Calculating Total Resistance and Branch Currents in Parallel Circuits

Kia ora. In the previous lesson, you used a voltage divider in a series circuit: the current was the same everywhere, while the supply voltage was shared between resistors. This lesson is the important contrast.

In a parallel circuit, every branch is connected across the same two points of the supply. Therefore, the voltage is the same across every branch, while the current splits between branches. You will use that fact, Ohm’s law, and the parallel-resistance formula to calculate total resistance and every branch current in an exam-ready format.

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Recognising a parallel circuit

A resistor is in parallel with another resistor when both of its ends connect to the same two circuit nodes. The branches may be drawn neatly underneath one another, as in the diagram below, or arranged differently on an exam diagram. Focus on the connection points, not the physical layout.

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The diagram shows the three rules that drive nearly every basic parallel-circuit calculation:

$$V_\mathrm{S}=V_1=V_2=V_3$$ $$I_\mathrm{T}=I_1+I_2+I_3$$ $$\frac{1}{R_\mathrm{T}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$

Read the first rule carefully:

The voltage across each parallel branch equals the supply voltage.

This is the opposite of a series circuit. In series, the supply voltage is divided among components. In parallel, each branch receives the full supply voltage.

The current, however, is not usually equal in each branch. Since each branch has the same voltage, Ohm’s law tells us:

$$I_\text{branch} = \frac{V_\mathrm{S}}{R_\text{branch}}$$

A branch with lower resistance carries higher current. A branch with higher resistance carries lower current.

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Total resistance in parallel

The total resistance, also called equivalent resistance, is the single resistance that would draw the same total current from the supply as the whole parallel network.

Parallel resistance is found using reciprocals:

$$\frac{1}{R_\mathrm{T}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} +\cdots$$

For calculator use, write the formula in its more direct form:

$$\boxed{ R_\mathrm{T} = \frac{1} { \left( \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right) } }$$

This format avoids a common formula-manipulation problem. You do not need to rearrange the reciprocal equation yourself. Enter the complete expression into the calculator with brackets around the entire denominator.

For three resistors, the calculator layout is:

$$1\div \left( 1\div R_1 + 1\div R_2 + 1\div R_3 \right)$$

The essential resistance check

For resistors in parallel:

$$\boxed{ R_\mathrm{T} < \text{the smallest branch resistance} }$$

For example, if the branch resistors are $6\mathrm{k\Omega}$, $12\mathrm{k\Omega}$, and $8\mathrm{k\Omega}$, the total resistance must be less than $6\mathrm{k\Omega}$.

This makes physical sense. Adding a branch provides another path for current, so the source sees less overall opposition to current flow.

If your calculated parallel total is larger than the smallest resistor, stop and check your formula. You may have accidentally used the series rule:

$$R_\mathrm{T}=R_1+R_2+R_3$$

That rule is for series only, not parallel.

The two-resistor shortcut

When there are exactly two resistors in parallel, this shortcut is often faster:

$$\boxed{ R_\mathrm{T} = \frac{R_1R_2}{R_1+R_2} }$$

For example, if $R_1=6\mathrm{\Omega}$ and $R_2=3\mathrm{\Omega}$:

$$R_\mathrm{T} = \frac{(6\mathrm{\Omega})(3\mathrm{\Omega})} {6\mathrm{\Omega}+3\mathrm{\Omega}}$$ $$R_\mathrm{T} = \frac{18}{9}\mathrm{\Omega}$$ $$\boxed{R_\mathrm{T}=2\mathrm{\Omega}}$$

The result is less than $3\mathrm{\Omega}$, the smallest branch resistance, so it passes the check.

Use this shortcut only for two parallel resistors. For three or more branches, use the reciprocal formula.

