my teacher said that in a step-up transformer(the transformer is ideal) ,a light
ID: 2028507 • Letter: M
Question
my teacher said that in a step-up transformer(the transformer is ideal) ,a light bulb connected to the secondary coil would be brighter than an identical bulb connected to the primary coil.As far as i know,this is because the voltage across the bulb in the secondary coil is higher,and by P=V^2/R,the power of that bulb is also higher.
However,if I use P=I^2R,the results obtained seem to be contradictory with the above.Since output power=input power in ideal transformers,the current in the secondary coil should be lower.Then,shouldn't the power of the bulb in the secondary coil be lower?
Note: the bulbs are connected in series with the coils,not across.
Explanation / Answer
to understand this, you must understand that there are two different places the power is going - some of it will be lost to the resistance of the wire, but whatever is remaining will reach the end and supply electricity ( voltage ) for the appliances at home.
when you use P = v2/R, you are using the wrong V. The V you need to use is the V lost to the resistance ( which can be found using V = IR, a really small current times R means a really small votage is lost to the resistor ) The V remaining after you subtract this small voltage is still rather large making it to the other end. Therefore, this formula P = v2/R is only solving for the power LOST in the resistor, not the power of the complete circuit.
P=I2R alos only solves for the power lost in the resisitor. A small current means a small amount of power is LOST.
To truly avoid this problem, only use P = IV which doesnt take any of the resitance into account, and only looks at how the current and voltage are affected.
Now, to answer why your teacher said that the lightbulb glows brighter is more complicated, but it turns out transformers don't just transform voltage and current, but also resistance. There is a mathmatical relationship that states V1 = I1 R / ( N2/N1)2
This shows that the resistance, at least mathmatically, of the circuit attached to the second coil of a step up transformer is LOWER than the resistance of primary. Changing the resistance of the secondary bulb ( it is actually changing the impedance, not the resistance ) will also affect how brightly the bulb burns.
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