Applying Kirchhoff’s Voltage Law to Single-Loop Circuits

Hello again. In the previous lesson, you used the series-circuit rule that the individual resistor voltage drops add to the supply voltage. Kirchhoff’s Voltage Law, usually shortened to KVL, is the formal version of that rule: it gives you a reliable way to write, solve, and check voltages around a complete circuit loop.

For your exam, KVL is particularly useful when one voltage is missing, when you need to prove that a set of calculated values is correct, or when you need to find circuit current by writing one loop equation. The key new skill is handling the signs of voltage rises and drops consistently.

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The idea behind Kirchhoff’s Voltage Law

A loop is any closed path that starts at one point in a circuit, travels through components, and returns to the same point.

KVL states:

$$\text{The algebraic sum of all voltage changes around a closed loop is zero.}$$

In exam form, write:

$$\sum V=0$$

“Algebraic” means the signs matter. A voltage rise is written as positive; a voltage drop is written as negative, provided you keep the same chosen direction all the way around the loop.

The physical reason is straightforward: after travelling around a complete loop and returning to where you began, you cannot be at a different electrical potential from the starting point. Any energy per unit charge supplied by the source has been accounted for by energy used in the loads.

For a common single-source resistor circuit, KVL can be written as:

$$V_S-V_1-V_2-V_3=0$$

This is equivalent to:

$$V_S=V_1+V_2+V_3$$

So the familiar series rule from the last lesson is not a separate rule: it is a simple application of KVL.

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A dependable sign convention

You may choose to travel around a loop clockwise or anticlockwise. The direction itself does not matter. What matters is that you use one direction consistently.

Use this rule every time:

Component crossed while travelling around the loop	Voltage change written in KVL
Source from $-$ terminal to $+$ terminal	$+V_S$
Source from $+$ terminal to $-$ terminal	$-V_S$
Resistor from $+$ side to $-$ side	$-V_R$ or $-IR$
Resistor from $-$ side to $+$ side	$+V_R$ or $+IR$

The resistor polarity normally follows the passive sign convention: conventional current enters the resistor’s $+$ side and leaves its $-$ side. When you travel with conventional current through a resistor, you travel from $+$ to $-$, so it is a voltage drop.

A useful memory prompt is:

• Negative to positive: voltage rises, so write $+$.
• Positive to negative: voltage falls, so write $-$.

Do not decide a sign merely because a component is a resistor or a battery. Decide it from the polarity you cross while following your selected route.

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Reading a KVL loop on a series circuit

The circuit below contains a $45\mathrm{V}$ source and three series resistors. The current is $2.0\mathrm{mA}$, flowing through all three resistors.

Choose a path that follows the marked conventional-current direction. You cross the source from $-$ to $+$, then cross each resistor from $+$ to $-$.

Therefore, the KVL equation is:

$$+V_S-V_{R1}-V_{R2}-V_{R3}=0$$

Substitute the source voltage:

$$+45\mathrm{V}-V_{R1}-V_{R2}-V_{R3}=0$$

To calculate the resistor drops, first find current. KVL lets you write the equation using $V=IR$:

$$+45\mathrm{V}-I(5\mathrm{k\Omega })-I(10\mathrm{k\Omega })-I(7.5\mathrm{k\Omega })=0$$

Group the resistance terms:

$$45\mathrm{V}-I(22.5\mathrm{k\Omega })=0$$

Move the current term to the other side:

$$45\mathrm{V}=I(22.5\mathrm{k\Omega })$$

Now divide by $22.5\mathrm{k\Omega }$:

$$I=\frac{45\mathrm{V}}{22.5\mathrm{k\Omega }}$$ $$I=2.0\mathrm{mA}$$

This is the same answer you would obtain by first calculating total resistance and then applying Ohm’s law. KVL shows why that method works.

Now find each voltage drop:

$$V_{R1}=I{R}_1$$ $$V_{R1}=(2.0\mathrm{mA})(5\mathrm{k\Omega })$$ $$V_{R1}=10\mathrm{V}$$ $$V_{R2}=I{R}_2$$ $$V_{R2}=(2.0\mathrm{mA})(10\mathrm{k\Omega })$$ $$V_{R2}=20\mathrm{V}$$ $$V_{R3}=I{R}_3$$ $$V_{R3}=(2.0\mathrm{mA})(7.5\mathrm{k\Omega })$$ $$V_{R3}=15\mathrm{V}$$

Finally, verify with KVL:

$$+45\mathrm{V}-10\mathrm{V}-20\mathrm{V}-15\mathrm{V}=0$$ $$0\mathrm{V}=0\mathrm{V}$$

That final line confirms the values are consistent.

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Solving for one missing voltage

A common written exam question gives the supply voltage and most of the voltage drops, then asks for the remaining drop.

Question: A $24\mathrm{V}$ source supplies three series loads. The measured voltage drops are:

$$V_1=5.5\mathrm{V}$$ $$V_3=9.0\mathrm{V}$$

Calculate $V_2$.

