(EAM) Making Calculations Using Ohm's Law Lesson

Making Calculations Using Ohm's Law

Current (amps) = $\frac{\text{Voltage (volts)}}{\text{Resistance(ohms)}}$

It is important to note that Current is represented by I, Voltage by V, and Resistance by R. You will need to be able to rearrange the equation to find the missing variable. You can do this algebraically or use the following triangle:

 

To begin with, let's practice using the equation with basic information.

Example Problem:

• There is a 22-ohm resistor in the heating element of a coffee percolator. It is plugged into a 110 Volt circuit. How much current passes through the heating element?

Exercise Problem

Looking For: Current	Solution:   $I=\frac{V}{R}$   $I=\frac{110V}{22ohms}$     I = 5 amps
Given: Voltage = 110 V Resistance = 22 ohms
Equation: You must rearrange the equation to find I: $I=\frac{V}{R}$

Now, let's use what we learned regarding series and parallel circuits and apply Ohm's Law to find the missing variable.  

Example Problem:

• A 5.00 Ω, a 10.0 Ω, and a 15.0 Ω resistor are connected in a series across a 90.0 V battery.  What is the current in the circuit?

Voltage, Current, Resistance triad

Looking For: Current	Solution:   $I=\frac{V}{R}$   $I=\frac{90V}{30.0\Omega}$     I = 3 amps
Given: Voltage = 90 V Resistance = 5.00 Ω, 10.0 Ω, and 15.0 Ω
Equation: You must first find the total resistance. Remember for a series circuit the current passes through each resistor. Therefore, the total resistance in the circuit is equal to the sum of the resistors. Total Resistance = 5.00 Ω + 10.0 Ω + 15.0 Ω = 30.0 Ω You must then rearrange the equation to find I: $I=\frac{V}{R}$

Example Problem: 

• Three lamps with 60 Ω , 30 Ω , and 30 Ω resistance are connected in parallel to a 90-volt circuit.

 

1. What current flows through each lamp?

Looking For: Current	Solution: $I_1=\frac{V}{R}$   $I_1=\frac{90V}{60\Omega}$   I1 = 1.5 amps ______________ $I_2=\frac{V}{R}$   $I_2=\frac{90V}{30\Omega}$   I2 = 3 amps ______________ $I_3=\frac{V}{R}$   $I_3=\frac{90V}{30\Omega}$   I3 = 3 amps
Given: Voltage = 90 V Resistance = 60 Ω, 30 Ω, and 30 Ω
Equation: For a parallel circuit, the voltage drop for the entire circuit is 90 V. Therefore, the voltage drop in each of the resistors is also 90 V. The current through each resistor can be found using the voltage drop and the resistance of each resistor. You also need to rearrange the equation to find I: $I=\frac{V}{R}$

b. What is the total current used by the three lamps?

The total current through the circuit would be the sum of the three currents in the individual resistors.

$I_T=I_1+I_2+I_3=1.5A+3.0A+3.0A=7.5A$

c. What is the total resistance of the three lamps?

Looking For: Resistance	Solution:   $R=\frac{V}{I}$   $I_2=\frac{90V}{7.5amps}$       R= 12 Ω
Given: Voltage = 90 V Current = 7.5 amps
Equation: The total resistance for the circuit can be found by using the total voltage drop and the total current.   You must rearrange the equation to find R: $I_2=\frac{V}{R}$

Practice Simulation of Ohm's Law

For a visual representation reviewing the principles of Ohm's Law, use the following simulation. Here you will see how the equation form of Ohm's law relates to a simple circuit. You will be able to adjust the voltage and resistance and see the current change according to Ohm's law. The sizes of the symbols in the equation change to match the circuit diagram.  

You want to make notes of how the current, voltage, and resistance change when you change one of the variables at a time.

[CC BY 4.0] UNLESS OTHERWISE NOTED | IMAGES: LICENSED AND USED ACCORDING TO TERMS OF SUBSCRIPTION