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Finding branch currents

Once you identify a resistor as a parallel branch, its voltage is already known:

$$V_\text{branch}=V_\mathrm{S}$$

Then apply Ohm’s law to that branch:

$$\boxed{ I_\text{branch} = \frac{V_\mathrm{S}}{R_\text{branch}} }$$

You calculate each branch separately. Do not divide the supply voltage between branches.

A useful unit relationship is:

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

So, for example:

$$\frac{24\mathrm{V}}{6\mathrm{k\Omega}} = 4\mathrm{mA}$$

This is convenient in exams, but only when the resistance is expressed in $\mathrm{k\Omega}$. If resistance is in $\mathrm{\Omega}$, the answer comes out in $\mathrm{A}$.

The total current drawn from the source is the sum of all branch currents:

$$\boxed{ I_\mathrm{T}=I_1+I_2+I_3+\cdots }$$

The total current must be greater than any single branch current because it includes the current supplied to every branch.

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Worked exam-style example

Question: Three resistors are connected in parallel across a $24\mathrm{V}$ DC supply.

$$R_1=6.0\mathrm{k\Omega}$$ $$R_2=12.0\mathrm{k\Omega}$$ $$R_3=8.0\mathrm{k\Omega}$$

Calculate:

1. the total resistance $R_\mathrm{T}$;
2. the current in each branch;
3. the total supply current.

Step 1: State the parallel voltage rule

Because the resistors are in parallel:

$$V_1=V_2=V_3=V_\mathrm{S}=24\mathrm{V}$$

Every branch has $24\mathrm{V}$ across it.

Step 2: Calculate total resistance

All resistance values are already in $\mathrm{k\Omega}$, so they can be used together.

$$R_\mathrm{T} = \frac{1} { \left( \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right) }$$ $$R_\mathrm{T} = \frac{1} { \left( \frac{1}{6.0} + \frac{1}{12.0} + \frac{1}{8.0} \right) } \mathrm{k\Omega}$$ $$R_\mathrm{T} = \frac{1}{0.375} \mathrm{k\Omega}$$ $$\boxed{ R_\mathrm{T}=2.67\mathrm{k\Omega} }$$

Check: $2.67\mathrm{k\Omega}$ is less than $6.0\mathrm{k\Omega}$, the smallest branch resistance. This is reasonable.

Step 3: Calculate the first branch current

$$I_1 = \frac{V_1}{R_1}$$ $$I_1 = \frac{24\mathrm{V}} {6.0\mathrm{k\Omega}}$$ $$\boxed{ I_1=4.0\mathrm{mA} }$$

Step 4: Calculate the second branch current

$$I_2 = \frac{V_2}{R_2}$$ $$I_2 = \frac{24\mathrm{V}} {12.0\mathrm{k\Omega}}$$ $$\boxed{ I_2=2.0\mathrm{mA} }$$

Step 5: Calculate the third branch current

$$I_3 = \frac{V_3}{R_3}$$ $$I_3 = \frac{24\mathrm{V}} {8.0\mathrm{k\Omega}}$$ $$\boxed{ I_3=3.0\mathrm{mA} }$$

Notice the pattern:

• $R_2$ is the largest resistance, so it has the smallest current.
• $R_1$ is the smallest resistance, so it has the largest current.

Step 6: Add the branch currents

$$I_\mathrm{T} = I_1+I_2+I_3$$ $$I_\mathrm{T} = 4.0\mathrm{mA} + 2.0\mathrm{mA} + 3.0\mathrm{mA}$$ $$\boxed{ I_\mathrm{T}=9.0\mathrm{mA} }$$

Step 7: Verify the total resistance using Ohm’s law

This is an excellent exam check:

$$R_\mathrm{T} = \frac{V_\mathrm{S}}{I_\mathrm{T}}$$ $$R_\mathrm{T} = \frac{24\mathrm{V}} {9.0\mathrm{mA}}$$ $$\boxed{ R_\mathrm{T}=2.67\mathrm{k\Omega} }$$

This agrees with the reciprocal-resistance calculation.

