Think i got it! Yes! So its 2-1.5= 0.5v following current direction. But when do i know to use current or electron flow direction? Above for potential divider i was told to look at electron flow
Should i put the 3v battery to the right or left of X and should i go from + terminal to -terminal from X to Y like current flow or electron flow?
Smiles :D
Thank you all for your inputs! I realised its the direction after thinking for so long. And yea it should be ohm instead of v.
I have another question regarding potential could you guys help me? Answer is A but i dont understand the solution as there are 'conflicting battery direction' and i...
Why is the potential at X , denoted by V, not
V/8 = 4.8/(4.8+7.6) ? Why is numerator of resistance part 7.6 and not 4.8 by potential divider principle?
Considering the train to be of 80 thousand kg i think that should be it. I dont know what's that principle lol, but i thought your own school tutorials and notes will suffice?
Why wouldnt it seem right? Hmm. You have to use 19.444 m/s though. More accurate
Because its negative? Your acceleration should be negative btw. Unless you put deceleration
It has to be operating :P i think by just looking at resistance i got it - Highest power cuz least resistance when 2 resistors in parallel
Least power when 2 resistors in series
But that's assuming I and V are not outweighing Which complicates the matter more. Well furthermore V and I are...
you need W= mg for weight
and to calculate the frictional force.
Calculate force in the direction of the retarding of the train through trigonometry then add them up to find total force slowing the train down. I thnk you can do the next part now.
Formulas:W=mg
F=ma...
Using method one is faster: v2=u2+2as
Kinematics equation to find s after finding acceleration thru F=ma.
Tedious part is to find resultant force in the direction that retards the train first.
"It did prove the potential at P fell since (increase in I supplied by cell due to fall in R total) is LESS than (fall in R total due to series to parallel change) which meant the v fell since IR fell, but well if no values were given how do i tell? Hmph "
Aww sorry i still dont get how to...
How do i get this? Filled in the answers for easy reference. Dont see why unless i try subbing values again. Seems like a high order application question to me. Least dissipated power: Series circuit with 2 resistors(P=V2/R V larger with 2 resistors by V=I(2R) why :( least dissipated power?)...
Too high order for me- that formula P
I guess thats the only way - to sub in values :(( which takes a whole lot of time for me and theres time limit for exams :(
It did prove the potential at P fell since (increase in I supplied by cell due to fall in R total) is LESS than (fall in R total due to series to parallel change) which meant the v fell since IR fell, but well if no values were given how do i tell? Hmph
The resistors have the same resistance. It is a general question without values but i think we got it. Except that im still unsure of the part about potential decrease for P- due to fall in R, I or both.:o I think now, that its quite ambiguous cuz fall in Total resistance actually increases I...
I think for potential at P it decreases cuz
Both R and I falls- R becomes R/2, I becomes I/2 is it?
For Q i think overall I has increased cuz of the decrease in total resistance with the switch from series to parallel for the left part---so the potential DROP after the earthing part at resistor...
Btw, Why would potential at Q decrease compared to original? Its earthed before that so potential at Q is pd across the Resistor R0 (answer is D)
Also, is it because R parallel is halved only or is it because BOTH R parallel being halved and split(meaning current decreases) of current that...
I think i got it from the 2nd last reply and the first reply to this topic :P But Except this case?Answer is C and i got it based on the 2nd last reply :)) Hmph.
Many millions of thanks to you 2 who helped! Hugs anyone? Hahaha.
Please pardon my drawing skills
Okay so lets say theres 5 bulbs, bulbs 2 and 4 broken. Counting From LHS. Emf is 12v.
Left hand side of bulb1: 12v
RHS of bulb 1: ( 12- 12/3 ) v ?
LHS of bulb 2 : (12- 12/3) v
RHS of bulb 2 : same
...
RHS of bulb 5: 0v
I read in a book p.d is zero across...
But i dont know when there's a voltage drop and when the potential is the same :( especially for the later parts about broken bulbs.
I do know how to draw circuit diagrams
But i dont know what happens at these points- tried using different colours too
Most importantly, i need to know part g...
ThIS IS NOT A HOMEWORK! I need to know the difference which is why i used simple setups to help ask a question. This helps me in doing more complicated circuits after understanding where a potential drop occurs and when it doesnt
I got the answer by comparing I2R of the circuit with resistor R but why do i have to use R/2 for power of resistor R?( P= (0.5I)^2 (R/2)) I know its parallel to Resistor Q but i want to find the power of that component so shouldnt it be P= (0.5I)^2 (R) instead?
I have covered newton's law and i understand that part. I think i got you now :) so a particle will be displaced downwards irregardless of movement when it is below the eqm position?( in a graph of sinusoidal waves diagram)
How can a particle move downwards and be displaced upwards? Sounds paradoxical. I understand the part about movement but not displacement. What's the difference? Isnt displacement like the amplitude so when a particle in a transverse wave move down doesnt it mean its displaced downward?
Why is the wire and ammeter not in series? Well i see the voltmeter as a separate entity, and its a complete rectangle circuit to me :0 i understand you :P
Parallel should be like on different lines right? ( in layman's term) one on a line in a small rectangle, one on a line in a big rectangle...
Circuit -- Why a more accurate value of the resistance?
Note: CURRENT CAN FLOW THROUGH THE VOLTMETER IN THESE CASES.
Why does the second setup as shown by the arrow in the picture give a more accurate value of the resistance of that wire?
It says the current in setup 1 on the left measured...