Series Circuit Analysis: Resistance, Current, and Voltage Drops

Hello, and welcome to the series-circuit revision module. In an exam, a series-circuit calculation is usually less about difficult mathematics than about keeping total values and individual component values separate. This lesson gives you a repeatable method for finding total resistance, circuit current, and each resistor’s voltage drop, with clear working for method marks.

By the end, you should be able to look at a single-loop resistor circuit and confidently decide which formula to use first, what to calculate next, and how to check whether your final answers make sense.

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Recognising a series circuit

A series circuit has only one continuous path for current. If the circuit has no branches or junctions where current can split, its components are in series.

That one-path arrangement gives three rules that you should treat as an exam checklist:

$$R_T=R_1+R_2+R_3+\dots$$ $$I_T=I_1=I_2=I_3=\dots$$ $$V_T=V_1+V_2+V_3+\dots$$

Where:

• $R$ is resistance, measured in ohms $\Omega$
• $I$ is current, measured in amperes $\mathrm{A}$
• $V$ is voltage, measured in volts $\mathrm{V}$
• the subscript $T$ means total
• the subscripts $1$, $2$, and $3$ refer to individual resistors

The rules mean:

1. Resistance adds. Each resistor makes it harder for current to move around the only available path.
2. Current is the same everywhere. Charge cannot choose another route or disappear between resistors.
3. Voltage drops add. The source voltage is shared among the resistors.

A useful physical check follows from these rules: in the same series circuit, the resistor with the largest resistance has the largest voltage drop, because every resistor carries the same current.

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The calculation order: total first, then individual values

For the usual exam question where the supply voltage and resistor values are given, follow this fixed order:

1. Add the resistors to find total resistance.
2. Use total voltage and total resistance to find circuit current.
3. State that this current is the current through every series resistor.
4. Use $V=IR$ for each resistor to find its voltage drop.
5. Add the voltage drops to check that they equal the supply voltage.

The order matters. You cannot correctly calculate current by dividing the total supply voltage by just one individual resistor. For example, $24\mathrm{V}$ divided by $R_2$ would mix a total voltage with an individual resistance. Those quantities do not describe the same part of the circuit.

Use total quantities together:

$$I_T=\frac{V_T}{R_T}$$

Then use individual quantities together:

$$V_1=I_1{R}_1$$

Since it is a series circuit, $I_1=I_T$.

The Ohm’s law triangle can help you recall the three arrangements, but write the formula before inserting numbers. That makes your working easier to follow and helps prevent calculator-entry mistakes.

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Worked example: full series-circuit solution

Question: A $24\mathrm{V}$ DC supply is connected to three resistors in series:

$$R_1=2\Omega$$ $$R_2=4\Omega$$ $$R_3=6\Omega$$

Calculate:

• total resistance $R_T$
• circuit current $I_T$
• voltage drops $V_1$, $V_2$, and $V_3$

Step 1: Identify the given and required quantities

Write the known values before calculating:

Quantity	Value
Total supply voltage $V_T$	$24\mathrm{V}$
Resistor $R_1$	$2\Omega$
Resistor $R_2$	$4\Omega$
Resistor $R_3$	$6\Omega$

The circuit is series because there is one path. Therefore, resistance adds and current is identical through all three resistors.

Step 2: Find total resistance

$$R_T=R_1+R_2+R_3$$ $$R_T=2\Omega +4\Omega +6\Omega$$ $$R_T=12\Omega$$

Step 3: Find circuit current

Use Ohm’s law with the total voltage and total resistance:

$$I_T=\frac{V_T}{R_T}$$ $$I_T=\frac{24\mathrm{V}}{12\Omega }$$ $$I_T=2\mathrm{A}$$

Because this is a series circuit:

$$I_1=I_2=I_3=2\mathrm{A}$$

Step 4: Find each voltage drop

For $R_1$:

$$V_1=I_1{R}_1$$ $$V_1=(2\mathrm{A})(2\Omega )$$ $$V_1=4\mathrm{V}$$

For $R_2$:

$$V_2=I_2{R}_2$$ $$V_2=(2\mathrm{A})(4\Omega )$$ $$V_2=8\mathrm{V}$$

For $R_3$:

$$V_3=I_3{R}_3$$ $$V_3=(2\mathrm{A})(6\Omega )$$ $$V_3=12\mathrm{V}$$

Step 5: Check the answer

Add the voltage drops:

$$V_1+V_2+V_3=4\mathrm{V}+8\mathrm{V}+12\mathrm{V}$$ $$V_1+V_2+V_3=24\mathrm{V}$$

This matches the $24\mathrm{V}$ supply, so the answer is consistent.

