or
Subelement T5
Electrical principles: math for electronics; electronic principles; Ohm's Law
Section T5D
Ohm's Law: formulas and usage
What formula is used to calculate current in a circuit?
• Current (I) equals voltage (E) multiplied by resistance (R)
Current (I) equals voltage (E) divided by resistance (R)
• Current (I) equals voltage (E) added to resistance (R)
• Current (I) equals voltage (E) minus resistance (R)

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 $R$)

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Tags: ohm's law electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 5

What formula is used to calculate voltage in a circuit?
Voltage (E) equals current (I) multiplied by resistance (R)
• Voltage (E) equals current (I) divided by resistance (R)
• Voltage (E) equals current (I) added to resistance (R)
• Voltage (E) equals current (I) minus resistance (R)

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 $R$)

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Tags: ohm's law electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 5

What formula is used to calculate resistance in a circuit?
• Resistance (R) equals voltage (E) multiplied by current (I)
Resistance (R) equals voltage (E) divided by current (I)
• Resistance (R) equals voltage (E) added to current (I)
• Resistance (R) equals voltage (E) minus current (I)

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 $R$)

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Tags: ohm's law resistance electromotive force (voltage) electrical current arrl chapter 3 arrl module 5

What is the resistance of a circuit in which a current of 3 amperes flows through a resistor connected to 90 volts?
• 3 ohms
30 ohms
• 93 ohms
• 270 ohms

$E = I \times R$

$R = \frac{E}{I}$ $=$ $\frac{90}{3}$ = 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

What is the resistance in a circuit for which the applied voltage is 12 volts and the current flow is 1.5 amperes?
• 18 ohms
• 0.125 ohms
8 ohms
• 13.5 ohms

$E = I \times R$

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

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Tags: math ohm's law resistance electromotive force (voltage) electrical current arrl chapter 3 arrl module 5

What is the resistance of a circuit that draws 4 amperes from a 12-volt source?
3 ohms
• 16 ohms
• 48 ohms
• 8 Ohms

$E = I \times R$

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

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Tags: ohm's law math electromotive force (voltage) resistance arrl chapter 3 arrl module 5

What is the current flow in a circuit with an applied voltage of 120 volts and a resistance of 80 ohms?
• 9600 amperes
• 200 amperes
• 0.667 amperes
1.5 amperes

$E = I \times R$

$I = \frac{E}{R}$ $=$ $\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

What is the current flowing through a 100-ohm resistor connected across 200 volts?
• 20,000 amperes
• 0.5 amperes
2 amperes
• 100 amperes

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}

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Tags: ohm's law electrical current math resistance electromotive force (voltage) arrl chapter 3 arrl module 5

What is the current flowing through a 24-ohm resistor connected across 240 volts?
• 24,000 amperes
• 0.1 amperes
10 amperes
• 216 amperes

$E = I \times R$

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

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Tags: ohm's law math electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 5

What is the voltage across a 2-ohm resistor if a current of 0.5 amperes flows through it?
1 volt
• 0.25 volts
• 2.5 volts
• 1.5 volts

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

What is the voltage across a 10-ohm resistor if a current of 1 ampere flows through it?
• 1 volt
10 volts
• 11 volts
• 9 volts

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:

• $E = ?$
• $I = 1A$
• $R = 10Ω$

$E = I \times R = 10Ω \times 1A = 10V$

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Tags: ohm's law math electrical current electromotive force (voltage) resistance arrl chapter 3 arrl module 5

What is the voltage across a 10-ohm resistor if a current of 2 amperes flows through it?
• 8 volts
• 0.2 volts
• 12 volts
$E = I \times R = 2 \times 10 = 20$ volts