Understanding Electrotechnics Problems: Explanation and Solutions

[TomTom1337]

Homework Statement

Ok, this is not actually a homework, but I still need help ... would somebody be so kind and explain me how to solve these 4 questions? Thanks! I already have answers on them (all answers are true/false type ... but I need to know WHY). Thanks!







[/quote]

Homework Equations

The Attempt at a Solution

1) LED light: When both inputs in AND gate are true (positive), current will flow through them to the LED. But since second AND gate input is hardwired to negative, how can switch S1 turn ON/OFF the LED? I really do not understand.

2) Not a clue on this one.

3) Same as above, I do not a have a clue. It there a similar problem with a complete solution so I can learn by "reverse engineering"?

4) a) First number in equation tells the effective voltage?
b) Looks like the answer is first number * 2/pi, but can somebody explain me why?

 

[berkeman]

Homework Statement

Ok, this is not actually a homework, but I still need help ... would somebody be so kind and explain me how to solve these 4 questions? Thanks! I already have answers on them (all answers are true/false type ... but I need to know WHY). Thanks!

Homework Equations

The Attempt at a Solution

1) LED light: When both inputs in AND gate are true (positive), current will flow through them to the LED. But since second AND gate input is hardwired to negative, how can switch S1 turn ON/OFF the LED? I really do not understand.

2) Not a clue on this one.

3) Same as above, I do not a have a clue. It there a similar problem with a complete solution so I can learn by "reverse engineering"?

4) a) First number in equation tells the effective voltage?
b) Looks like the answer is first number * 2/pi, but can somebody explain me why?

Welcome to the PF.

You said you have the T/F answers for all of them. Can you post those please?

 

[TomTom1337]

Sure. All of them are true.

 

[berkeman]

Sure. All of them are true.

Well, the first one is clearly false...

 

[TomTom1337]

What do you mean? No matter which position the switch S1 is, LED will always be OFF?

 

[berkeman]

What do you mean? No matter which position the switch S1 is, LED will always be OFF?

You are correct in your analysis. And the question asks if the position of the switch affects the operation of the LED. That is clearly not true.

 

[TomTom1337]

Official statement is: "LED is ON regardless of position the switch S1 is in". True or false?

So the catch is in the first part of the statement - LED is never ON? If yes, I understand and do not need any explanation why.

 

[berkeman]

Official statement is: "LED is ON regardless of position the switch S1 is in". True or false?

So the catch is in the first part of the statement - LED is never ON? If yes, I understand and do not need any explanation why.

Well that's not what you posted. Are the wordings of any of the other problems different from what you posted? Not giving you a bad time here, just wanting to get things straight.

 

[TomTom1337]

No, I just changed statement in question #1. All others are "intact".

 

[berkeman]

No, I just changed statement in question #1. All others are "intact".

Well, the answer to #2 appears false as well, or at least there is a typo. Can you describe how you should go about approaching this problem? What are the impedances of the resistor and capacitor at the given frequency? That's why we ask for the "Relevant Equations"...

 

[TomTom1337]

I really have no idea right now ...

 

[TomTom1337]

... and I also doubt the answer to the #2 is false, since official answer is "true".

 

[berkeman]

I really have no idea right now ...

Then you should go back and re-read that section of your course materials. You can also read about "phasors" at wikipedia. You need to show some effort in figuring out these problems. That's part of the PF rules (see the Site Info link at the top of the page).

... and I also doubt the answer to the #2 is false, since official answer is "true".

Then it may just be a typo in the answer that is listed...

 

[TomTom1337]

Then you should go back and re-read that section of your course materials. You can also read about "phasors" at wikipedia. You need to show some effort in figuring out these problems. That's part of the PF rules (see the Site Info link at the top of the page).

I have read the wiki and my materials, but I still cannot figure it out.

I need to find impedance (Z) of the circuit first. In my case, I have 2 elements in serial connection (resistor + capacitor), so:

Z = R + 1/jωC = 1Ω + 1/j*50Hz*(10/3pi)mF = ...

Is my equation correct? How to solve it (what to do with that (10/3pi)mF?)?

After that I can find the I by solving equation I = U / Z?

 

[TomTom1337]

Can somebody provide some help?

 

[gneill]

I have read the wiki and my materials, but I still cannot figure it out.

I need to find impedance (Z) of the circuit first. In my case, I have 2 elements in serial connection (resistor + capacitor), so:

Z = R + 1/jωC = 1Ω + 1/j*50Hz*(10/3pi)mF = ...

Is my equation correct? How to solve it (what to do with that (10/3pi)mF?)?

After that I can find the I by solving equation I = U / Z?

You're on the right track. Convert the mF units to F (1000 mF = 1 F). Convert the frequency f to angular frequency ω. Then the units for the capacitor term will reduce to Ohms.

When you go to solve for the current, be SURE to pay attention to the directions of the potential and current indicated on the diagram

 

[TomTom1337]

Ok, so:

Z = 1Ω + 1/j*50Hz*(10/3∏)mF = 1Ω + 1000*3∏/j*2∏*50Hz*10F = 1Ω + 3/j Ω

Is this solution correct? How to pay attention on direction, can you explain? Thanks!

 

[TomTom1337]

... and then:

I = U/Z = 230V/1Ω+(3/j)Ω = 230A + j(230/3)A

Correct?

 

[gneill]

Ok, so:

Z = 1Ω + 1/j*50Hz*(10/3∏)mF = 1Ω + 1000*3∏/j*2∏*50Hz*10F = 1Ω + 3/j Ω

Is this solution correct?

Yes, although it is common practice to move the "j" to the numerator to render the value into the canonical form of a complex number: a + bj .

How to pay attention on direction, can you explain? Thanks!

Look at the diagram. Which direction is the voltage source polarity? Which direction should that source drive current around the loop? What direction is indicated for the assumed current? Do the two match?

 

[TomTom1337]

Yes, although it is common practice to move the "j" to the numerator to render the value into the canonical form of a complex number: a + bj .

... so the result in "correct" form in my case is?

Look at the diagram. Which direction is the voltage source polarity? Which direction should that source drive current around the loop? What direction is indicated for the assumed current? Do the two match?

I see. I tought that does not matter since we have alternating current. So the final answer (current I) should be negative (but is otherwise correct)?

 

[gneill]

... so the result in "correct" form in my case is?

Normalize the imaginary term: get rid of the "j" in the denominator by multiplying its top and bottom by j.

I see. I tought that does not matter since we have alternating current. So the final answer (current I) should be negative (but is otherwise correct)?

Even AC sources have a polarity in order to keep track of the phase relationships throughout the circuit. You'll see voltage measurements with phase angles in a circuit, such as: 240V ∠17° . Those angles have to have some reference. So if the power supply is taken as the reference and it's a sinusoidal signal, then it needs to have a polarity (direction) since -sin(x) is 180° offset from sin(x), and -cos(x) is 180° offset from cos(x).

If the current you calculate is for the opposite direction of the assumed current on the diagram, then yes, multiply by -1 to "flip" the direction to match.

 

[TomTom1337]

OK, everything clear now. Big thanks!

 

[TomTom1337]

Now I need help with next one, case #3. Any hint?

 

[gneill]

Now I need help with next one, case #3. Any hint?

I don't understand the question, so I have no hints to give! Perhaps someone else will recognize the type of question and be able to interpret it...