Can You Solve These Simple Circuits? LER Christmas Challenge Quiz. Test Your Electronics Knowledge
HTML-код
- Опубликовано: 16 сен 2024
- Christmas is just around the corner so let's start the festive fun right here right now. Here are three simple circuits and three simple questions. But the answers may be a bit more difficult. Join in the fun and test your electronics knowledge
I work in collaboration with:
The Electronics Channel (with Carlos and Detlef)
/ @theelectronicschannel
Gran Canaria Uncovered
/ @grancanariauncovered (with Detlef and Julie. Not electronics related)
Det Builds Stuff - / @detbuildsstuff (Detlef)
Retro Upgrade - / @retroupgrade (Carlos)
For All Your PCB needs: free $5 discount coupon
www.pcbway.com...
Equipment used in my videos. These are affiliate links, you pay the normal price and I make a small commission.
TEST METERS
ANENG AN8009 MULITIMETER
s.click.aliexp...
amzn.to/4da4Q7h
amzn.to/4fsjkkm
amzn.to/3yuf03q
amzn.to/4fwQOOD
www.banggood.c...
KAIWEETS HT118E MULTIMETER
s.click.aliexp...
amzn.to/3yC0ZRe
amzn.to/3SHIEcy
www.banggood.c...
VC480C+ MILLIOHM METER
s.click.aliexp...
amzn.to/46Gg1Cm
amzn.to/3YwUjyn
MESR-100 ESR METER
amzn.to/3yxQ0Zd
amzn.to/3WSr73A
s.click.aliexp...
XC6013L CAPACITOR METER
s.click.aliexp...
amzn.to/4cf2lzm
amzn.to/4dw9cp5
TM-902C TEMPERATURE METER
s.click.aliexp...
amzn.to/4fAz44L
amzn.to/4fBxD69
LCR-T4 COMPONENT ANALYZER
s.click.aliexp...
www.banggood.c...
amzn.to/3X4rOHv
amzn.to/3Agcj6h
FNB58 USB ANALYZER
s.click.aliexp...
amzn.to/3MgBFnn
amzn.to/4dyLU1H
www.banggood.c...
PCI POST ANALYZER
s.click.aliexp...
amzn.to/4dd8ihw
amzn.to/3LTpeO7
TL460S PLUS PCI_E ANALYZER
s.click.aliexp...
amzn.to/3WEsdi6
TOOLTOP ET120MC2 SCOPE
s.click.aliexp...
amzn.to/4dCaz5L
amzn.to/4dxOYLt
FNIRSI 1014D SCOPE
s.click.aliexp...
amzn.to/3AdZqK1
www.banggood.c...
NPS3010W 30V 10A PSU
s.click.aliexp...
amzn.to/4dzl2yM
T12 STATION WITH M8 9501 HANDLE
s.click.aliexp...
amzn.to/46E73W6
M8 9501 HANDLE
s.click.aliexp...
YIHUA 982 Soldering Station C210/C245
s.click.aliexp...
FNIRSI HS-02 PORTABLE SOLDERING C210/C245
s.click.aliexp...
amzn.to/3Yv7Ma5
QUICK 861DW
s.click.aliexp...
amzn.to/3WDF4B5
amzn.to/4ci7rdW
PROS'KIT SS-331H
s.click.aliexp...
INFIRAY P2 PRO IR CAMERA + MACRO LENS
s.click.aliexp...
www.banggood.c...
amzn.to/4coSRle
amzn.to/4coSV4s
TOOLTOP ET13S THERMAL CAMERA METER
s.click.aliexp...
ET13S MACRO LENS
s.click.aliexp...
OPTICAL MICROSCOPE
s.click.aliexp...
usa.banggood.c...
amzn.to/4dzBA9G
TOMLOV TM4K AF FLEX
amzn.to/3LY0YKs
tomlov.com/pro...
TL866 II+ PROGRAMMER
s.click.aliexp...
amzn.to/3SGKL05
CH341 PROGRAMMER
s.click.aliexp...
NC-559-ASM FLUX
s.click.aliexp...
DESOLDER BRAID (I use size 8045)
s.click.aliexp...
s.click.aliexp...
