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F2.2 · Analyse series, parallel, and mixed circuits with Ohm’s and Kirchhoff’s laws

Learn to analyse series, parallel, and mixed circuits with ohm’s and kirchhoff’s laws through clear examples and targeted practice.

Ontario Grade 11 Physics

Electricity and Magnetism

Using Ohm’s law and Kirchhoff’s laws

A circuit is a connected path that allows electric charge to move. In this lesson, the system is the complete circuit, including its source and resistors. We use conventional current: its indicated direction is the direction positive charge would move, from the source’s positive terminal through the circuit toward its negative terminal. Electron flow is opposite to conventional current, but circuit calculations here use conventional current. Current is measured in amperes (A), voltage in volts (V), and resistance in ohms (Ω). Voltage, current, and resistance are scalar quantities; they have size but no direction. A direction arrow on a circuit shows the chosen current direction, not a vector quantity.

What you will learn

1. Prerequisites and circuit patterns

A circuit diagram uses symbols and lines to show electrical connections. A resistor is a component that opposes current. A source, such as a battery, provides a potential difference, or voltage, that can drive current. For this lesson, assume ideal connecting wires: their resistance is negligible.
In a series connection, components are joined one after another along a single path. The same current passes through every component in that path. In a parallel connection, components connect across the same two points, creating separate branches. Each branch has the same voltage across it. A mixed circuit contains both kinds of connection.
Resistance values in series add. For parallel resistors, the reciprocals of the resistances add. A reciprocal is one divided by a value. These rules let us replace a group of resistors with one equivalent resistance, which has the same overall effect on the source as that group.
Rseries=R1+R2R_{\mathrm{series}}=R_1+R_2

2. Ohm’s law and Kirchhoff’s laws

Ohm’s law relates voltage, current, and resistance for a resistor. Use it when two of those quantities are known and the third is needed. Keep units in the substitution so that the result can be checked.
Kirchhoff’s current law describes a junction, where a circuit path splits or joins. The total current entering a junction equals the total current leaving it. This follows from charge not building up at the junction in a steady circuit.
Kirchhoff’s voltage law describes a complete loop. A loop is a closed path through the circuit. The voltage rises and drops around one complete loop balance: the source’s voltage rise equals the total voltage drops. For a loop traced in the conventional-current direction, a resistor has a voltage drop. If you trace through a source from its negative to positive terminal, it is a voltage rise.
Choose a current direction before calculating. If a calculation gives a negative current, the actual current is opposite to the chosen arrow. In the examples, the source’s positive terminal sets the conventional-current direction through the external circuit.
V=IRV=IR

3. A reliable analysis method

First, mark the source voltage, each resistance, and any known current. Identify the paths: trace whether there is one route or more than one between connection points. Then decide which resistors are in series and which are in parallel.
Next, simplify one series or parallel group at a time. Find the total resistance seen by the source, then use Ohm’s law to find the total current. For a mixed circuit, work backward through the groups: use voltage drops in series sections and the shared voltage across parallel branches. Use Kirchhoff’s laws to check that currents and voltages agree at the junctions and around loops.
A labelled circuit sketch can prevent a common error: treating components as parallel just because they are drawn beside each other. Components are parallel only when both ends connect to the same two points. Components are in series when they share a single path and no branch splits between them.
1Rparallel=1R1+1R2\frac{1}{R_{\mathrm{parallel}}}=\frac{1}{R_1}+\frac{1}{R_2}

4. Units and reasonableness checks

Keep units through each substitution. For example, dividing volts by ohms gives amperes. Adding resistances gives a result in ohms. Round only the final reported values, using a number of significant figures that suits the given values.
Check whether your result fits the circuit. In a series path, each resistor’s voltage drop should add to the source voltage. In parallel branches, branch currents should add to the total current. For a fixed source voltage, adding a parallel branch provides another path and lowers the equivalent resistance; the total current then increases.
A negative answer is not automatically an arithmetic error. It can show that the actual direction is opposite to the arrow you selected. State that direction clearly. Also check that a current in a branch is not larger than the total current entering the split, and that a branch voltage agrees with the voltage across its two shared endpoints.
\sum I_{in}=\sum I_{out\mathrm{out}}

Series and parallel patterns

FeatureSeriesParallel
PathsOne pathSeparate branches
Same quantityCurrentVoltage
Equivalent resistanceResistances addReciprocals add
Junction checkNo split in a single pathEntering current equals leaving current

