Say, I am using a Red LED which takes 20mA current from 2.0V battery. (I don't know if its the right specifications but I checked from this PDF I found on Google.)
So, since I will be using a 9V battery so, I will need at least a 350 ohms resistor to resist the 7V coming from the battery.
What happens to this extra 7V of power? Does it go unused, dissipates as heat, or simply gets blocked from resistor such that it draws only 2V from battery?
Best Answer
You are getting your terminology all muddled up. Firstly, V = Volts which is a measure of potential difference (voltage), not power. So saying "extra 7V of power" is incorrect. Secondly, the resistor doesn't "resist voltage", it resists the flow of current. Thirdly, a device doesn't "draw 2V from the battery", it draws a current from the battery.
So lets go through and express what is happening. You have a \$9\mathrm{V}\$ power supply (e.g. battery) which is connected to a series circuit of an LED and a resistor. The battery causes a potential difference across the circuit of \$9\mathrm{V}\$.
Now we know that the LED will drop \$2\mathrm{V}\$ over a fairly wide range of currents, so that means that the remaining \$7\mathrm{V}\$ must be dropped across the resistor. Think of it like this, the battery is like taking the lift, and you are the current. You go through the battery and take the lift from the ground floor up to the 9th floor. To get back down to the ground floor, you go through the LED which drops you down two floors, the resistor must then take you down 7 floors back to the ground.
When a voltage is dropped across a resistor, it will cause a current to flow, a relationship governed by Ohms law (\$V = IR\$). With a voltage drop of \$7\mathrm{V}\$, and a resistor of \$350\Omega\$, you will have a current of \$\frac{V}{R}=\frac{7}{350}=20\mathrm{mA}\$. This current must also be equal to the current through the LED (go back to the lift example, you are the current and there is the same number of you passing through both components).
In any device, when a current flows it causes heating as the electrons bounce around in the material. This heating is the dissipation of power. Power dissipation can be calculated as \$P = IV\$. So for your resistor you have a voltage drop of \$7\mathrm{V}\$ and a current flow of \$20\mathrm{mA}\$ which means you are dissipating \$P=7\mathrm{V}\times20\mathrm{mA}=140\mathrm{mW}\$ of power. Similarly, for the LED you have a voltage drop of \$2\mathrm{V}\$ and a current of \$20\mathrm{mA}\$, so are dissipating \$40\mathrm{mW}\$ of power (much of which is converted to light by the LED).
As a result, 78% of the power is dissipated in the resistor as explained by @Mario. The resistor is effectively acting as a regulator to reduce the voltage across the LED and limit the current flowing through it.
If you use a battery with a lower voltage, you will require less voltage to be dropped across the resistor and hence need a smaller resistor for the same current flow. For example if you use a \$3\mathrm{V}\$ battery, you would need a resistor of \$R = \frac{V}{I} = \frac{1\mathrm{V}}{20\mathrm{mA}} = 50\Omega\$. The lower resistance means less power is being dissipated and so less power is being wasted.