Isolating a Squared Variable in Electrical Formulas

Kia ora. In the previous lesson, you used inverse operations to rearrange formulas such as $V=IR$ and $P=IV$. The reliable rule was: keep the equation balanced, and undo multiplication with division or undo division with multiplication.

This lesson adds one operation that appears often on electrical formula sheets: a square. You will learn to isolate an unsquared variable from formulas such as $P=I^{2}R$ and $P=\frac{V^2}{R}$. These are common power relationships, and the method is also useful wherever an electrical formula contains a squared quantity.

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A square needs a square root

A squared term means a quantity multiplied by itself:

$$I^2=I\times I$$

For example:

$$5^2=25$$

The inverse operation of squaring is taking a square root:

$$\sqrt{25}=5$$

So, if an equation has:

$$I^2=25$$

and the question asks for $I$, take the square root of both sides:

$$\sqrt{I^2}=\sqrt{25}$$ $$I=5$$

For the electrical formulas used in basic calculations, current, voltage, and resistance are usually treated as positive magnitudes. Therefore, use the positive square root unless the question specifically deals with a chosen polarity or direction convention.

This differs slightly from a pure algebra question, where $x^2=25$ can have two mathematical solutions, $x=5$ and $x=-5$. In a power formula such as $P=I^{2}R$, both positive and negative current directions produce the same heating power. When an exam asks for the current magnitude, report the positive value.

The key rule is:

To remove a square from the required variable, first isolate the squared term. Then take the square root of both sides.

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The order matters: clear other operations first

Consider this power relationship:

$$P=I^{2}R$$

Suppose you need to find current $I$. It may be tempting to take the square root immediately, but $I^2$ is still multiplied by $R$. First isolate $I^2$.

Since $R$ multiplies $I^2$, divide both sides by $R$:

$$\frac{P}{R}=\frac{I^{2}R}{R}$$ $$\frac{P}{R}=I^2$$

Now the squared term is alone. Take the square root of both sides:

$$\sqrt{\frac{P}{R}}=\sqrt{I^2}$$ $$I=\sqrt{\frac{P}{R}}$$

This is the familiar current form of the power equation:

$$I=\sqrt{\frac{P}{R}}$$

Notice that the square applies to current only in $I^{2}R$. It does not mean that both $I$ and $R$ are squared.

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Pattern 1: $P=I^{2}R$

The formula

$$P=I^{2}R$$

is used for the power dissipated in a resistor when current and resistance are known. You may also see it when power and resistance are known and you need current.

Rearranging for current

Start with:

$$P=I^{2}R$$

Divide by $R$:

$$\frac{P}{R}=I^2$$

Take the square root:

$$I=\sqrt{\frac{P}{R}}$$

The square root must cover the whole fraction:

$$I=\sqrt{\frac{P}{R}}$$

not

$$I=\frac{\sqrt{P}}{R}$$

Those expressions are not generally equal.

Exam-style setup

A resistor dissipates power $P$ and has resistance $R$. Write a formula for current.

A clear method-mark answer is:

$$P=I^{2}R$$ $$\frac{P}{R}=I^2$$ $$I=\sqrt{\frac{P}{R}}$$

Do the symbolic rearrangement before entering numbers into the calculator. That makes it easier to identify whether the square root should apply to a fraction, a product, or something else.

Rearranging the same formula for resistance

If the required quantity is $R$, no square root is needed because $R$ is not squared:

$$P=I^{2}R$$

Divide both sides by $I^2$:

$$R=\frac{P}{I^2}$$

Be careful to square the current before dividing:

$$R=\frac{P}{I^2}$$

not

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

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Pattern 2: $P=\frac{V^2}{R}$

Another power relationship is:

$$P=\frac{V^2}{R}$$

This form is particularly useful when voltage and resistance are involved.

Rearranging for voltage

Suppose the question asks for voltage $V$.

Start with:

$$P=\frac{V^2}{R}$$

The $R$ is a denominator, so multiply both sides by $R$:

$$PR=\frac{V^2}{R}R$$ $$PR=V^2$$

Now take the square root of both sides:

$$\sqrt{PR}=\sqrt{V^2}$$ $$V=\sqrt{PR}$$

So the rearranged formula is:

$$V=\sqrt{PR}$$

The brackets are understood by the square-root bar, but when you use a calculator, enter the product as one complete expression:

$$\sqrt{P\times R}$$

Do not take the square root of $P$ and then multiply by $R$.

