# Quickies on electric current

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## 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

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

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## Answers and Replies

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Delphi51
Homework Helper
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.

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.

Delphi51
Homework Helper
Most welcome! We worked really well together.

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,,,,