Hi guys I have a Minn Kota 35 its a 17lb thrust motor. Would anyone have any idea how many amps this will draw at full power? Thanks

Set the motor at the full power setting.  Disconnect the motor leads at the battery.  Measure the resistance between the leads.  Divide that number into your battery voltage (12V.)  That's how many Amps it will draw.  This includes the leads leading from the battery to the motor, which have to be taken into account.

A general rule of thumb is a 12V motor will draw 1 amp of current per 1lb of thrust.

here is some info everyone should read.

http://www.minnkotamotors.com/support/faq.asp?pg=general

Set the motor at the full power setting. Disconnect the motor leads at the battery. Measure the resistance between the leads. Divide that number into your battery voltage (12V.) That's how many Amps it will draw. This includes the leads leading from the battery to the motor, which have to be taken into account.

There are a number of flaws in this suggestion. It will not give you the correct result. It will perhaps give you a rough guide to maximum possible current under stalled rotor conditions but it will not answer the question of amps drawn at full power. So you could use it for sizing the cable and circuit breaker but it is useless to estimate running time at full load using battery capacity.

Technical details for those who aren't quite confused sufficiently:-

The faster the motor rotates, the less current it will draw due to "back emf" generated in the rotor. This is the voltage generated by the rotor that gets subtracted from the 12 volt supply. At full speed but no load (in air) the back emf voltage could be as much as 90% or more of the supply voltage so instead of 12 volts available to flow through the Resistance it may be less than 1 volt. As you place a load on the motor, it slows down, produces less back emf, so more voltage is available to flow through the windings and it draws more current to provide power for the load you applied.

This is further complicated depending on how the speed control works.  In the case of electronic controls, there will be no voltage supplied to the electronics when measuring resistance so the motor will never turn on and you will get an artificially high result.

Set the motor at the full power setting. Disconnect the motor leads at the battery. Measure the resistance between the leads. Divide that number into your battery voltage (12V.) That's how many Amps it will draw. This includes the leads leading from the battery to the motor, which have to be taken into account.

There are a number of flaws in this suggestion. It will not give you the correct result. It will perhaps give you a rough guide to maximum possible current under stalled rotor conditions but it will not answer the question of amps drawn at full power. So you could use it for sizing the cable and circuit breaker but it is useless to estimate running time at full load using battery capacity.

Technical details for those who aren't quite confused sufficiently:-

The faster the motor rotates, the less current it will draw due to "back emf" generated in the rotor. This is the voltage generated by the rotor that gets subtracted from the 12 volt supply. At full speed but no load (in air) the back emf voltage could be as much as 90% or more of the supply voltage so instead of 12 volts available to flow through the Resistance it may be less than 1 volt. As you place a load on the motor, it slows down, produces less back emf, so more voltage is available to flow through the windings and it draws more current to provide power for the load you applied.

This is further complicated depending on how the speed control works. In the case of electronic controls, there will be no voltage supplied to the electronics when measuring resistance so the motor will never turn on and you will get an artificially high result.

WOW!! Now that is pure electrical genius talking.

Ann-Marie, I forgot about the back emf.  Not surprising, since the last time I studied any electronics was forty years ago.  :-[