HX-T100 SOLDER (0.6MM)
s.click.aliexp...
ESD-11 TWEEZERS
s.click.aliexp...
Learn Electronics Repair is now on Discord! Come and join the fun, it's free.
/ discord
If you would like to support this channel
You can send donations
www.paypal.com...
You can subscribe to Patreon
/ learnelectronicsrepair
You can click Join to become a channel member
Thank you
Richard
Lots of answers so a few comments to add:
With regard to the LED challenge the answer is the same regardless of daylight, artificial lighting or full darkenss - it makes no difference.
With the two switch and lights challenge you can *not* change the original wiring as I showed it, only add components to it
1 dual diodes common a, 1dual diodes common k,
Can you mix a video on repairing a CUI in a hifi system like the lg
First question 666 ohms
My thinking:
1) As you drew it on paper, we can ignore the vertical resistors. The voltage on either end of them will be the same, so no current can flow. They might just as well not be there. Then the series pairs can the seen as one 2K resistor each and we end up with 3 2K resistors in parallel and that'll be ~667 ohm.
2) I think it depends on where you put it. That wire could act as an antenna. And bright lights could cause the LED to act like tiny solar panels. But excluding external influences, I think the voltage is 0. There's no energy source in the circuit.
3) There's a reason you sneaky fella chose AC and incandescent bulbs. So the idea is to use 3 diodes to direct one half-wave to one bulb and the other half-wave to the other bulb. That way, we can separately control the bulbs even if they're on the same circuit. And it'll *look* like they're both lit at the same time when both switches are on. So one diode across the first switch, another between the junction of the switches and the bulbs, and a third across the first bulb.
Did I get anything right?
I hope so because I got the same as you for 1 and 3.
I can't thank you enough for teaching us everything we need to know about electronics.
1. 2/3 kOhm (because the vertical resistors carry no current, and thus can be left out)
2. Depends :
- if there's light the LED works as solar cells. The voltage you measure depends on the strength of the light.
- If there's no light the loop might catch strong changing (electro)magnetic fields and the LEDS might rectify the signal. The voltage you measure depends on the strength of the change of the magnetic field (and the area of the loop).
- If there is neither a light source nor electromagnetic field you might measure some very small thermoelectric effects in the LEDS (the voltage you measure will be nearly zero)
3. use diodes
D1 parrallel over switch 1, cathode right,
D2 between common point between SW1 and SW2 and common point between bulbs, cathode up,
D3 parallel over bulb1 cathode left.
A lot of interesting answers - tomorrow I will reveal all. Well actually I will explain the solutions to the puzzles, you really wouldn't want me to reveal anything else LOL. *I just want to add there appears to be at least two solutions to Q3. Det got it to work adding the same three components as I did - but he put them in different positions in the circuit to my solution*
To assist Det with the 3rd one, add 2 carrots and a potato to the 2 broccoli and you have a stew 😆
Hope that helps. 😉
Damit, Warwick - now I'm hungry. Hmmmmm - stew....
realmente desafiador, anotei e tentarei resolver amanha e só entao checar as respostas. Mesmo o video já tendo 4 meses, vou entrar na brincadeira com seriedade! Bravo Richard!
Hi Richard🤗🤗 🤗🤗, what a great idea
1.) The value is 666 ohms,,, 1/ (r1+r2) + 1/(r3+r4) +1/(r5+r6). R7,8 are negligible.
2.) After a while it is 2-3Volt , put it in the near of an 1Kw transmitter, then it is holiday🤣
3.) 3x diodes,,,
first diode parallel to switch 1(upper left), cathode right,,,,
Second one,,, between switch 1 and switch two the cathode, anode between the bulps one,two
Third one,,, parallel to bulp one(buttom left) cathode left, anode between bulp one and two .
Many thx for your challenge🤗🤗👍👍🥳🥳
As you say 3 diodes, and all the diodes can be mounted the opposite way around to how you have configured them and it still works due to being fed from AC.
Thank you Richard . I love this challenge for a Christmas special its really good fun.
1) If all the resistors have random values, you'll have to use star to delta resistor network transformations 2 times, then calculate some serial and parallel sets of resistors.
