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Subelement T5
Electrical principles: math for electronics; electronic principles; Ohm's Law
Section T5D
Ohm's Law: formulas and usage; components in series and parallel
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 4

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 4

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 4

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 4

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 4

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 4

What is the current 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 4

What is the current 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 4

What is the current 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 4

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 4

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 4

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
• 20 volts

$E = I \times R = 2 \times 10 = 20$ volts

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

What happens to current at the junction of two components in series?
• It divides equally between them
• It is unchanged
• It divides based on the on the value of the components
• The current in the second component is zero

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.

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What happens to current at the junction of two components in parallel?
• It divides between them dependent on the value of the components
• It is the same in both components
• Its value doubles
• Its value is halved

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.

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What is the voltage across each of two components in series with a voltage source?
• The same voltage as the source
• Half the source voltage
• It is determined by the type and value of the components
• Twice the source voltage

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.

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What is the voltage across each of two components in parallel with a voltage source?
• It is determined by the type and value of the components
• Half the source voltage
• Twice the source voltage
• The same voltage as the source

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.

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Tags: arrl chapter 3 arrl module 3