Starting from a different arrangement

A formula sheet may instead show:

$$R=\frac{V^2}{P}$$

This is the same relationship, just rearranged. If you need $V$, use the same method:

$$R=\frac{V^2}{P}$$

Multiply both sides by $P$:

$$PR=V^2$$

Take the square root:

$$V=\sqrt{PR}$$

The starting layout can look different, but the operations are identical:

1. Remove the denominator.
2. Isolate the squared term.
3. Take the square root.

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The formula wheel uses $E$ instead of $V$ for voltage. In many electrical texts, $E$ means electromotive force, but for these basic formula-sheet calculations it is treated as voltage:

$$E=V$$

Copy the symbol used in the exam question or supplied formula sheet, but recognise that the rearrangement method is unchanged.

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A general squared-variable template

Many of these formulas fit a general pattern:

$$A=\frac{B^2}{C}$$

If the question asks for $B$, proceed in two stages.

First, multiply both sides by $C$:

$$AC=B^2$$

Then take the square root:

$$B=\sqrt{AC}$$

Similarly, for the pattern:

$$A=B^{2}C$$

first divide by $C$:

$$\frac{A}{C}=B^2$$

then take the square root:

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

Original pattern	Required variable	Rearranged form
$A=B^{2}C$	$B$	$B=\sqrt{\frac{A}{C}}$
$A=\frac{B^2}{C}$	$B$	$B=\sqrt{AC}$
$A=B^{2}C$	$C$	$C=\frac{A}{B^2}$
$A=\frac{B^2}{C}$	$C$	$C=\frac{B^2}{A}$

This table is useful for revision, but in an exam the safer approach is still to write the supplied formula and apply inverse operations one line at a time.

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Use units as a quick formula check

Even before calculating, units can help you spot a misplaced square root.

For current:

$$I=\sqrt{\frac{P}{R}}$$

Power has unit watts and resistance has unit ohms. Since:

$$\frac{\text{W}}{\Omega }={\text{A}}^2$$

taking the square root gives amperes:

$$\sqrt{\frac{\text{W}}{\Omega }}=\text{A}$$

For voltage:

$$V=\sqrt{PR}$$

The units become:

$$\sqrt{\text{W}\Omega }=\text{V}$$

You do not need to show this unit algebra in every exam answer. But if you accidentally write:

$$V=\frac{\sqrt{P}}{R}$$

the units will not simplify to volts, which is a warning that the rearrangement is wrong.

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Common errors with squared terms

Error	What goes wrong	Correct approach
Taking the square root before isolating the squared variable	The other factors remain mixed into the expression	Isolate $I^2$ or $V^2$ first, then take the root
From $P=I^{2}R$, writing $I=\sqrt{PR}$	Resistance should reduce current for a fixed power, not increase it	$I=\sqrt{\frac{P}{R}}$
From $P=\frac{V^2}{R}$, writing $V=\sqrt{\frac{P}{R}}$	The resistance was handled in the wrong direction	$V=\sqrt{PR}$
Forgetting the square	Writing $R=\frac{P}{I}$ from $P=I^{2}R$	Preserve the exponent: $R=\frac{P}{I^2}$
Rooting only one part of an expression	Calculator entry no longer matches the formula	Use brackets or the full root bar: $\sqrt{P/R}$ or $\sqrt{PR}$

A useful physical reasonableness check is this:

• At fixed resistance, more power requires more current and more voltage.
• At fixed power, greater resistance gives less current but more voltage.

The correct formulas show that behaviour:

$$I=\sqrt{\frac{P}{R}}$$ $$V=\sqrt{PR}$$

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A short exam routine

When the required variable has a square on it, use this routine:

1. Write the original formula from the formula sheet.
2. Identify the required variable and confirm that it is squared.
3. Clear any denominator first by multiplying both sides.
4. Divide away any factor multiplying the squared term.
5. Stop when the form is $x^2=\text{expression}$.
6. Take the square root of both sides.
7. Put the final formula in the form $x=\text{expression}$.
8. Only then substitute values and use the calculator.

For example, if asked for voltage from $P=\frac{V^2}{R}$, your symbolic working should end as:

$$P=\frac{V^2}{R}$$ $$PR=V^2$$ $$V=\sqrt{PR}$$

That is enough structure to earn method marks and makes calculator use much safer.

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Key takeaways

A square is undone by a square root, but only after the squared term has been isolated.

The two high-value electrical rearrangements are:

$$P=I^{2}R$$ $$I=\sqrt{\frac{P}{R}}$$

and

$$P=\frac{V^2}{R}$$ $$V=\sqrt{PR}$$

Keep the full product or fraction inside the square root, retain squared terms such as $I^2$ when rearranging for another variable, and use the expected unit or physical behaviour as a final check.

Next, you will combine formula rearrangement with careful substitution and calculator entry, including fractions, squares, and square roots.