0.667kOhms. 3x 2k in parallel. The north-south resistors are irrelevant as potential currents are offsetting each other.
that what i thought
I had to solve a resistor network like this for my City and Guilds Telecommunication Principles exam in the early 1980's. All of the resistor values were different so there were no shortcuts.
We had to solve it using Kirchoff's Laws and it was not a very pleasant experience.
However I did manage to get a Pass with Distinction in the test.
I doubt that I would pass the test if I had to do it now though.
Really nice. Thank you.
Have no clue. Lol! Maybe these challenges should be later in the course after diodes have been covered? Now, I know how they work, but I'm still too much of a beginner to solve these puzzles. Great puzzles though.
Hello , first challenge is 666.6667 ohms , second 0 volt and third is 3 diodes , Merry christmas !!
1) 667 Ohms. The two resistors across are irrelevant as they connect to points with the same voltage and no current flows through them. Hence, 3 parallel strings of 1K+1K resistors remain. 2K / 3 = 667
2) 0V. I wasn't sure. Maybe the LEDs produce a voltage via the photo effect or the wire and the capacitor form a resonant circuit which is stimulated by radio waves. Finally I build it on a breadboard with 4 white LEDs and measured 0.04V.
3) Add 3 diodes. In order to make the right switch working independent from the left you have to bridge the left switch and the left lamp each with a diode and both need to have the same polarity to conduct in the same half cycle. For the left switch to work you have to bridge the right switch and right lamp which is possible by just one diode. Connect it from the middle of the switches to the middle of the lamps and in opposite polarity to the first two diodes. As both lamps work only in one half cycle they are not at full brightness.
The first one will be a bit over 500 ohms because it's basically three 2k circuits in parallel. Number two I'm going to guess zero because there's is no voltage injected in the circuit. Number three I'm going to say diodes but I haven't worked out the exact configuration. Nice challenge makes you think.
Great idea, i don't like theory but i will try my best
🎄🎄🎄🎄
Here are my answers:
1 . 667 ohms. The resistors which are vertical will have same potential on both pins because of circuit cimetry ,therefore no current will flow thru them we ca ignore them and remove . The others are 3 groups of 2 in series
2 . Blues LED have forward drop of around 3 V . Any exposed to light semiconductor will produce some current even if not called "photo" and this current will be in one direction - diode, which will charge the small cap to the forward Vf of 4 series Led.
3. Need to add 3 diodes. One over left switch kathod to trafo pin , second diode bypass to left bulb anod to trafo pin , thirth diode between mid points of switches and mid point of bulbs kathod toward bulbs. It may work reversing all diodes, most likly...
Merry Christmas!
Challenge 2 im guessing it depends on the supply voltage, the capacitor and led's are in series, capacitors block dc current. I think maybe the capacitor charges fully to the supply voltage and doesnt allow current to flow through led's, so no voltage drops across led's
Hi Rick, 1) 1k x 2 resistors in series / 3 lines in parallel = 2/3 = 666 ohms the last two are there just for beauty. 2) It is too long to explain, but let me say it that way: when the capacitor is full and there is light the voltage will be 0 and vice versa when the capacitor is empty and it is dark then the voltage will be 0 again . 3) 2 components (of the same type - different polarity) only and no rewire and no cuts and no matter the voltage and no matter the bulb wattage.
667 ohms - depends on how much light is on the LEDs, - 1 microcontroller.
You have three thumbs up so show us how to solve Q3 with one microcontroller and no other components (OK so to help you, you can add two more components if you really need to) I can't see how your solution would work regardless but it would not be the first time I was wrong
@@LearnElectronicsRepair Arduino uno r3, thats one component.
Great idea to test knowledge 👍
No 2: Between 2.7 and 3.7 V. However without an resistor or unless this is a constant cirrent supply, the LEDs won‘t live long
0.66k
A very fun challenge. Amusing, yet not too simple. I know I am a tad late to the party, but here are my results nonetheless:
1st challenge: after some delta to wye conversions, you obtain an equivalent resistance of 666.66 ohms.
2nd challenge: a bit tricky, as the network of 4 diodes in series would make you think of a potential difference of 2.8V (4×0.7), however there is no energizing source in the circuit. This means that the diodes will in fact never be forward biased, making the voltage across the capacitor 0 V.
