circuits and internal resistance?

Question

A battery with an internal resistance is connected to two resistors in series.

pic: http://i182.photobucket.com/albums/x255/manphysics/quiz9number2.jpg

A) What is the emf of the battery?

B)What is the magnitude of the potential difference across the terminals y and x of the battery?

C) What power is dissipated by the internal resistance of the battery?


help please :)

Answer 1

a) 9.315V (a nine volt battery)
b)8.586V
c)0.19683 watt

These should be pretty obvious... using the formulas:
V=IR and P=(I^2)R

in case you still don't get it (And to make sure u chose this as best answer)

A) EMF is the total voltage of the batt not counting internal drop in potential. That is equal to the voltage drop of all circuit elements. Since the resistors are in series, add them up and multiply by the current.

B) The potential of the battery is the EMF - drop across the internal resistor. So find the voltage drop across the resistor (current*resistance) and subtract from part a.

C)Power is the voltage multiplied by the current. (P=IE). Since the voltage is given by current*resistance (V=IR), substituting you get current squared *resistance P=(I^2)R.
So use that to get the power of the resistor.

Answer 2

A) total resistance of the circuit: Rt = 2.7 + 20.7 + 11.1

EMF = I x R (ohm's law)

So EMF = 0.27 * Rt

B) V = 0.27 * (20.7 + 11.1)

C) P = V * I = I^2R = (0.27)^2 (2.7)