A table keeps the information organised and earns clearer method marks in a written calculation:

Quantity	Branch 1	Branch 2	Branch 3	Total
Resistance	$6.0\mathrm{k\Omega}$	$12.0\mathrm{k\Omega}$	$8.0\mathrm{k\Omega}$	$2.67\mathrm{k\Omega}$
Voltage	$24\mathrm{V}$	$24\mathrm{V}$	$24\mathrm{V}$	$24\mathrm{V}$
Current	$4.0\mathrm{mA}$	$2.0\mathrm{mA}$	$3.0\mathrm{mA}$	$9.0\mathrm{mA}$

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Unit handling and common exam errors

Use matching resistance units

The branch currents can be calculated using $\mathrm{\Omega}$ or $\mathrm{k\Omega}$, but use a matching unit consistently.

For example:

$$470\mathrm{\Omega}=0.470\mathrm{k\Omega}$$

If one branch is $470\mathrm{\Omega}$ and another is $1.0\mathrm{k\Omega}$, convert before using the two-resistor resistance formula or reciprocal formula.

Using $\mathrm{k\Omega}$:

$$R_\mathrm{T} = \frac{(0.470)(1.00)} {0.470+1.00} \mathrm{k\Omega}$$ $$R_\mathrm{T} = 0.320\mathrm{k\Omega}$$ $$\boxed{ R_\mathrm{T}=320\mathrm{\Omega} }$$

Avoid these mistakes

Mistake	Why it is wrong	Better exam habit
Adding branch resistances directly	That is the series-resistance rule.	Use reciprocals, or product-over-sum for exactly two branches.
Dividing supply voltage between branches	In parallel, each branch is connected directly across the supply.	Write $V_1=V_2=V_3=V_\mathrm{S}$ first.
Assuming all branch currents are equal	Equal current occurs only when branch resistances are equal.	Use $I=V/R$ for each branch.
Giving $R_\mathrm{T}$ greater than the smallest branch resistor	More parallel paths reduce total resistance.	Check that $R_\mathrm{T}$ is the smallest resistance in the table.
Forgetting a branch when adding current	The supply current feeds every branch.	List $I_1$, $I_2$, and $I_3$ before adding.
Rounding each intermediate answer too early	Small rounding errors can make the total-current check look incorrect.	Keep extra calculator digits until the final answer.

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A repeatable exam method

When a question gives a parallel network and asks for total resistance and branch currents, use this order:

1. Confirm it is parallel: each branch connects to the same two nodes.
2. Write the voltage rule:

$$V_1=V_2=V_3=V_\mathrm{S}$$

3. Calculate total resistance using the reciprocal formula, or product-over-sum if there are exactly two resistors.
4. Calculate each branch current using:

$$I_\text{branch} = \frac{V_\mathrm{S}}{R_\text{branch}}$$

5. Add branch currents to find total current:

$$I_\mathrm{T}=I_1+I_2+I_3$$

6. Check the result:
   • $R_\mathrm{T}$ is less than the smallest branch resistance.
   • The smallest branch resistance has the largest branch current.
   • $I_\mathrm{T}$ is greater than any individual branch current.
   • If required, verify with:

$$R_\mathrm{T} = \frac{V_\mathrm{S}}{I_\mathrm{T}}$$

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Key takeaways

In a parallel circuit:

$$\boxed{ V_\mathrm{S}=V_1=V_2=V_3 }$$ $$\boxed{ I_\mathrm{T}=I_1+I_2+I_3 }$$ $$\boxed{ R_\mathrm{T} = \frac{1} { \left( \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \right) } }$$

The practical rules to remember are:

• Voltage is the same across every parallel branch.
• Current divides between branches.
• The lowest resistance branch takes the highest current.
• Total parallel resistance is always less than the smallest branch resistance.

Next, you will formalise the total-current rule at a circuit junction using Kirchhoff’s Current Law and use it to solve and check unknown currents.