Maybe we should be asking this question "What are you trying to accomplish with this circuit, and that particular measurement?".
If you have a motor that draws 30A at 12V, then putting a 4700 ohm resistor in series with it will probably not even allow the motor to start turning! Think about the following scenario.
At t=0, when 12V is applied to an idle motor and a 4700 ohm resistor, the motor looks like a simple wire (R=0, short-circuit). That means you will have a 4700 ohm resistor with 12V across it. That will result in a maximum current of (V=I*R), 12V / 4700 ohms = 2.55 mA! If that is a 30A motor, then there isn't a snowball's chance that it'll develop enough of a field to get turning.
If you want to measure a 30A motor's current, you typically use something called a current shunt, which is simply a very low resistance strip of metal with a tight tolerance (like 0.01 ohms, with a 1% tolerance). In this case, if you had a 12V @ 30A motor and a 0.01 ohm current shunt, then 30A through 0.01 ohms = 0.3V drop across the current shunt (so the motor would see 11.7V, instead of it's desired 12V... it should still run fine). Also, the current shunt would be dissipating power (recall Power = V*I = I²R = V²/R). So (30^2)*0.01 = 9 watts. Most shunts will handle that pretty easily. Find some that look like this:
**broken link removed**
or
these.
So, what you are trying to do/measure, or was this simply an academic exercise?