| 
  • If you are citizen of an European Union member nation, you may not use this service unless you are at least 16 years old.

  • You already know Dokkio is an AI-powered assistant to organize & manage your digital files & messages. Very soon, Dokkio will support Outlook as well as One Drive. Check it out today!

View
 

13_faq

Page history last edited by johnlittlephysics 10 years, 1 month ago

13 Current Electricity FAQ


  •  Post your questions, comments, ideas etc. below using Add a comment.
  • Use Reply to continue a tread of ideas or discussion.   

 

Comments (4)

johnlittlephysics said

at 9:59 pm on Apr 8, 2014

Yi Xuan: If the question asks "Why is tungsten commonly used in light bulbs?", how should I answer with respect to tungsten's high resistance and the conversion of electrical energy to light energy?

johnlittlephysics said

at 10:00 pm on Apr 8, 2014

The key reason is tungsten has a very high melting point 3422 deg C, so it does not melt when the filament is very hot and bright.

References:
http://home.howstuffworks.com/light-bulb2.htm
http://scienceline.ucsb.edu/getkey.php?key=2548

johnlittlephysics said

at 8:17 pm on May 8, 2014

405 Celeste: Could I make a clarification on Current of Electricity? With reference to a circuit with two resistors in series with a battery,
1. is the amount of current flowing through the two resistors the same as they are in series, regardless of whether the resistance they possess is different or not?

2. Also, in a series circuit, how is emf split amongst resistors? Is it according to proportion? If so, does a resistor of greater resistance have great pd across its ends?

johnlittlephysics said

at 8:22 pm on May 8, 2014

1. When resistors are in series, the same current will flow through them. This is shown by ammeters placed at different points along the circuit showing the same current reading.
2. V = IR, where I is the same but R is different, V is directly proportional to R
The p.d. V across each resistor would be different depending on its resistance R, the greater its R, the greater its p.d. V.

If two resistors R1 and R2 are in series with a battery with e.m.f., and their p.d.s are V1 and V2, e.m.f. = V1 + V2
Also e.m.f. = IR1 + IR2 (same current I) = I (R1 + R2) = I R(effective)
so current I = e.m.f. / R(effective)

You don't have permission to comment on this page.