3rd challenge: I believe I was able to do it with two resistors and 1 NPN transistor. I implemented the first resistor in such a way that one of its terminals is between the switches and the other is between the light sources. This will ensure that the first lightbulb turns on with the activation of the fist switch regardless of the state of the second switch; additionally, this will not short circuit the other luminaire in case the second switch is activated. Lastly, I used the remaining resistor in conjunction with the transistor in order to make a sort of a not gate that will ensure the activation of the second light source even if the first switch is open. The base of the transistor is connected to the node in between the switches(which also contains 1 of the terminals of the previous resistor). The collector is connected to positive terminal of the transformer's secondary through the 2nd resistor which is used as a pull up resistor. Finally, the emitter is connected to a ground rail which is isolated from the rest of the system.
I am new to electronics so I will give it a go
1, is 666.66 ohms
2, If I remember right a multimeter injects 5 volts to test resistance and blue leds have a forward voltage of 3.4 I think the capacitor has 1.6 volts on it
3, I think you can make that circuit work with a rectifier and a diode or just three diodes however you want to look at it.
0,66 k
The second one could either be a small voltage from RF pickup which would be very small or the small voltage produced by an led when exposed to light. You can prove which it is by putting the circuit in a cardboard box to eliminate the light and measure the voltage then. I think it's the light effect causing the voltage.
ok last one add 3 resistors of 1 meg in series from the middel (between the 2 switches and the 2 lightbulbs) should behave the same still
666,67 ohm, no voltage if there is nothing to give tension. The third one, the only thing I would do, is connect the out of switch one to the in of resistor 1 (light left) connect the in of switch two to the in of switch one as to make a parallel connection. Lastly, connect resistor 2 out to resistor one out. I don't know about adding much components but the only viable option seems to be a serial connection of switch and lamp in a parallel connection of both serial connections. So, I'd be messing around with switching 2 calbles and adding one jumpercable.
Challenge 1: 2/3kOhm or 666 Ohms assuming ideal resistors. The two extra resistors connecting the mid points do nothing at all. The could be left out or even be replaced with a jumper wire with 0 ohms.
Challenge 2: Could be anything. Can't really be answered without knowing more about the LEDs. The usage of LEDs as a kinda solar cell depends on the size and material used and if it's a low current LED or not. Clear LEDs like used here work better than diffused ones.
Challenge 3: I'm sure it's some trickery with diodes, but right now I'm too tired ..
667 ohms.
About 1 volt depending on the room brightness.
A diode across each switch in opposite directions and one across the left bulb in the same direction as the diode across the first (left) switch.
Well, I hope I got these right. I love these kinds of challenges! More please!
Crap, update... Instead of the diode across the second switch, it has to go from between the switches to between the lights.
The circuit is AC.
@@andymouse that's correct.
Should this video follow Lesson 5 ("What is a potential divider")? Capacitors & switches etc have not been covered to date.
number 3 i think what you tried to show was 2 switches connected to 2 bulbs by 2 wires .ie the switches should seperate from the lamps with just 2 wires connecting the transformer/ switches to the bulbs
Can you elaborate on your solution to this Rob? The circuit I showed can not be changed, you have to add at least three components to make it work as I showed. My circuit does contain those components they are just hidden (otherwise this would not be a challenge of course)
@@LearnElectronicsRepair
one leg of AC fed to 2 back to back diodes
each diode goes a switch the other end of the switches are conected together the wire from the switches along with the other ac wire go to the bulbs each bulb has a diode to one vside one to the anode the over to the cathode.
the other end of the bulbs join together and connect to the other ac lead
Ill go with 1.) 666.7 ohms (but it seems like im missing something I learned) 2.) 0 volts (no power source as shown) Need more info on this one, and 3.) I was thinking relays and all kinds of stuff, now my brain hurts. Ill go with everyone else and use diodes
The vertical resistors are irrelevant, since they connect points of equal potential (they can be replaced by whatever value you like). So the answer is 2/3k. The solution to the second problem depends on the initial voltage condition of the capacitor. If biased in forward direction to the diodes, the voltage can be any value
between 0 and 4x(forward voltage of the diodes). Otherwise the diodes prevent the capacitor from discharging. But for all practical purposes I believe 0V is a save bet.
