Semiconductor devices do not work in the ashion ui have in mind!!! To save the led from burning out, u have to limit the current using the resistor, the excess voltage will be sunk on the resistor automatically, the resistance value will determine the current that flows through the circuit whenever a voltage above 2v is applied, since the led will act as a short circuitted path once it has the voltage across its terminals. The actual current handling capability will be given in the datasheet of the led,
however, for 9V, a resistor of 8.2k would bring the current down to about 1mA which, if the LED is like the one usually found in appliances for indications, should be enough to drive the led yet keep it from blowing. U can experiment by using a potentiometer to reduce the resistance till u get the desired brightness & dont smell any odor that shouldnt be coming from circuits.
An 8.2k resistor with 5% tolerance would have the bands from left to right:
Grey Red Red Golden
The tolerance band is more spaced out than the bands depicting the actual value.