Ohms law involves 3 variables - Voltage (\(E\), for Electromotive force) Resistance (\(R\)), and Current (\(I\)). In a recent webcast, Gordon West suggested a simple method for remembering their order. He suggests that you think of \(E\) as an Eagle, \(I\) as an Igloo, and \(R\) as a Rabbit.

Any time \(E\) is on the left side, \(I\) and \(R\) are on the right side; the Igloo and the Rabbit are both on the ground, so they go next to each other (multiplication, or \(E=I\times R\)). If \(E\) is on the right side, then the Eagle is always on top (in the air). So Resistance is \(\frac{E}{I}\), because the Eagle is always above the Igloo. Current (\(I\)) is \(\frac{E}{R}\), because the eagle is always above the Rabbit. (Remember that \(\frac{E}{R}\) means \(E\) divided by \(R\))

This might help you remember the formulae for Ohms Law.

Last edited by kd7bbc. Register to edit

Tags: ohm's law electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 4

Ohms law involves 3 variables - Voltage (\(E\), for Electromotive force) Resistance (\(R\)), and Current (\(I\)). In a recent webcast, Gordon West suggested a simple method for remembering their order. He suggests that you think of \(E\) as an Eagle, \(I\) as an Igloo, and \(R\) as a Rabbit.

Any time \(E\) is on the left side, \(I\) and \(R\) are on the right side; the Igloo and the Rabbit are both on the ground, so they go next to each other (multiplication, or \(E=I\times R\)). If \(E\) is on the right side, then the Eagle is always on top (in the air). So Resistance is \(\frac{E}{I}\), because the Eagle is always above the Igloo. Current (\(I\)) is \(\frac{E}{R}\), because the eagle is always above the Rabbit. (Remember that \(\frac{E}{R}\) means \(E\) divided by \(R\))

This might help you remember the formulae for Ohms Law.

Last edited by kd7bbc. Register to edit

Tags: ohm's law electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 4

Ohms law involves 3 variables - Voltage (\(E\), for Electromotive force) Resistance (\(R\)), and Current (\(I\)). In a recent webcast, Gordon West suggested a simple method for remembering their order. He suggests that you think of \(E\) as an Eagle, \(I\) as an Igloo, and \(R\) as a Rabbit.

Any time \(E\) is on the left side, \(I\) and \(R\) are on the right side; the Igloo and the Rabbit are both on the ground, so they go next to each other (multiplication, or \(E=I\times R\)). If \(E\) is on the right side, then the Eagle is always on top (in the air). So Resistance is \(\frac{E}{I}\), because the Eagle is always above the Igloo. Current (\(I\)) is \(\frac{E}{R}\), because the eagle is always above the Rabbit. (Remember that \(\frac{E}{R}\) means \(E\) divided by \(R\))

This might help you remember the formulae for Ohms Law.

Last edited by kd7bbc. Register to edit

Tags: ohm's law resistance electromotive force (voltage) electrical current arrl chapter 3 arrl module 4

Ohms law involves 3 variables - Voltage (\(E\), for Electromotive force) Resistance (\(R\)), and Current (\(I\)). In a recent webcast, Gordon West suggested a simple method for remembering their order. He suggests that you think of \(E\) as an Eagle, \(I\) as an Igloo, and \(R\) as a Rabbit.

Any time \(E\) is on the left side, \(I\) and \(R\) are on the right side; the Igloo and the Rabbit are both on the ground, so they go next to each other (multiplication, or \(E=I\times R\)). If \(E\) is on the right side, then the Eagle is always on top (in the air). So Resistance is \(\frac{E}{I}\), because the Eagle is always above the Igloo. Current (\(I\)) is \(\frac{E}{R}\), because the eagle is always above the Rabbit. (Remember that \(\frac{E}{R}\) means \(E\) divided by \(R\))

This might help you remember the formulae for Ohms Law.

Last edited by kd7bbc. Register to edit

Tags: math ohm's law resistance electrical current electromotive force (voltage) arrl chapter 3 arrl module 4

\(E = I \times R\)

\(R = \frac{E}{I}\) \(=\) \(\frac{12}{1.5}\) = \(8\) ohms

See Ohm's Law on Wikipedia

Last edited by kd7bbc. Register to edit

Tags: math ohm's law resistance electromotive force (voltage) electrical current arrl chapter 3 arrl module 4

\(E = I \times R\)

\(R = \frac{E}{I}\) \(=\) \(\frac{12}{4}\) \(=\) \(3\) ohms

See Ohm's Law on Wikipedia

Last edited by kd7bbc. Register to edit

Tags: ohm's law math electromotive force (voltage) resistance arrl chapter 3 arrl module 4

\(E = I \times R\)

\(I = \frac{E}{R} = \frac{120}{80} = 1.5 \text{ amperes}\)