The solution to problem 3 is obtained by 3 diodes placed in parallel to the first switch, the bulb beneath and vertically between the diodes in order || and downward, respectively.
can you elaborate as I can't follow the logic. if you put a diode parallel to the first switch it's like only partly disconnecting the line so the light will still shine. Or am I mistaken
@@wallygrandpa
This might go terribly wrong, bur for the sake of ASCII art, here is the schematics:
D1
------||-----
If S1 is open, current can't flow through B1, since it get's blocked by D2.
@@Nepenth I tried to draw it in falstad circuit simulator. But although the first light gets less amperage, I still see a stream that can slightly light up the bulb. Is that about right?
wallycompelectro.blogspot.com/2023/12/diodes-in-serial-switch-and-light.html if you copy this url, you should see what I think you mean.
@@wallygrandpa
You have to be a little more precise. There are 8 cases to consider depending on whether AC is + or - at the input switch S1 (2 cases) and whether
S1 and S2 are open or closed (4 cases, so 2x4 = 8 altogether). So which case is it that you are still concerned with?
Also, the 3rd can be done with 3 diodes
Love the theory
circuit 2... between 1 and 3 volts unless you put it next to a 200 watts fm transmitter then a flash and de the led's blow LOL
2K
1 is .667k
Need clarification. In ref to the Voltage for the Diode challenge. Will Darkness effect your challenge? In ref to the 12VAC lamps circuit. Must the wiring not be cut? Are we only tapping the added items?
With regard to the LED challenge the solution would work in daylight or full darkenss - it makes no difference.
With the two switch challenge you can not change the original wiring as I showed it, only add components
👌👌👍👍
The total resistance in challenge #1 is 1500 ohms.
circuit 1.... ITS THE DEVIL yup yup the devil
No ...😱😱 .is Voodoo..🤣🤣🤣
My 2 cents} CKT #1 is 66.6 ohms CKT #2 is 0.0 VDC CKT #3 is pending clarification if ckt wires can be cut.
1) 667 ohms
Are the 4 leds covered up or open to a light source
2k resistance
, One/1.32K / 3 = 667v ish, Two/ 2.7v to 2.8
Richard, I am beginner in electronics and want to learn how to repair any electronic device.
Where should I begin?
Waiting for tomorrow
no1. 500 ohms
no2. zero volts
no3. 3diodes
might not be understanding number 2.
The sum of the sultanas entering a node will equal the sum of sultanas leaving. Or is it Currents.
PMSL
Question 3 can probably be done with diodes, if you accept half brightness.
There was no indication of the voltage rating of the light bulbs. If you use 6 V bulbs, they lit up with almost full brightness since they only use either positive or negative portion of the AC. However, the rightmost bulb suffers from two diode drops, whereas the left one only from one diode.
@@harrilumme1875he actually said 12volt bulbs
@@joeds3775 OK, I missed it🙄
2nd is 0 volts caps block dc
Challenge 1 2k, challenge 2: 0 v , change 3: 2 MOSFET, 1 driver
Second challenge thé answer IS 0 VOLT if the capacitor is connected to ground
666.666 0.0000 and the therd is very easy
Just wire it in series parallel
The answers to the forts test IS 1000 ohms
now I do not understand how 2 switches in series and one is off it still makes the lightbulb glow there is no complete loop for current to flow it doesnt make sense and what do you mean with make it behave as it does now ... i mean it behaves like that already so how can we make it behave as it already does ???????? so I am with Det I do not get it !!!! I have to ask Det if you have gone completely bonkers LOL
He has already included his solution. The circuit components are complete BEHIND the cardboard that is holding the visible components. He wants YOU to design the components and place them in his circuit to see if yours will work. Get to work.
I'll give you a clue or at least something to focus your attention on, the output of the transformer is AC.
@@robs905 Yeas exactly that - add three components to the visible circuit so it behaves like the one I showed. Mine has the three components hidden behind the cardboard.
666ohms i said 400 ohms thats not correct