Timeline for Confirmation on Diode in parallel with a Resistor
Current License: CC BY-SA 4.0
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| Jan 27, 2020 at 15:48 | comment | added | User323693 | I get the loop currents as 17.6 mA and 26 mA. Did you solve? | |
| Jan 27, 2020 at 2:34 | comment | added | User323693 | I had done a mistake in the loop equations. Have updated the image now. We get two simultaneous equations which can be solved easily.. | |
| Jan 27, 2020 at 2:33 | history | edited | User323693 | CC BY-SA 4.0 |
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| Jan 27, 2020 at 0:19 | comment | added | Ciribear |
@User323693 I think so. I am using the equations under your schematic: V = I1R3 + (I1-I2).......etc and substituting my values for I1 and I2 in there. I calculated I1 = 0.02A and I2=0.019A then I put those values in your equation and tried to solve it. But i got V = 1v + 0.2v + 2v = 3.2v How do you use your equations in your photo to solve this? and How would you find i1 and i2? Sorry to be a bother. I have added some values to the schematic
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| Jan 27, 2020 at 0:08 | comment | added | User323693 | @Ciribear are you solving for I1 and I2? | |
| Jan 26, 2020 at 23:00 | comment | added | Ciribear |
@User323693 Thank you for adding that schematic. To be honest I feel I understood things a lot less then I thought I had. But I don't quite understand your working. If you wouldn't mind explaining further. Assuming Supply = 5v, R1 = 200Ω, R2 = 210Ω, R3= 50Ω, LED = 2v then my workings in your formula is: V = 1v + 0.2v + 2v = 3.2v which is 2 less then the supply? or the LED and 0= -0.2v + 2v +3.99v = 6v which isnt 0. I1= (5v -2v) ÷ 200Ω + 50Ω = 0.02A and i2= 5v ÷ (210Ω + 50Ω) 210Ω being an arbitrary value.
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| Jan 26, 2020 at 16:48 | comment | added | Peter Bennett | R1 and R2 ARE NOT in parallel. R1+LED is in parallel with R2. | |
| Jan 26, 2020 at 9:41 | history | edited | User323693 | CC BY-SA 4.0 |
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| Jan 26, 2020 at 9:28 | history | edited | User323693 | CC BY-SA 4.0 |
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| Jan 26, 2020 at 9:18 | vote | accept | Ciribear | ||
| Jan 26, 2020 at 9:13 | comment | added | Ciribear | Sorry with a resistor not in the original schematic. I have added a new one. If R1-LED and R2 was in series with R3. So to confirm, in the case of a third resistor added in series, would you calculate total current by using the supply voltage minus the LED 2v. And find the total resistance the normal way for the parallel resistors? (1 ÷ (1/R1 + 1/R2) )+ R3 even though the LED with the parallel resistors? | |
| Jan 26, 2020 at 7:14 | comment | added | User323693 | @Ciribear resistor in series with that .. with which component? R1? | |
| Jan 26, 2020 at 7:04 | comment | added | Ciribear | Thank you I understand that. What happens if there is a resistor in series with that. Would you calculate the total current using supply minus 2v. And still calculate the total resistance of the parallel as 1 ÷ 1/r1 + 1/r2 (then plus the 3rd R)...... Then divide both? | |
| Jan 26, 2020 at 6:34 | comment | added | User323693 | @Ciribear R1 and R2 are not in parallel because they won't share same nodes | |
| Jan 26, 2020 at 6:28 | comment | added | User323693 | @Ciribear there are two branches. One branch is the series combination of R1 and LED. The second branch is R2 alone. Both branches are parallel to each other. They share common nodes. The voltage across both the branches is 5 V. First Branch had two components.. hence the LED will take its 2V and the remaining (3V) will be dropped across R1. The R2 will see full voltage as it is the only component in its branch | |
| Jan 26, 2020 at 6:18 | comment | added | Ciribear | Thank you for your in-depth reply. Sorry I made a mistake i meant minus 2v for the led I was just thinking of normal diodes. But how does that work with kerchoffs law? If say the supply is 5v and it's a red LED of 2V to forward bias. And let's say both resistors are 10 ohms. Wouldn't there be a voltage of supply - 2v across both resistors? As the 5v is Across the resistor and also the resistor and led? And both would need to use the 5v so the resultant is 0v back to ground. | |
| Jan 26, 2020 at 2:25 | history | edited | User323693 | CC BY-SA 4.0 |
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| Jan 26, 2020 at 2:20 | history | answered | User323693 | CC BY-SA 4.0 |