See Ohm's Law on Wikipedia

• E = Electromotive force (Volts)
• I = Intensity (Amperes)
• R = Resistance (Ohms)

Last edited by kd7bbc. Register to edit

Tags: math ohm's law electromotive force (voltage) resistance electrical current arrl chapter 3 arrl module 4

Voltage = Current \(\times\) Resistance

Which we can write as: \[E = I \times R\]

Where:

• \(E\) is the voltage applied to the circuit, in volts (V)
• \(I\) is the current flowing in the circuit, in amperes (A)
• \(R\) is the resistance in the circuit, in ohms (\(\Omega\))

In this case we have voltage and resistance and need to find current, so rearrange the equation: \begin{align} E &= I \times R\\ I &= \frac{E}{R}\\ \end{align}

Time to plug and chug! \begin{align} I &= \frac{200 \text{ V}}{100\ \Omega}\\ I &= 2 \text{ A}\\ \end{align}

---

See Ohm's Law on Wikipedia

Last edited by qubit. Register to edit

Tags: ohm's law electrical current math resistance electromotive force (voltage) arrl chapter 3 arrl module 4

\(E = I \times R\)

\(I = \frac{E}{R} = \frac{240}{24} = 10\) amperes

See Ohm's Law on Wikipedia

Last edited by kd7bbc. Register to edit

Tags: ohm's law math electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 4

This is an Ohm's law question. Remember

\(E = I \times R\)

\(E = 0.5 \text{ A} \times 2\ \Omega = 1 \text{ V}\)

See Ohm's Law on Wikipedia

Last edited by qubit. Register to edit

Tags: ohm's law math electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 4

Ohm's Law is the relationship between voltage, current, and the resistance in a DC circuit. It can be represented in the equation:

\(E = I \times R\)

Where \(E\) is the voltage/electromotive force (\(E\)), \(I\) is the current (\(A\)), and \(R\) is the resistance (Ω).

If you have any two values in the equation you can find the third:

• \(E = I \times R\)
• \(R = \frac{E}{I}\) (obtained by dividing both sides by I)
• \(I = \frac{E}{R}\) (obtained by dividing both sides by R)

So for our equation:

• \(E = ?\)
• \(I = 1A\)
• \(R = 10Ω\)

\(E = I \times R = 10Ω \times 1A = 10V\)

Last edited by kd7bbc. Register to edit

Tags: ohm's law math electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 4

\(E = I \times R = 2 \times 10 = 20\) volts

Last edited by kd7bbc. Register to edit

Tags: ohm's law math electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 4

In the animation below, the amount of voltage is indicated by the darkness of the green, and the current is represented by the "walking ant" animation.

The current is determined by the total resistance through the whole series so the current at the junction is unchanged as can be seen from the fact that all three test points show 10mA, and the dots move at the same speed through the whole series.

HINT

Current at junction in series unchanged.

(Current at junction in parallel depends.)

Voltage across two components in series depends.

(Voltage across two components in parallel is the same.)

Last edited by mike881. Register to edit

Tags: arrl chapter 3 arrl module 3

In the animation below, the amount of voltage is indicated by the darkness of the green, and the current is represented by the "walking ant" animation.

Notice that across the components in parallel the current (85mA) is divided between the components based on the value of the components.

In this case the components are resistors and their values are resistance in ohms. The dots are moving faster through the \(100 \Omega\) resistor than they are through the \(500\Omega\) resistor. The speed of the dots indicates the amount of current, indicating less current through the higher value resistors.

HINT

Current at junction in series unchanged.

(Current at junction in parallel depends.)

Voltage across two components in series depends.

(Voltage across two components in parallel is the same.)

Last edited by mike881. Register to edit

Tags: arrl chapter 3 arrl module 3

In the animation below, the amount of voltage is indicated by the darkness of the green, and the current is represented by the "walking ant" animation.

The current in the series doesn't change, but the voltage across each of the two components does change. So it is determined by the type and value of the components, in this case the resistance of the resistors.

There are many similar questions regarding current vs voltage using circuits in series or in parallel.

HINT

Current at junction in series unchanged.

(Current at junction in parallel depends.)

Voltage across two components in series depends.

(Voltage across two components in parallel is the same.)

Last edited by mike881. Register to edit

Tags: arrl chapter 3 arrl module 3

In the animation below, the amount of voltage is indicated by the darkness of the green, and the current is represented by the "walking ant" animation.

Notice that across the components in parallel the voltage (5V) is the same as the voltage source.. The values of the resistors have no effect.

HINT

Current at junction in series unchanged.

(Current at junction in parallel depends.)

Voltage across two components in series depends.

(Voltage across two components in parallel is the same.)

Last edited by mike881. Register to edit

Tags: arrl chapter 3 arrl module 3