles complex numbers and are forced to do it all by hand (reciprocate the individual impedances in polar form, then convert them all to rectangular form for addition, then convert back to polar form for the final inversion, then invert). The second way to calculate total current and total impedance is to add up all the branch currents to arrive at total current (total current in a parallel circuit -- AC or DC -- is equal to the sum of the branch currents), then use Ohm's Law to determine total impedance from total voltage and total current (Z=E/I).

This is an excerpt from a JPG file I found online.

These calculations are commonplace in the electronic world. If this Casio cannot perform them, it is a MAJOR shortcomming and will make me regret buying this product if I cannot find a way to do this.

regards,

Charles

This is an excerpt from a JPG file I found online.

These calculations are commonplace in the electronic world. If this Casio cannot perform them, it is a MAJOR shortcomming and will make me regret buying this product if I cannot find a way to do this.

regards,

Charles

### #5 Guest_Guest_Chuckster7_*_*

Posted 26 November 2007 - 05:38 PM

A similar logic applies in the case of the relationship I=V/Z, where, on the (correct) assumption that the reciprocal of a vector (i.e., 1/Z) is also a vector,. we can identify the pseudoscalar voltage V as the reference vector against which the phase of I will be determined:

V = V( |V| , 0? ) = +|V|

there it is........

*Also taken from a JPG online.*

I know it is "possible" to invert a vector in this way. I am beginning to think it is impossible with a CP.

Webmaster....sorry about the duplicates above. I dont see an option to delete posts.

V = V( |V| , 0? ) = +|V|

there it is........

*Also taken from a JPG online.*

I know it is "possible" to invert a vector in this way. I am beginning to think it is impossible with a CP.

Webmaster....sorry about the duplicates above. I dont see an option to delete posts.

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