• Support PF! Buy your school textbooks, materials and every day products Here!

Quickies on electric current

  • #1
1,137
0

Homework Statement


<pic>

The Attempt at a Solution



1) Ummm ... well ...no idea :tongue2:

2) I guess no. Because if a capacitor is connected with battery, work done by batery is QE (EMF) but thermal energy is just QE/2. Rest QE/2energy is stored in capacitor.

3) Same as (2)

4) i = E/(R+r)
--- r is internal resistance of battery
--- R is external resistance.
--- E is EMF
Work done by battery = QV = Q(E - ir) where r is internal resistance of battery
Thermal energy across in resistor = i2Rt = E2RT/(R+r)2

Now i dont know what to do.
But i still guess that for non-ideal its no and for ideal its yes ... don't know why

5) Well i thought of some explanation before i started typing but now i cant remember it :biggrin:

6) Yes, only for non ideal battery. At time of charging, V = E + ir
 

Attachments

Answers and Replies

  • #2
Delphi51
Homework Helper
3,407
10
In #1, the U = i*V*t formula does not tell the story of current dependence because the V across the resistor depends on current, V = i*R

I disagree with your answer for #2, especially when the resistor is connected. The work is converted to heat but we still say the work is "done" by the battery.

In #3 I would argue that the work done by the battery includes both heat lost in the circuit and electrical energy stored in the capacitor. Could be wrong, though.

#4 could be a matter of definition. To me, "the battery" includes its internal resistance. The work done by the battery is the i*V*t for the current coming out of the combination. I would answer YES to the first part and no change for the ideal battery.

No idea on #5! Good thinking on #6 - I missed it.
 
  • #3
1,384
0
I think #5 is technically incorrect. Its same like using E.M.F. when we know that there is no such 'force'.
Heat is developed in a circuit (or resistor) due to the collision of moving electrons with the ions of the conductor and not due to the difference in temperature across the resistor.
 
  • #4
Delphi51
Homework Helper
3,407
10
Most welcome! We worked really well together.
 
  • #5
1,137
0
Thank you guys for help!!!!

But for #3, wouldn't energy stored in capacitor + heat developed be equal to work done by battery.

for#4, heat will also be developed in battery if its non-ideal due to its resistance, so shouldn't the heat developed in the resistor + battery be equal to battery's work?
and for ideal one, due to no resistance, no heat will develop in battery,,,,
 

Related Threads on Quickies on electric current

  • Last Post
Replies
2
Views
989
  • Last Post
Replies
0
Views
653
  • Last Post
Replies
3
Views
831
  • Last Post
Replies
2
Views
968
  • Last Post
Replies
2
Views
875
  • Last Post
Replies
5
Views
924
  • Last Post
Replies
2
Views
1K
Replies
8
Views
5K
  • Last Post
Replies
4
Views
710
  • Last Post
Replies
4
Views
1K
Top