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Electronics help?

FarmerGiles97
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So ive just started a Mechanical And Electrical apprenticeship, and i missed one lesson due to a doctors appointment. 

I need some help on a question i didnt learn, is there anyone that could potentially link me to a video or a forum that could help? 

 

This is the question, i dont need people to do the work, i just needa find a video that will help me do the questions. 

Im fine with doing circuits without diodes, its just i wasn't at the lesson that involved diodes and now im completely confused 

 

PSDi8rj.jpg

 

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Diodes are one-way electric "valves" so to speak. In the ideal sense, they let current flow one way, but not the other.

 

But since they have a voltage drop of 0.7 and there's current going through it, then it stands that Ohm's law should apply to get the apparent resistance.

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2 minutes ago, M.Yurizaki said:

Diodes are one-way electric "valves" so to speak. In the ideal sense, they let current flow one way, but not the other.

 

But since they have a voltage drop of 0.7 and there's current going through it, then it stands that Ohm's law should apply to get the apparent resistance.

Soo ill just do ohms law, but take 0.7v of the 12v? 

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2 minutes ago, M.Yurizaki said:

No. Voltage in Ohm's law for a component is voltage across the component. The voltage across the component is what the voltage drop is.

Now im just confused lmao 

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1 minute ago, M.Yurizaki said:

What's confusing?

Right. 

 

So i needa do: 12v / Rt (42) = 0.28A

 

R1= 0.28A x 19Ohm = 5.32V

And so on ect? Or am i doing it wrong? 

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10 minutes ago, JordanHopkins97 said:

<snip>

The voltage drop across each diode is given as 0.7V

 

  • Subtract the voltage drop for each diode from the supply voltage...
  • The remaining voltage is across the resistors, solve as you claim you're fine with... (ohm's law)...
  • Once you solved the current flowing trough the resistors you also know the current trough the diodes (elements in series carry the same current)...
  • Now you know the current trough the diodes and the voltage across them you can calculate their apparent resistance (ohm's law)...
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2 minutes ago, Unimportant said:

The voltage drop across each diode is given as 0.7V

 

  • Subtract the voltage drop for each diode from the supply voltage...
  • The remaining voltage is across the resistors, solve as you claim you're fine with... (ohm's law)...
  • Once you solved the current flowing trough the resistors you also know the current trough the diodes (elements in series carry the same current)...
  • Now you know the current trough the diodes and the voltage across them you can calculate their apparent resistance (ohm's law)...

Thank you! Thats what i needed to know whas the first part!

 

I i just use ohms law, with 10.6 as the voltage? 

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1 minute ago, JordanHopkins97 said:

Thank you! Thats what i needed to know whas the first part!

 

I i just use ohms law, with 10.6 as the voltage? 

No

 

The voltage part in Ohm's Law is voltage difference between the input and the output of the part itself, not just the input voltage. The voltage difference across a diode is 0.7V, regardless of what the input voltage is (to a certain extent)

Edited by M.Yurizaki
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Just now, Unimportant said:

The picture says 100V supply voltage. I'm not sure where you get 12V from?

Omg sorry, it was from another question i wqas doing lmao 

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Can someone actually show me the calculation because now im getting confused

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22 minutes ago, JordanHopkins97 said:

Can someone actually show me the calculation because now im getting confused

Since you missed the lesson I'll give you this one to learn from:

  • 100V supply voltage minus two 0.7V drops, one for each diode, equals 98.6V across the 3 series resistors...
  • The resistors are in series, so their resistances add up. 19 Ohm + 14 Ohm + 9 Ohm equals 42 Ohms...
  • Current equals voltage over resistance. 98.6V / 42 Ohm equals 2.35 A...
  • Voltage equals current times resistance...
    • For R1: 2.35 A x 19 Ohm = 44.65V
    • For R2: 2.35 A x 14 Ohm = 32.9V
    • For R3: 2.35 A x 9 Ohm = 21.15V
  • Apparent resistance for a single diode is voltage over current: 0.7 V / 2.35 A =  0.3 Ohm
  • Total resistance is Rd1 + Rd2 + R1 + R2 + R3 = 0.3 + 0.3 + 19 + 14 + 9 = 42.6 Ohm.

(Calculations rounded)

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Diodes are like one way valves, as others said. 

 

So you have 100v input voltage (E1), and you have voltage acros diodes at 0.7v , which means the resistors "see" a voltage of 100v - 2 x 0.7v = 98.6v

 

You have Ohm's law : V = I x R  so  in this case 98.6v = I x R =  I x (R1 +R2 + R3) 

 

If those are the values for each resistor, then 19 ohm + 14 ohm + 9 ohm is 100% of the total resistance, so now you can calculate how much voltage drop across resistor you would have:

 

42 ohm ..... 100 %

n ohm ...... ? %

 

? % = n ohm x 100 / 42 = n ohm x 2.3809 = 2.381 x n

 

so :

19 ohm ::  % = 19x2.381  = 45.239 %  ... so the voltage drop across the 19ohm resistor is  98.6v x 45.239 / 100 = 44.6v

14 ohm :: % = 14x2.381 = 33.334% .. so voltage drop across the 14 ohm resistor is 98.6v x 33.334 / 100 = 32.867v

9 ohm :: % = 9x2.381 = 21.429% .. so voltage drop across the 9 ohm resistor is 98.6v x 21.429 / 100 = 21.128v

 

they add up to around 100.02% ... that's due to me rounding up the number to 2.381 when it's actually 2.38095 something... it's within tolerances for an exercise, i think.