Notice the pattern: $R_3$ is the largest resistance and has the largest voltage drop. That is what you should expect in a series circuit with the same current flowing through every resistor.

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Handling kilo-ohms and milliamps

In pre-trade questions, resistors are often given in kilo-ohms $\mathrm{k\Omega }$, and the resulting current may be in milliamps $\mathrm{mA}$.

A convenient pairing is:

$$\mathrm{mA}\times \mathrm{k\Omega }=\mathrm{V}$$

For example, suppose:

$$V_T=18\mathrm{V}$$ $$R_1=1.2\mathrm{k\Omega }$$ $$R_2=2.2\mathrm{k\Omega }$$ $$R_3=2.6\mathrm{k\Omega }$$

First find total resistance:

$$R_T=1.2\mathrm{k\Omega }+2.2\mathrm{k\Omega }+2.6\mathrm{k\Omega }$$ $$R_T=6.0\mathrm{k\Omega }$$

Then calculate current:

$$I_T=\frac{18\mathrm{V}}{6.0\mathrm{k\Omega }}$$ $$I_T=3.0\mathrm{mA}$$

Now calculate the voltage drops:

$$V_1=(3.0\mathrm{mA})(1.2\mathrm{k\Omega })=3.6\mathrm{V}$$ $$V_2=(3.0\mathrm{mA})(2.2\mathrm{k\Omega })=6.6\mathrm{V}$$ $$V_3=(3.0\mathrm{mA})(2.6\mathrm{k\Omega })=7.8\mathrm{V}$$

Check:

$$3.6\mathrm{V}+6.6\mathrm{V}+7.8\mathrm{V}=18.0\mathrm{V}$$

If you instead convert resistance to ohms, the same current is:

$$I_T=\frac{18\mathrm{V}}{6000\Omega }$$ $$I_T=0.003\mathrm{A}$$ $$I_T=3.0\mathrm{mA}$$

Both approaches are correct. The important point is to keep units consistent and write the unit on every line.

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Common exam errors and quick checks

Most wrong answers in series-circuit questions come from a small number of avoidable errors.

Error	Why it is wrong	Better approach
Dividing total voltage by one resistor	Total voltage is across the whole network, not one resistor	Add resistances first, then use $I_T=V_T/R_T$
Giving a different current for each resistor	A series circuit has only one path	Write $I_T=I_1=I_2=I_3$
Adding currents in a series circuit	Current does not split in one path	Add resistances and voltage drops, not currents
Forgetting units	A correct number without a unit can lose marks	Write $\Omega$, $\mathrm{A}$, $\mathrm{mA}$, or $\mathrm{V}$ every time
Not checking voltage drops	A calculator error may go unnoticed	Add all drops and compare with supply voltage

Before moving on, use this short desk check on every answer:

• Is $R_T$ greater than the largest individual resistor? It should be, if all resistors are positive.
• Is there only one current value throughout the circuit?
• Do the voltage drops add to the supply voltage, allowing for small rounding differences?
• Does the largest resistor have the largest voltage drop?
• Have total voltage and total resistance been used together?

For written calculations, a clear answer layout earns method marks even if you make a final arithmetic error:

1. Write the formula.
2. Substitute values with units.
3. Show the calculator result.
4. State the final answer with the correct unit.
5. Show a voltage-sum check.

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Key takeaways

For a resistor-only series circuit:

$$R_T=R_1+R_2+R_3+\dots$$ $$I_T=\frac{V_T}{R_T}$$ $$I_T=I_1=I_2=I_3$$ $$V_n=I_T{R}_n$$ $$V_T=V_1+V_2+V_3+\dots$$

The reliable exam routine is: add resistance, calculate one circuit current, calculate each voltage drop, then check the voltage total.

Next, you will use the voltage-sum rule more formally through Kirchhoff’s Voltage Law, which is especially useful for checking a loop or finding one missing voltage.