OHM'S LAW

In this simulation we are going to test Ohm's law V = I · R.

To do this we are going to use the Ohm's equation, solving for the current I = V/R, since it is easier to check with our circuit how the current is affected if we change any of the other two variables.
In that equation, we see that V is in the numerator. If we keep R constant and increase the voltage provided by the battery, I will increase. We say that they are directly proportional quantities. On the other hand, R is in the denominator. So if we increase the resistance of the circuit, maintaining the voltage constant, the current will decrease, since we are dividing V in the equation by a larger number.
Let's do two experiments to check this.

In our first experiment, we have a circuit with a light bulb with a resistance of 10 ohms and a battery that provides a voltage of 10 V. The current would be I = V/R = 1. Let's check it with the ammeter. It does indeed read 1 A.

Next we are going to increase the battery voltage to 20 V. According to Ohm's law, how do you think the current will vary?

As you can see, the bulb is now brighter. As the battery voltage increases, the energy of the electrons increases and they can move faster. More electrons per second will pass through the circuit and the intensity increases. According to Ohm's law, we would have that I = V/R = 20/10 = 2 A. By measuring I with the ammeter we verify it. If we decrease the voltage, for example to 5 V, what do you think would happen to the intensity? If the energy that the electrons receive is decreased, the speed at which they move decreases as well, and therefore the intensity. If we calculate it: I = 5/10 = 0.5 A, and when we measure it with the ammeter we obtain the same result.

For the second experience we are maintaining the voltage of the battery at 10 V and what we are going to change is the value of the resistance. If we change it from 10 to 20 ohms, how do you think the value of the current will vary? When the value of the resistance is increased, the electrons will find more difficulty to pass through the bulb and will go more slowly. We see that the bulb lights less. If we calculate it with Ohm's law we get I = V/R = 10/20 = 0.5 A, the same as with the ammeter.
Now we are going to put a value to the resistance of 5 ohms. As the resistance decreases, the electrons have less trouble passing through it and more electrons per second will pass through the bulb. What will the ammeter measure? I = 10/5 = 2 A.