Introduction: Ohm's Balloons

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From switching a light bulb in our house On and Off to sending and receiving signals to the Mars rover, electrical circuits are ubiquitous in our day-to-day lives and Ohm's Law (V = IR) is as fundamentally important in electronics as Einstein's Relativity equation (E = mc²) is to physicists.

Ohm's law is the basis of all electrical systems. Electrical engineers use this equation to guide the design of electrical systems. Students need a strong foundation in Ohm's law while designing circuits on their own but it is observed that students often get confused between Voltage(V) and Ampere(I).

As a teacher, I have always been an advocate of Montessori education. I always try to include hands-on activities, role-plays, and games in my lesson plan to amplify the engagement, understanding, and learning of my students. So, here in my Instructables today, I'm sharing an activity that my students love doing while understanding the difference between Volt and Amps.

Supplies

  1. 4 Large balloons
  2. Electric/Insulation Tape
  3. Scissors
  4. Screwdriver or any pointed object to poke a hole on balloons

Step 1: Case I: High Voltage

Procedure:

  1. Take two balloons and fill them with water until it's about 90 percent full so that we have two balloons that are of the same size.
  2. Tape the top portion of both the balloons leaving different-sized spaces(5 mm and 10 mm, respectively) in between.
  3. Using a pointed object, poke a hole between the untaped square area.

Observations:

When we poke holes in the balloons we observe that:

  • water starts flowing out of the balloons like a fountain.
  • amount/volume of water coming out is different.
  • water is coming out at the same rate/speed.

Step 2: Case II: Low Voltage

Procedure:

  1. Take two balloons and fill them with water until it's about 40 percent full so that we have two balloons that are of the same size.
  2. Now, tape the top part of the balloon leaving a small square space(hole) of about 4-5mm in the middle.
  3. Similarly, tape the other balloon leaving a slightly bigger space(hole) of about 10 mm in the middle.
  4. Using a pointed object, poke a hole between the untaped square area.

Observations:

When we poke holes in the balloons we observe:

  • water starts flowing out of the balloons like a fountain.
  • amount of water coming out is different.
  • water is coming out at the same rate/speed.
  • the height of the water fountain coming out of the balloon is higher in the balloon with a bigger hole as compared to the balloon with a smaller hole.

Step 3: Understanding Observations

In this activity, we have assumed the size of the balloons as the Voltage and the size of the holes as Ampere.

In Case I, both balloons have around 9V, but one balloon has 10 A and the other has 5 A, ie the same amount of voltage but different amounts of current. Due to the bigger size of the balloons and hence higher pressure when we poke holes in the balloons it creates a lot of water outflow. We can see that water is actually coming out at the same rate/speed but the amount that is coming out is different.

Checking the result mathematically using Ohm's law:

Ohm's Law: V = IR or R = V/I

where V = Voltage, I = Current and R = Resistance (SI unit = Ohms)

For balloon with larger hole:

R = V/I

R = 9/10 = 0.9 Ohms


For balloon with smaller hole:

R = V/I

R = 9/5 = 1.8 Ohms

In Case II, with lower voltage or less water in the balloon, the pressure in the balloons is less, therefore water is not coming out that high or shooting up. Also, the water coming out from the balloon with a smaller hole is even less because Amp is different(less).

Checking the result mathematically using Ohm's law.

Ohm's Law: V = IR or R = V/I

where V = Voltage, I = Current and R = Resistance (SI unit = Ohms)

For balloon with a larger hole:

R = V/I

R = 2/10 = 0.2 Ohms


For balloon with smaller hole:

R = V/I

R = 2/5 = 0.4 Ohms


In both cases, we can clearly see the higher the value of R, lower the amount of water coming out of the balloon, ie R resists the flow of water from the balloon.

Step 4: Result

According to Ohm's law, the current flowing in a circuit is directly proportional to the applied potential difference(V) and inversely proportional to the resistance(R) in the circuit.

In our activity, we observed that:

  • the higher the voltage or water in the balloon, higher the pressure in the balloons.
  • the amount of water coming out of the larger hole(higher Ampere) is more because it has less resistance as compared to the balloon with a smaller hole.

Therefore, the water flowing out of the balloons is directly proportional to the size of the balloons(V) and inversely proportional to the Resistance(R) generated because of the size of the balloon(V) and the size of the hole(A) in the balloons.