Step 1: Write KVL before inserting numbers

Follow the current direction through the resistors:

$$+V_S-V_1-V_2-V_3=0$$

Step 2: Substitute values

$$+24\mathrm{V}-5.5\mathrm{V}-V_2-9.0\mathrm{V}=0$$

Step 3: Combine the known terms

$$9.5\mathrm{V}-V_2=0$$

Step 4: Isolate the unknown

$$-V_2=-9.5\mathrm{V}$$ $$V_2=9.5\mathrm{V}$$

The voltage across the second load is:

$$\boxed{V_2=9.5\mathrm{V}}$$

For a basic series circuit with one supply, you could calculate it more quickly as:

$$V_2=V_S-V_1-V_3$$

However, writing the full KVL equation first is safer in an exam because it shows your method and makes the signs visible.

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Verifying a set of values

KVL is also a fast error check. Suppose you are given:

$$V_S=24\mathrm{V}$$ $$V_1=5.5\mathrm{V}$$ $$V_2=9.5\mathrm{V}$$ $$V_3=9.0\mathrm{V}$$

Write the signed KVL check:

$$+24\mathrm{V}-5.5\mathrm{V}-9.5\mathrm{V}-9.0\mathrm{V}=0$$ $$24\mathrm{V}-24\mathrm{V}=0$$ $$0\mathrm{V}=0\mathrm{V}$$

The values pass the KVL check.

If you instead obtained:

$$+24\mathrm{V}-5.5\mathrm{V}-8.5\mathrm{V}-9.0\mathrm{V}=1.0\mathrm{V}$$

then the values do not satisfy KVL. In an ideal exam circuit, that means at least one value, calculation, sign, or stated polarity is wrong.

In practical testing, a small mismatch can arise from meter accuracy, source variation, resistor tolerance, or rounded readings. But only accept a tolerance explanation if the question gives enough practical context. For a calculation question, the expected answer should normally balance exactly, apart from rounding.

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Avoiding the common KVL mistakes

Mixing positive magnitudes with signed terms

A resistor drop can be described as “$10\mathrm{V}$” in magnitude, but in a KVL equation it may appear as either $+10\mathrm{V}$ or $-10\mathrm{V}$, depending on the direction you travel across it.

For the $45\mathrm{V}$ circuit, travelling with current gives:

$$+45\mathrm{V}-10\mathrm{V}-20\mathrm{V}-15\mathrm{V}=0$$

Travelling the opposite way is also correct:

$$-45\mathrm{V}+10\mathrm{V}+20\mathrm{V}+15\mathrm{V}=0$$

Both equations say exactly the same thing. The mistake is reversing direction halfway through an equation.

Forgetting the source voltage

A loop equation must include every component crossed on the selected route, including the source. If you write only the resistor drops, they cannot add to zero unless the circuit has no applied voltage.

Using resistance values directly in a voltage sum

KVL sums voltages, not resistances. If only resistor values are given, convert each resistor term to voltage using:

$$V_R=IR$$

For a one-loop circuit, this gives a useful general form:

$$V_S-I{R}_1-I{R}_2-I{R}_3=0$$

You can then factor out $I$:

$$V_S-I(R_1+R_2+R_3)=0$$

This leads directly to:

$$I=\frac{V_S}{R_1+R_2+R_3}$$

Losing units

Every term in a KVL equation must be in volts. Write $\mathrm{V}$ on your working, especially when checking a final answer.

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An exam routine for a single-loop KVL question

Use this sequence when you see a circuit with one closed loop:

1. Mark or identify the source polarity and current direction. If the direction is not given, choose one for the purpose of writing the equation.
2. Choose one travel direction around the entire loop.
3. Write each voltage change with a sign. Crossing from $-$ to $+$ is positive; crossing from $+$ to $-$ is negative.
4. Set the signed total equal to zero.
5. Use $V=IR$ when a resistor voltage must be expressed in terms of current and resistance.
6. Rearrange one line at a time, keeping units beside values.
7. Substitute the answer back into the original KVL equation and confirm that the result is $0\mathrm{V}$.

For method marks, a strong final layout is:

$$\sum V=0$$ $$+V_S-V_1-V_2-V_3=0$$ $$\text{Substitute known values}$$ $$\text{Solve for the unknown}$$ $$\text{Verify that the signed voltage sum is }0\mathrm{V}$$

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Key takeaways

Kirchhoff’s Voltage Law states:

$$\sum V=0$$

Around a complete loop:

• A crossing from $-$ to $+$ is a voltage rise and is written positive.
• A crossing from $+$ to $-$ is a voltage drop and is written negative.
• In a one-source series circuit, supply voltage equals the total of the resistor voltage drops.
• KVL can find an unknown voltage, calculate loop current when combined with Ohm’s law, and check whether calculated or measured values are consistent.

The main habit is simple: choose one direction around the loop, assign signs from the polarity you cross, and do not change direction halfway through.

Next, you will use the same series-voltage relationship in a faster form: the voltage-divider relationship, which lets you calculate a resistor’s voltage drop directly from its share of total series resistance.