Best video ever on KCL! Very neat, organised and very educational, unlike some others which are poorly illustrated and drawn on the spot. Now I understand this, thanks!
MintEEV
Brilliant... no need to even attend school now... all the videos are here!
JB
Yes, I think you’ve got it. Add a new series resistor, and current instantantly drops. The same new current flows through the whole series circuit.
Work through this:
12V battery in series with 2Ω and 4Ω resistors. TOTAL resistance = 2+4=6Ω.
Current I = V/R = 12/6 = 2A.
A current of 2A flows through the battery and each resistor.
Add a 3rd resistor of 6Ω in series. New TOTAL resistance =2+4+6= 12Ω.
Current I = V/R = 12/12 = 1A.
A new current of 1A flows through the battery and each resistor.
StevePhysics
Hi. If calculating the algebraic sum of currents at a junction, the signs used ONLY apply to that junction.
If the 1.8A has come OUT of another junction, then when you do the calculation for the OTHER junction, you would use –1.8A.
The + or – signs ONLY tell you if the current is entering (+) or leaving (-) a particular junction. If a current leaves one junction and enters the next, you can’t use the same sign at the 2 junctions. (In fact you need the opposite sign). Sorry it’s confusing!
StevePhysics
Thanks man.. You're a life saver..
AliRaza-qctz
Hi TheDaVincci.
If N electrons per second enter a resistor, then N electrons per second leave it. Increasing the resistance (and keeping voltage fixed) just reduces N.
Watch the 1st 4 minutes of my video called:
GCE (A level) Physics E03 Emf and voltage. Part 1 of 2
(YouTube prevents me giving the actual link).
Increasing the resistance is equivalent to increasing the friction in the tube (it will make sense when you watch the video).
Hope that helps.
StevePhysics
I really enjoyed this video, even though i am not doing a level physics
pandorat
@javedb2
Sorry if it's a bit slow for you. But I know some students can struggle a bit with the material if they haven't really understood current in the past.
StevePhysics
2) You need to watch the videos on resistors in series and in parallel. I can’t go through the details here – you will need to watch the videos. But here is a quick example:
A 6V battery, a 4Ω resistor and an 8 Ω resistor are in series.
Total resistance for series combination = 4+8=12 Ω. Current I = V/R = 6/12 =0.5A.
The same current of 0.5A flows through the cell, the 4Ω resistor and the 8Ω resistor.
It’s my bedtime now so I won’t be answering any more queries till tomorrow!
StevePhysics
I haven’t enough space to answer. I’ll try 2 replies.
1) I don’t know what you mean by ‘already’. I made up a random value for illustration purposes.
Suppose the cell is 7V and the resistance of the bulb is 10Ω, the current I = V/R = 7/10 = 0.7A. This is the current in and out of the cell, and in and out of the bulb.
If the resistance is doubled (20Ω) the new current is I = V/R = 7/20 = 0.35A.
Have you watched the videos on resistance?
StevePhysics
If the lamp has double the resistance, the current is halved: 0.35A would go in and 0.35A leave.
Current is the rate of flow of charge - you can’t really talk about the ‘rate of flow of current’.
Think of current as the number of electrons per second (eps). One amp means 6.25x10 ¹⁸ eps.
If 1A enters a lamp, 6.25x10 ¹⁸ electrons enters the lamp each second. The electrons can’t build up or disappear, so 6.25x10 ¹⁸ electrons must leave the lamp each second. 1A in means 1A out.
StevePhysics
@JB27611
I hope you dont mean that! Your teachers wil prepare you for the specific syllabus you are using, give you problems and feedback, monitor your progress, and then there's the practical work....
StevePhysics
The 2 bulbs are not identical – they have different resistances. I should have made that clearer. If the bulbs had the same resistance, then you would be correct – the current would split equally into 0.9A and 0.9A.
In the example, the right hand bulb has a slightly lower resistance than the left hand one. As a result, when the current splits, more current flows through the right hand bulb. Watch 'E13 Resistors in parallel' for more information.
StevePhysics
@Steve4Physics ;) - Don't worry.... I wouldn't dream of missing a lesson
