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.
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Tags: ohm's law electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 5
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 5
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 5
\(E = I \times R\)
\(R = \frac{E}{I}\) \(=\) \(\frac{90}{3}\) = 30 ohms
\(R\) = Resistance (ohms), \(E\) = Voltage (volts), \(I\) = Current (amperes)
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Tags: math ohm's law resistance electrical current electromotive force (voltage) arrl chapter 3 arrl module 5
\(E = I \times R\)
\(R = \frac{E}{I}\) \(=\) \(\frac{12}{1.5}\) = \(8\) ohms
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Tags: math ohm's law resistance electromotive force (voltage) electrical current arrl chapter 3 arrl module 5
\(E = I \times R\)
\(R = \frac{E}{I}\) \(=\) \(\frac{12}{4}\) \(=\) \(3\) ohms
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Tags: ohm's law math electromotive force (voltage) resistance arrl chapter 3 arrl module 5
\(E = I \times R\)
\(I = \frac{E}{R}\) \(=\) $\frac{120}{80} $\(= 1.5\) amperes
E = Electromotive force (Volts) I = Intensity (Amperes) R = Resistance (Ohms)
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Tags: math ohm's law electromotive force (voltage) resistance electrical current arrl chapter 3 arrl module 5
Voltage = Current \(\times\) Resistance
Which we can write as: \[E = I \times R\]
Where:
Time to plug and chug! \begin{align} I &= \frac{200 \text{ V}}{100\ \Omega}\\ I &= 2 \text{ A}\\ \end{align}
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Tags: ohm's law electrical current math resistance electromotive force (voltage) arrl chapter 3 arrl module 5
\(E = I \times R\)
\(I = \frac{E}{R} = \frac{240}{24} = 10\) amperes
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Tags: ohm's law math electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 5
This is an Ohm's law question. Remember
\(E = I \times R\)
\(E = 0.5 \text{ A} \times 2\ \Omega = 1 \text{ V}\)
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Tags: ohm's law math electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 5
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:
So for our equation:
\(E = I \times R = 10Ω \times 1A = 10V\)
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Tags: ohm's law math electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 5
\(E = I \times R = 2 \times 10 = 20\) volts
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Tags: ohm's law math electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 5