A battery labled 12V supplies 1.90A to a 6.00ohm resistor.?

Question

What is the terminal voltage of the battery.
What is its internal resistance.

I am having a crappy time with these problems if any one could help that would be great!

Answer 1

The first three answers are correct if you are modeling the battery as an ideal voltage source of 12.0V with a series resistance (internal resistance). However, a standard 12V automotive lead-acid battery actually has a nominal voltage of 12.6V. That's probably not relevant to your homework assignment, though.

Answer 2

The current must be the same through both the internal and external resistances. The voltage drop acorss the internal resistance is
Vl = 12 - 1.9*6 = 0.6
Subtracting this from the nominal 12 V gives terminal voltage.
Vt = 12 - 0.6 = 11.4 V
Dividing internal voltage drop by the current yields
Ri = 0.6/1.9 = 0.316 Ω

Answer 3

Sketch two resistors in series( one being the internal resistance of the battery), being supplied by a 12 volt source. Put the internal resistance at the positive terminal of the battery. The six ohm going to ground(negative).
The terminal voltage of the battery is;
E = IR = (1.9)(6) = 11.4 volts
The internal resistance of the battery is:
R = E/i = 12/1.9 = 6.316 ohms
6.316 - 6 = .316 ohms

Answer 4

V = I*R [Ohms Law]

R [total] = V/I = 12/1.9 = 6.316 ohms

Internal resistance = 6.316 - 6.00 = 0.316 ohms

Terminal voltage = 1.9 * 6.00 = 11.4 volts