Rearranging Three-Variable Multiplication and Division Formulas

Kia ora. In the previous lesson, you practised choosing a formula by identifying the required quantity, the known quantities, and the expected unit. That prevents a common error: doing correct arithmetic with the wrong equation.

This lesson handles the next step. Sometimes the formula sheet gives the relationship you need, but the required quantity is not alone on the left-hand side. You will learn a reliable way to rearrange any three-variable multiplication or division formula before substituting numbers. The main electrical examples will be Ohm’s law and power, since these are frequent exam questions.

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An equation is a balanced statement

Think of the equals sign as saying “the value on both sides is the same.” Rearranging does not mean a letter mysteriously “moves across” the equals sign. It means you perform the same operation to both sides so the equation remains balanced.

There are two rules:

1. You may swap the left and right sides of an equation.
2. You must do the same operation to both sides.

For example, begin with the power formula:

$$P=IV$$

Suppose the question asks for voltage $V$. The $I$ is multiplying $V$, so divide both sides by $I$:

$$\frac{P}{I}=\frac{IV}{I}$$

On the right, $I$ divided by $I$ equals $1$, leaving:

$$\frac{P}{I}=V$$

Swap the sides to put the required quantity first:

$$V=\frac{P}{I}$$

The key idea is simple:

• Multiplication attached to the quantity you want is undone by division.
• Division attached to the quantity you want is undone by multiplication.

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The multiplication pattern

Many basic electrical formulas have this form:

$$A=BC$$

Here, $B$ and $C$ are multiplied together.

Required quantity	What you do to both sides	Rearranged formula
$A$	Nothing; it is already isolated	$A=BC$
$B$	Divide by $C$	$B=\dfrac{A}{C}$
$C$	Divide by $B$	$C=\dfrac{A}{B}$

Do not try to memorise the table as three unrelated formulas. Instead, inspect the original formula and ask:

“What operation is currently attached to the letter I need?”

If it is being multiplied by something, divide by that something.

Power formula: $P=IV$

The power formula is a direct example of the multiplication pattern:

$$P=IV$$

Its three useful forms are:

$$P=IV$$ $$I=\frac{P}{V}$$ $$V=\frac{P}{I}$$

Worked example: find voltage

A heater takes $4.0\ \text{A}$ and uses $960\ \text{W}$. Find its supply voltage.

The question asks for voltage, so start with the original power relationship:

$$P=IV$$

Divide both sides by $I$:

$$\frac{P}{I}=\frac{IV}{I}$$ $$V=\frac{P}{I}$$

Now substitute values:

$$V=\frac{960\ \text{W}}{4.0\ \text{A}}$$ $$V=240\ \text{V}$$

Writing the rearranged formula before numbers is worthwhile in an exam. It shows the marker your method and gives you a place to spot an error before you calculate.

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The division pattern

Now consider a formula written as a fraction:

$$A=\frac{B}{C}$$

This time, $B$ is divided by $C$.

Required quantity	What you do	Rearranged formula
$A$	Nothing; it is already isolated	$A=\dfrac{B}{C}$
$B$	Multiply both sides by $C$	$B=AC$
$C$	Multiply by $C$, then divide by $A$	$C=\dfrac{B}{A}$

The most important point is that a denominator is removed by multiplying.

To isolate $B$, start with:

$$A=\frac{B}{C}$$

Multiply both sides by $C$:

$$AC=\frac{B}{C}C$$

The $C$ terms on the right cancel:

$$AC=B$$

Then write the required quantity first:

$$B=AC$$

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Ohm’s law: one relationship, three forms

For basic DC calculations, Ohm’s law is usually written:

$$V=IR$$

Some formula sheets use $E$ rather than $V$ for voltage:

$$E=IR$$

For the purposes of these basic calculations, treat $E$ and $V$ as voltage notation, but copy whichever notation the exam question or supplied formula sheet uses.

The three forms of Ohm’s law are:

$$V=IR$$ $$I=\frac{V}{R}$$ $$R=\frac{V}{I}$$

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The triangle can be useful under time pressure, particularly for Ohm’s law. Cover the required letter:

• Cover $V$: $I$ and $R$ are side by side, so multiply.
• Cover $I$: $V$ is above $R$, so divide $V$ by $R$.
• Cover $R$: $V$ is above $I$, so divide $V$ by $I$.

However, the triangle is only a mnemonic for this particular three-variable relationship. The balanced-equation method works even when a triangle does not.

Worked example: find current

A $24\ \text{V}$ supply is connected across a $6\ \Omega$ resistor. Find the circuit current.

Start from Ohm’s law:

$$V=IR$$

The required quantity is $I$, which is multiplied by $R$. Divide both sides by $R$:

$$\frac{V}{R}=\frac{IR}{R}$$ $$I=\frac{V}{R}$$

Substitute values:

$$I=\frac{24\ \text{V}}{6\ \Omega}$$ $$I=4\ \text{A}$$

Worked example: find resistance

A load has $12\ \text{V}$ across it and draws $3.0\ \text{A}$. Find its resistance.

Start with:

$$V=IR$$

The $I$ multiplies $R$, so divide both sides by $I$:

$$\frac{V}{I}=\frac{IR}{I}$$ $$R=\frac{V}{I}$$

Then substitute:

$$R=\frac{12\ \text{V}}{3.0\ \text{A}}$$ $$R=4.0\ \Omega$$

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A dependable exam routine

When the needed formula is not already arranged for the quantity asked, use this sequence:

1. Write the original formula from the formula sheet.
2. Circle the required variable in your written formula.
3. Identify whether that variable is being multiplied or divided.
4. Apply the inverse operation to both sides.
5. Cancel only identical factors, such as $R/R$ or $I/I$.
6. Write the final rearranged formula with the required quantity on the left.
7. Substitute values, including any required unit conversions.

For example, if a question asks for resistance, do not write numbers immediately. Your working should first establish:

$$V=IR$$ $$R=\frac{V}{I}$$

Only then insert the voltage and current values.

Errors to watch for

Error	Why it is wrong	Better approach
“Moving” a letter without changing the operation	It hides the balanced-equation step and often produces a wrong formula	State the inverse operation applied to both sides
From $V=IR$, writing $I=VR$	$R$ is multiplying $I$, so it must be cancelled by division	$I=\dfrac{V}{R}$
From $I=\dfrac{V}{R}$, writing $V=\dfrac{I}{R}$	$R$ is a denominator, so multiply by $R$ to remove it	$V=IR$
Substituting values before rearranging	It makes the method difficult to check and can lose method marks	Rearrange symbolically first
Relying on a triangle for every formula	Not every electrical formula fits that layout	Use inverse operations as the general method

A quick sense check can help. In the formula

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

increasing resistance while voltage stays fixed should reduce current. The resistance therefore belongs in the denominator, not the numerator.

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Key takeaways

A three-variable formula can be rearranged without guessing:

• An equation stays true when you perform the same operation on both sides.
• To undo multiplication, divide.
• To undo division, multiply.
• For $P=IV$, the alternative forms are $I=\dfrac{P}{V}$ and $V=\dfrac{P}{I}$.
• For $V=IR$, the alternative forms are $I=\dfrac{V}{R}$ and $R=\dfrac{V}{I}$.
• The Ohm’s law triangle can help recall these three forms, but the balanced-equation method is the more reliable exam skill.

Next, you will extend this method to formulas containing a squared variable. That requires one additional operation: taking a square root after isolating the squared term.