Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
The 4th problem can be solved via LOGs. I didn't try to solve it. The first thing you do is to take the LOG of both sides.
To multiply, you add exponents.
If you can do this problem USING LOGS, we are there:
[latex]1.6^{3}+2.7{}^{3.2}[/latex]
This one is a really tough problem and there are two ways that I know to do it. I'd have to brush up on my skills to do it. I don't expect you to be able to do it.
Well, my wall isn't going to stand for long if I don't put the foundation in, I know exponent rules, but it is good to practise & have some one say yes, you've understood or no, youve got that wrong
If I haven't finished the answers, it's because I might have to get off computer, not that I've given up **broken link removed**
What I'm not understanding is why the answers are not matching the log table **broken link removed** which means I should have got these wrong **broken link removed**
That is where I'm getting confused, I'm not making up answers, I simply aren't relating to it if they are wrong
This is the first time I have actually tried to answer something like this from my own (wrong, by the looks of it) understanding
I understood that there was a logarithm for each & every number & by using two numbers together, they can give us the sine, cosine etc
Is this not true?
Ok, let's put you in my shoes so you can see my picture for a moment, then you can tell me where I went wrong, after reading all your links & remembering a link a fair while back, I have this picture that looks like a ball.
The centre of this ball = 0 (which in turn = 1, the anything & everything object) I'm seeing this as base e or log2
The outer radius of the ball = 2 radians (any way around its shape) or again 360degrees or log10
Anything outside of this is the magnitude of the ball (*by)
So I'm seeing logarithms as giving us any angle for a given number inside this ball
Now, is it possible to correct me with a picture?
This is the first thing I've learnt & probably why I'm intrigued by this subject as everything I've learnt to date in my life has been by ear,feel or respect, never by reading data
Either you a very kind soul to stick with me or did something really bad in your last life.lol
OK, the 100^(1/2) is really √100 or 10. The whole point is (1/3) power is cube root, 1/2 power is square root, 1/6 power is the sixth root etc.
Exponent rules. The bases MUST be the same.
MULTIPLY: Add exponents. Yea I tricked you. (10^2)*(10^2) may have been easier. That answer is: 10^4. 2+2 = 4.
I gave you exponents of-1 and 3, so -1+3=2 and the answer is 10^2.
DIVISION is Subtract exponents. Yea, I tricked you again. (3-(-2)) = 5, so the answer is 10^5. It's 1000/0.01=10^5
You could have used your trusty calculator to get these same answers. Try raising 100 to the 1/2 power and see if it's not the same as the sqrt(100) or 10.
Ok, im going to have to reply to each one seperately as im onphone & its easier
so fractions cannot be converted to decimals
& two bases side by side are *
Ok, i clearly missed that one completely but it gets worse, not only did I miss exponent, I see 10^5 as first part of your answer but where did the /0.01 come from
Dare I ask about the 'n' question.lol
Not very good at this am I........I hasten to add..........yet
You could have used your trusty calculator to get these same answers. Try raising 100 to the 1/2 power and see if it's not the same as the sqrt(100) or 10.
Hey, what does 100 * 100 equal? It's not 1000. You'll find your extra zero.
---
10^3 is 1 with 3 zeros or 1000 or 1e3
10^(-2) is 1E-2 or 0.01, remember moving the decimal two digits to the left. Iused the 1000/0.01 as ANOTHER way to solve the problem.
The EASY way was to subtract the exponents. the HARD way was to do each calculation separately and do the divide.
Dividing by 10/0.1 moves the decimal place of 10 to the right 1, so you get 100. Guess what dividing by 0.01 does?
And yes 10^2 is equal to 10 * 10 and 1E2 because the "En" means multiply by 10^n or in pictures: Means move the decimal point.
I forgot the decimal moves with zeros, so if I add a picture of a decimal point connected to AC, I now have an oscillation so it has the ability to move back & forth
You may have just solved my problem with the decimal point
So am I still alright to use my ball if I adapt it, if I change 1E to negative (inside the ball) & 1e to magnitude (outside the ball)
But that does leave my log theory in a mess as I was seeing 1e or base e as centre
You want a theory for LOGS? How about LOGS are like planets, base camps,countries, Languages (examples: 10, 2, e) Each place uses it's own language or BASE, but just like Bleistift is the word for Pencil in English, 10 decimal (English-a universal or common language) is like Pencil as 1010 is like Blesiteff is to German. The number 1 or zero might be words which are the same in ALL Languages.
You can convert from any base (language) to any other base. [latex]LOG{}_{2}, LOG{}_{10}, LOG{}_{e}[/latex] are Languages. Bases are planets, countries or languages.
You ready to try Multiplying again. Bases the same. Add exponents.
[latex]10^{5}10^{2}[/latex]
5+2 = 7 so it's 10
It's like having 2 tens (10)(10) multiplied by 5 tens (10)(10)(10)(10)(10).