 

Answering the 2nd question ... you know V is 98.6, you know there's 42 ohm in the resistors ... ohm's law is V = I x R so I = V / R = 98.6 / 42 = 2.3476 A

 

If there's 0.7v drop across diode and current is 2.3476A then V = I x R => 0.7v = 2.3476 R => R = 0.7v / 2.3476 =  0.2981 ohm 

 

So the total resistance would be 42 ohm + 2 x 0.2981 ohm?

 

Here's the circuit to play around in a simulator, move the mouse over the resistors and diodes to get info voltage and current across them : http://tinyurl.com/y9kszqrr

(note circuit simulator only lets me specify voltage drop across diode at 1A, so at higher current it simulates diode with ~ 0.745v drop that's why the numbers are slightly different in simulator)

 

If you're anal about in real world, even the wires between the components would have some resistance, but for the exercise's sake this is good enough and wires can be considered "ideal"

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23 minutes ago, Unimportant said:

Since you missed the lesson I'll give you this one to learn from:

  • 100V supply voltage minus two 0.7V drops, one for each diode, equals 98.6V across the 3 series resistors...
  • The resistors are in series, so their resistances add up. 19 Ohm + 14 Ohm + 9 Ohm equals 42 Ohms...
  • Current equals voltage over resistance. 98.6V / 42 Ohm equals 2.35 A...
  • Voltage equals current times resistance...
    • For R1: 2.35 A x 19 Ohm = 44.65V
    • For R2: 2.35 A x 14 Ohm = 32.9V
    • For R3: 2.35 A x 9 Ohm = 21.15V
  • Apparent resistance for a single diode is voltage over current: 0.7 V / 2.35 A =  0.3 Ohm
  • Total resistance is Rd1 + Rd2 + R1 + R2 + R3 = 0.3 + 0.3 + 19 + 14 + 9 = 42.6 Ohm.

(Calculations rounded)

 

18 minutes ago, mariushm said:

Diodes are like one way valves, as others said. 

 

So you have 100v input voltage (E1), and you have voltage acros diodes at 0.7v , which means the resistors "see" a voltage of 100v - 2 x 0.7v = 98.6v

 

You have Ohm's law : V = I x R  so  in this case 98.6v = I x R =  I x (R1 +R2 + R3) 

 

If those are the values for each resistor, then 19 ohm + 14 ohm + 9 ohm is 100% of the total resistance, so now you can calculate how much voltage drop across resistor you would have:

 

42 ohm ..... 100 %

n ohm ...... ? %

 

? % = n ohm x 100 / 42 = n ohm x 2.3809 = 2.381 x n

 

so :

19 ohm ::  % = 19x2.381  = 45.239 %  ... so the voltage drop across the 19ohm resistor is  98.6v x 45.239 / 100 = 44.6v

14 ohm :: % = 14x2.381 = 33.334% .. so voltage drop across the 14 ohm resistor is 98.6v x 33.334 / 100 = 32.867v

9 ohm :: % = 9x2.381 = 21.429% .. so voltage drop across the 9 ohm resistor is 98.6v x 21.429 / 100 = 21.128v

 

they add up to around 100.02% ... that's due to me rounding up the number to 2.381 when it's actually 2.38095 something... it's within tolerances for an exercise, i think.

 

Answering the 2nd question ... you know V is 98.6, you know there's 42 ohm in the resistors ... ohm's law is V = I x R so I = V / R = 98.6 / 42 = 2.3476 A

 

If there's 0.7v drop across diode and current is 2.3476A then V = I x R => 0.7v = 2.3476 R => R = 0.7v / 2.3476 =  0.2981 ohm 

 

So the total resistance would be 42 ohm + 2 x 0.2981 ohm?

 

Here's the circuit to play around in a simulator, move the mouse over the resistors and diodes to get info voltage and current across them : http://tinyurl.com/y9kszqrr

(note circuit simulator only lets me specify voltage drop across diode at 1A, so at higher current it simulates diode with ~ 0.745v drop that's why the numbers are slightly different in simulator)

 

If you're anal about in real world, even the wires between the components would have some resistance, but for the exercise's sake this is good enough and wires can be considered "ideal"

Thank you both! 

 

Ive done a mechanical engineering apprenticeship already, but want to become multiskilled, and tbh i am finding electircs hard?? Its all new to me tbh! Ive never done any of this before so thank you?

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