The discussion revolves around understanding the behavior of current (I), voltage (V), and resistance (R) in electrical circuits, specifically comparing series and parallel configurations with bulbs of varying brightness. Participants explore how to remember the relationships between these quantities and the implications of bulb brightness on power dissipation.
Participants express conflicting views on the relationships between brightness, resistance, and voltage in series circuits. While some assert that the brightest bulb has less resistance, others challenge this by stating that higher brightness indicates higher power dissipation, which should correlate with higher resistance. The discussion remains unresolved with multiple competing interpretations.
Participants reference specific laws and equations but do not reach a consensus on their application to the problem of bulb brightness in series versus parallel circuits. The discussion highlights the complexity of these relationships and the need for careful analysis of the underlying principles.
In terms of your question, the total voltage drops in the two bulbs = the emf of the cell, and the current flowing through a series circuit is constant.sophiecentaur said:I'm not too keen on the idea of rote learning stuff like this. There are some very straightforward 'Laws' which you can use and you can guarantee they will give you the right answers.
Firstly, we all know that V=I/R (and the other two combinations) and that Power = VI. Then:
Kirchoff's Voltage Law (KVL) says that the total voltage drop around a series circuit (loop) is zero. That means that the total supplied volts adds up to the total of voltage drops across all the other elements. (aka Volts are 'shared' in series)
Kirchoff's Current Law (KCL) says that the total current into a node (connection point for two or more components) is Zero. (aka Current is shared in parallel)
So, using KVL, what can you say about the volts across the two bulbs and the Cell voltage in a series circuit? What can you say about the Current flowing through the series circuit?
Work it out with ant particular example you can think of - then another one. That will show you the trend.
Greater, so...:robphy said:Brightness is related to the power dissipated: ##P=I_{through}V_{across}##.
As @sophiecentaur , asks
in series, how are the currents through each resistor related?
Then, if one bulb is brighter than the other, what can you say about the voltage across the brighter bulb?
Power dissipated in the bulb will be I2Rbulb (this comes from P=Vacross*Ithrough).IDK10 said:In series:
The brightest bulb has less resistance, greatest voltage and same current as the dimmest bulb.
In parallel:
The brightest bulb has less resistance, greater current and the same voltage as the dimmer bulb?
Hang on there! Will they have different currents?IDK10 said:in series, what would be an easy way to visualise which one has the highest current and voltage
IDK10 said:But what I mean is, what if the two bulbs are of different brightness, and are in series, what would be an easy way to visualise which one has the highest current and voltage
IDK10 said:the total voltage drops in the two bulbs = the emf of the cell, and the current flowing through a series circuit is constant.
DaveIDK10 said:In series:
The brightest bulb has less resistance, greatest voltage and same current as the dimmest bulb.
How can that be true? If the bulb is brighter then it is dissipating more power. For the same current (series circuit), the VI must be greater, which means the V must be higher so R (=V/I) must be higher. The higher resistance bulb 'steals' more of a share of the volts.IDK10 said:In series: The brightest bulb has less resistance, greatest voltage and same current as the dimmest bulb.
NO, was just kind of quickly typing that, sorry.sophiecentaur said:Hang on there! Will they have different currents?
Had shift on for too long, the o in 'NO' was meant to be lower case.IDK10 said:NO, was just kind of quickly typing that, sorry.