So you have 7 tens
Negative exponents are a little tricky, BUT there is another IDENTiTY that states that staes that
[latex]\frac{1}{x^{n}}= x{}^{-n}[/latex]
or that dividing by (stuff) is the same as multiplying by it's INVERSE or 1/(stuff)
so [latex]\frac{x^{3}}{x^{-2}} = x^{3}x^{2}[/latex]
That's all that's happening.
---
Calculator:
You should have a button on your calculator labeled: [latex]x^{n}[/latex]
I only own an RPN (Reverse Polish Notation Calculator) so I am of little help.
So, I use something like 2 enter 3 enter [latex]x^{n}[/latex]
There are no parenthesis on my calculator. To evaluate: 5-(1+3)/10
I would use 5 enter; 1 enter 3 enter + 10 divide -
There is a 4 level Stack in the calculator and one element is the display.
An enter "PUSHES" an item on the stack.
An operator like "+" operates on the last two items on the stack and puts the result in the display.
----
Past life comments: Basically I enjoy a challenge.
ok, throw some easy ones at me, whats the worse that can happen **broken link removed** I get them wrong & look stupid, well, you all know I mastered that long ago, so the only way is forward **broken link removed**
To evaluate: 5-(1+3)/10
I would use 5 enter; 1 enter 3 enter + 10 divide -
There is a 4 level Stack in the calculator and one element is the display.
An enter "PUSHES" an item on the stack.
An operator like "+" operates on the last two items on the stack and puts the result in the display.
Just watchin' and learnin' from ya'll and playin' with tubes here.
I've got a feeling that TINA (maybe other sims as well) just treat tubes (rectifiers like the 5U4GB) as high voltage, relatively high current solid state diodes. Maybe not. Here's the pspice definition (note the stuff NOT modelled):
--------------------------------------------------------------
* 5U4GB pspice rectifier model
*
* The following parameters are not modelled:
*
* (1) Filament and filament warmup time
* (2) Reverse voltage breakdown
* (3) Saturation
* (4) Inter-electrode capacitance
*
* Please note that this model is provided "as is" and
* no warranty is provided in respect of its suitability
* for any application.
*
* 5U4GB maximum ratings:
*
* PIV 1550V
* Max in 550V RMS
* Ia pk 1000mA repetitive (4600mA transient)
*
*
* .----- Anode
* | .--- Cathode
* | |
.SUBCKT 5U4GB A K
GP A K VALUE={7.55E-4*(PWR(V(A,K),1.5)+PWRS(V(A,K),1.5))/2}
.ENDS 5U4GB ----------------------------------------------------------------------
I'm going to see what they do with a triode. And here it is:
----------------------------------------------------------------------
* 12AU7A Triode PSpice Model 8/96, Rev. 1.0 (fp)
*
* -------------------------------------------------------------------
* This model is provided "as is", with no warranty of any kind,
* either expressed or implied, about the suitability or fitness
* of this model for any particular purpose. Use of this model
* shall be entirely at the user's own risk.
*
* For a discussion about vacuum tube modeling please refer to:
* W. Marshall Leach, jr: "SPICE Models for Vacuum-Tube Amplifiers";
* J. Audio Eng. Soc., Vol 43, No 3, March 1995.
* -------------------------------------------------------------------
*
* This model is valid for the following tubes:
* 12AU7A/ECC82, 6680, 6C4/EC90, 12DW7/7247 (Unit #2);
* at the following conditions:
* Plate voltage : 0..400V
* Grid voltage : 0..-18V
* Cathode current: 0..30mA
*
*
* Connections: Plate
* | Grid
* | | Cathode
* | | |
.SUBCKT 12AU7A P G K
E1 2 0 VALUE={V(P,K)+18.28*V(G,K)}
R1 2 0 1.0K
Gp P K VALUE={10.88E-6*(PWR(V(2),1.5)+PWRS(V(2),1.5))/2}
Cgk G K 1.6P
Cgp G P 1.5P
Cpk P K 0.5P
.ENDS 12AU7A ---------------------------------------------------------------------------
This one does include inter-electrode capacitances, so I guess they really do model tubes.
I don't know how to edit or create a pspice model. Doesn't look all that complicated but, as I said, just don't know how. Definitely looking up the Leach book referenced above.
Marshal Leach passed away a few years ago. My audio amp is basically designed by him. https://users.ece.gatech.edu/mleach/lowtim/faq.html The web didn't exist then and of course I made changes like 40,000 uF of capacitance and a custom 4 winding toroidal transformer. I had to design a slow turn-on ckt because of the large capacitance.
Anyway, I was outside for a couple of hours collecting a 30 gal trash bag of tree pollen.
This site uses cookies to help personalise content, tailor your experience and to keep you logged in if you register.
By continuing to use this site, you are consenting to our use of cookies.
