# How to use Ohm’s law

Now you know what is Ohm’s law. If you haven’t read about Ohm’s law, just click here and read. Let us see how it affects the things connected in a circuit.

Look at the circuit given below

3V Circuit

According to Ohm’s law,

V = I x R

To calculate current flowing through the resistor, we can use this formula (remember the V, I, R triangle)

I = V / R

I = 3 / 330 = 0.009 Amps

Multiply this value with 1000 to obtain the current in a smaller unit, milli Amperes.

I = 9 milliAmps or 9mA

Now, look at this circuit. Can you spot the difference between this circuit and the last one?

9V Circuit

The only change is that the battery is now of 9 Volts.

Now, let us calculate the current flowing through the resistor.

Using same formula i.e. I = V / R

We get,

I = 9 / 330 = 0.0272 Amps

0.0272 Amps multiplied by 1000 = 27.2 mA

We can just ignore the decimal part, and take this as,

I = 27 mA

Can you now compare the results in the two cases?

When we increased voltage from 3 volts to 9 volts, the current increases from 9 mA to 27 mA.

So, we can say that when we increased voltage by 3 times (3 x 3 = 9), the current flowing through the resistor also got increased by 3 times (9 x 3 = 27).

**This analysis proves that Ohm’s Law is indeed true, which is**

Current flowing in a circuit is directly proportional to the voltage given to that circuit.

Wondering what if we change the resistance of the circuit?

Remember Ohm’s law is valid, only if, the resistance in circuit remains the same.

# Making circuits with LED!

After going through the basics of electronic circuits and resistors, we can now look at some simple circuits and understand how things happen practically.

# Circuit 1 – Voltage

We are going to connect LED in our next circuit. There is a strange thing about LEDs that they light up only if voltage above a certain voltage is given to them. That certain voltage is known as Forward Voltage.

We can see this practically.

Collect these –

- A 1.5 V AA Battery
- A 3V coin shaped battery or two 1.5 V AA batteries
- A Red LED

Now, take 1.5 V battery and connect the longer leg of LED to positive(+ve) terminal of battery and shorter leg to negative(-ve) terminal of battery like this.

Does the LED glow?

No, it won’t because red LED needs 2.2V or more than that to glow and you have supplied only 1.5V.

Now, take the 3V battery or two 1.5 V batteries and connect the same LED to it.

What do you notice now?

It glows. 😀

**NOTE 1**: The above activity dealt with voltages below 5V. This is the reason there was no need to connect any external resistor with the LED. For other circuits where you use voltages more than 5V you have to use an external resistor with the LED.

**NOTE 2**: It will be discussed further how to choose which resistor is to be connected with the LED.

# Circuit 2 – Current

Now, let us calculate current which flows through the LED when we connect it to a resistor and a battery.

Look at the circuit given below

It consists of a battery, an LED and a resistor. The resistor used here is 1000ohm or 1Kohm. Here we used a red LED and it has a forward voltage drop of 2.2V. Total voltage supplied to the circuit is 9V.

Now, how would you find out the voltage across that resistor? It’s easy:

9 (System Voltage) = 2.2 (LED) + Resistor (because the total voltage you supply is used up by the resistor and LED)

NOTE: Total Voltage Supplied = Total Voltage Used

Resistor = 9 – 2.2 = 6.8V

Voltage across resistor is 6.8V and value of resistor is 1000 ohm.

So, current will be I = V/R = 6.8/1000 = 0.0068Amps = 6.8mA

The resistor current is 6.8mA and that current is also going through the LED, so the LED current is 6.8mA.

Now let us replace 1k Ohm (1000 ohm) resistor by 680 ohm resistor.

Using same formula for current, I = 6.8/680 = 0.0100Amps = 10mA.

But why are we calculating current through LED?

Because brightness of LED depends upon the current flowing through it.

**NOTE**: The amount of current flowing through an LED is directly proportional to how bright it glows.So, the LED through which 10mA current passes glows brighter than the LED through which 6.8mA current passes.

You can see in this image how increase in current increases brightness of the LED.

More current = More brightness!

# Circuit 3 – Resistance

In this circuit, you will learn how to calculate the resistor value that should be connected with your LED.

While making a circuit,

Collect the following things-

- A 9V battery
- An LED (any colour you want, here we will use a blue LED)
- A resistor (the value you are going to find out)

Here we are going to use a blue LED. The forward voltage drop of blue LED is 3.3V(V*F*).

The voltage we are going to supply is 9V(Vs). The maximum forward current is usually around 20mA for basic LEDs. The current flowing through the LED should always be equal to or less than that current rating. So, we will limit the current to 12mA. So, I*F* for us is 12mA since we want the current to be 12mA. We have to convert current in Amperes first before using because Amperes is standard unit. So, in amperes current is 0.012.

The current to the LED can be limited by choosing a proper resistor. To calculate the value of resistor, we can use the formula below:

After calculating we get, 475 ohm. So, the resistor we should use is 475 ohm.

But 475 ohm resistor is not available in the market. So, we can approximate it’s value to 470 ohm, which is available in the market.

The change of 5 ohm will not affect much the current flow in the circuit.

So, you can use 470 ohm resistor to limit the current to 12mA in the circuit.

But why are we calculating and connecting resistor with LED?

Because it prevents your LED from burning when the amount of current that the LED can’t handle passes through it.

470 Ω Resistor

This is what happens when you don’t connect any resistor while connecting LED to the battery.