Finding current using EMF & internal resistance What exactly is the difference between internal resistance and resistance?

This came up in the context of a homework problem I have been given:

The circuit shown in the figure contains two batteries, each with an
  emf and an internal resistance, and two resistors.


I need to find the magnitude of the current in this circuit.
I believe I'm supposed to be using the equation:
$I = \frac E{(R + r)}$
where $E = 24.0V, R = 17 \Omega$.
So how do I identify the internal resistance.
 A: 
what exactly is the difference between internal resistance and
resistance?

An ideal voltage source can provide unlimited current to an external circuit such that the source voltage is maintained.
But, there are no ideal voltage sources, i.e., all real voltage sources have some maximum current delivered into a short circuit.
This is modelled by placing a resistor in series with an ideal voltage source and this resistance is the "internal resistance" or "source resistance".
Clearly, the maximum current from such a source is:
$i_{max} = v_{oc} / r_s$
where $v_{oc}$ is the "open circuit voltage", i.e., the voltage across the source when the source provides no current.
This is related to the concepts of Thevenin and Norton equivalent circuits.
(Note: the original link is broken - now linking to the most recent archive)
A: Internal resistance usually refers to the resistance associated with batteries. 
So, $1.6$ $\Omega$ and $1.4$ $\Omega$ are related to $16.0 V$ and $8.0 V$ batteries respectively. 
It is different from resistance in the sense that resistance is related to resistors in the circuit.
$R = 5.0 + 9.0 = 14 \Omega$
$r = 1.6 + 1.4 = 3 \Omega$
Total Resistance  = $R + r = 17 \Omega$.
Then use $I = \mathcal E / (R+r)$
$I = (16.0 - 8.0)/ (17) = 8/17 A$.
Direction of current is from positive polarity of $16 V$ to its negative polarity.
Why $\mathcal E = (16.0 - 8.0)$? It follows from the Kirchhoff's Circuit Laws.
For your question "How would I find the internal resistance", the way question was presented, internal resistance $r$ is figured out from the yellow-coloured part in the figure. That part contains both battery symbol and resistance symbol. They pertain to batteries.
A: Hey you're getting wrong, the equivalent emf(voltage) of system will be 16V-8V because both have opposite poles facing each other, so their will be net flor of current (-ve to +ve) according to the cell of greater emf(16V cell).Then your total resistance is $$5+1.6+1.4 = 8 \Omega$$(all are in series) . $$I(Current) = E(e.m.f or Voltage)/R(Resistance) = 8/8 = 1 A(Ampere)$$
