your battery never determine the amount of current throw to the load, rather the load resistance and operating voltage of the load determine the amount of current. For two or more load resistance (Vs= Vr1+Vr2+Vr3...+Vrn) and each voltage drop (Vr1=IR1, Vr2=IR2, ..., Vrn=IRn). Therefore; for a closed loop of single load resistance, the voltage drop(Vr) is the same as source voltage/battery voltage(Vs) and current can be determined by the equation (Vr/Vs=IR or Vs=IR). this implies if the operating load resistance is 4.5ohm it is possible to draw 2 amp unless it will below or above the required amount. To draw the desired amount of current,current; the load resistance (designed resistance or designed load of an equipment) multiplied by desired current should equal with the battery voltage. Le's assume the load resistance is 4.5ohm and battery voltage is 9v, so current flow through the loop is 2amp; for the same load resistance(not be changed in any variation of voltage and current), if the battery voltage is 18v the current flow through the loop becomes 18v/4.5ohm=4amp. if I am wrong please give me feed back.
