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VCE Physics Question Thread!
sin0001:
For part b of the attached question, I get that the resistor, parallel to the LED, will have the same potential difference as the LED-2.5 V, so doesn't that mean the Resistor should have the same current flowing through it as the LED-11 mA; so can't we then use these values for current & voltage, to work out the resistance of Resistor-'R'?
But in the worked solutions they haven't assumed that current across 'R' will be same as I(LED), instead this is what how they solved resistance:
Vtherm. = 10 – 2.5 = 7.5 V, Itherm. = 7.5/500 = 0.015 A
since ILED = 0.011, IR = 0.004 A and RR = 2.5/0.004 = 625 Ω
Alwin:
--- Quote from: Homer on May 02, 2013, 07:08:52 pm ---Hey guys how would you calculate the value of R?
Thanks ;D
ANS: 309.1ohms
--- End quote ---
Well, first you look at the diodes:
Diode 1:
Since the current is 20 mA, V = 1. Hence, R = 1/0.020 = 50 ohm. Thus, you have a 50ohm 'resistor' and a 500 ohm resistor in parallel
Diode 2:
This diode is in reverse bias, hence no current flows. There are only the 2 100ohm resistors in parallel
Total resistance:
Total current:
From Ohm's law, R = V/I
Hope this makes sense!
Alwin:
--- Quote from: sin0001 on May 02, 2013, 07:48:45 pm ---For part b of the attached question, I get that the resistor, parallel to the LED, will have the same potential difference as the LED-2.5 V, so doesn't that mean the Resistor should have the same current flowing through it as the LED-11 mA; so can't we then use these values for current & voltage, to work out the resistance of Resistor-'R'?
But in the worked solutions they haven't assumed that current across 'R' will be same as I(LED), instead this is what how they solved resistance:
Vtherm. = 10 – 2.5 = 7.5 V, Itherm. = 7.5/500 = 0.015 A
since ILED = 0.011, IR = 0.004 A and RR = 2.5/0.004 = 625 Ω
--- End quote ---
I think you got a bit confused between parallel and series circuits. In parallel circuits, the potential difference across both branches is the same, BUT the current in each branches is different. The sum of the currents in each of the branches is equal to the total current, in this case 0.015A. This is different from series circuits.
Hopefully you can see how the solutions got the answer from here!
sin0001:
--- Quote from: Alwin on May 02, 2013, 08:08:37 pm ---I think you got a bit confused between parallel and series circuits. In parallel circuits, the potential difference across both branches is the same, BUT the current in each branches is different. The sum of the currents in each of the branches is equal to the total current, in this case 0.015A. This is different from series circuits.
Hopefully you can see how the solutions got the answer from here!
--- End quote ---
Ahhh yeah I had a mental blank lol
Also, how would you work out the resistance of these diodes, would you just plug it into the formula: R = V/I?
Thanks!
Alwin:
--- Quote from: sin0001 on May 02, 2013, 08:16:53 pm ---Ahhh yeah I had a mental blank lol
Also, how would you work out the resistance of these diodes, would you just plug it into the formula: R = V/I?
Thanks!
--- End quote ---
It's okay! just try not to have a mental blank in the exam ;)
Yeah, just R = V/I. I have an example in a previous post:
Re: Physics [3/4] Question Thread!
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