Recent content by JordanHood

Homework Statement A parallel plate capacitor has a capacitance of 1.5µF with air between the plates. The capacitor is connected a 12V battery and charged. The battery is then removed. When a dielectric is placed between the plates, a potential difference of 5.0V is measured across the plates...

That gives me 625mW, which is one of the answers, so assuming that is right. Bit concerned about not using the 1800mAh at all though, possibly there to confuse you?

Ok so I set the internal resistance of the battery to 10 Ohms and work from there?

So do I need to calculate the current per second?

Is this when the source resistance and load resistance are equal?

Homework Statement A 5V battery rated at 1800mAh has a soucre resistance of 10 Ohms. What is the maximum power that can be extracted from the battery? Homework EquationsThe Attempt at a Solution I understand that had the battery not had any source resistance then the power out would have been...

ah yes, gosh that was stupid P=i^2*R not divided like I was doing

I used I=P/V to get 1x10^9/400x10^3 = 2500A. I then substituted this into Power loss= I^2/R to get 2500^2/5 = 1.25x10^6 W.

Homework Statement A 400kV line is carrying 1GW of power and experiences a line resistance of 5 ohms. What is the power loss in the cable due to resistive losses? Homework EquationsThe Attempt at a Solution You are given 4 potential answers to this question a) 3.2x10^7 W b) 78 W c) 5000 W d)...