Worked example

Two resistors in series

A 12 V source is connected to a 2.0 Ω resistor and a 4.0 Ω resistor in series. Find the total resistance, circuit current, and voltage drop across each resistor.
  1. Set up
    The system is the source and both resistors. Conventional current is chosen from the positive terminal through the 2.0 Ω and 4.0 Ω resistors toward the negative terminal. The unknowns are the total resistance, current, and two drops.
  2. Combine resistance
    There is one path, so the series resistances add. This gives the resistance seen by the source.
    Rtotal=2.0 Ω+4.0 Ω=6.0 ΩR_{\mathrm{total}}=2.0\,\Omega+4.0\,\Omega=6.0\,\Omega
  3. Find current
    Use Ohm’s law for the whole circuit. The source voltage divided by total resistance gives the current, which is the same through both series resistors.
    I=12 V6.0 Ω=2.0 AI=\frac{12\,\mathrm{V}}{6.0\,\Omega}=2.0\,\mathrm{A}
  4. Find voltage drops
    Apply Ohm’s law to each resistor. Current flows from the source’s positive terminal through each resistor, so each listed value is a voltage drop in that direction.
    V1=(2.0 A)(2.0 Ω)=4.0 V,V2=(2.0 A)(4.0 Ω)=8.0 VV_1=(2.0\,\mathrm{A})(2.0\,\Omega)=4.0\,\mathrm{V},\quad V_2=(2.0\,\mathrm{A})(4.0\,\Omega)=8.0\,\mathrm{V}
Answer: The total resistance is 6.0 Ω, the current is 2.0 A, and the drops are 4.0 V and 8.0 V.
Check: The units are Ω, A, and V as required. The drops add to 12.0 V, matching the source. The larger resistance has the larger voltage drop, which is reasonable because the series current is shared.

Worked example

Two resistors in parallel

A 12 V source is connected across a 6.0 Ω resistor and a 3.0 Ω resistor in parallel. Find the equivalent resistance, total current, and current in each branch.
  1. Set up
    The system is the source and two branches connected across the same pair of points. Choose conventional current from the positive terminal toward the junction, then into each branch. Both resistors have the source voltage across them.
  2. Find equivalent resistance
    For parallel resistors, add the reciprocals and then take the reciprocal of the result. The equivalent resistance is less than either branch resistance.
    1Req=16.0 Ω+13.0 Ω=12.0 Ω,Req=2.0 Ω\frac{1}{R_{\mathrm{eq}}}=\frac{1}{6.0\,\Omega}+\frac{1}{3.0\,\Omega}=\frac{1}{2.0\,\Omega},\quad R_{\mathrm{eq}}=2.0\,\Omega
  3. Find total current
    Use the source voltage and equivalent resistance in Ohm’s law. This is the current entering the junction before it divides between branches.
    Itotal=12 V2.0 Ω=6.0 AI_{\mathrm{total}}=\frac{12\,\mathrm{V}}{2.0\,\Omega}=6.0\,\mathrm{A}
  4. Find branch currents
    Each branch has 12 V across it. Apply Ohm’s law separately to each branch; the lower resistance carries the greater current.
    I6.0 Ω=12 V6.0 Ω=2.0 A,I3.0 Ω=12 V3.0 Ω=4.0 AI_{6.0\,\Omega}=\frac{12\,\mathrm{V}}{6.0\,\Omega}=2.0\,\mathrm{A},\quad I_{3.0\,\Omega}=\frac{12\,\mathrm{V}}{3.0\,\Omega}=4.0\,\mathrm{A}
Answer: The equivalent resistance is 2.0 Ω. The total current is 6.0 A; the branch currents are 2.0 A and 4.0 A.
Check: The branch currents add to 6.0 A, as required at the junction. Each branch has 12 V across it. The units are consistent, and the larger current is in the lower-resistance branch.

Worked example

A mixed circuit

A 18 V source is connected in series with a 2.0 Ω resistor, followed by a parallel pair of 6.0 Ω and 3.0 Ω resistors. Find the total current, the voltage across each part, and the branch currents.
  1. Set up
    The system includes the source, a series resistor, and a parallel pair. Choose conventional current from the positive terminal through the 2.0 Ω resistor, then into the parallel junction. The unknowns are total current, voltage drops, and branch currents.
  2. Simplify the parallel pair
    Find the pair’s equivalent resistance using the parallel rule. Then add the 2.0 Ω series resistance to get total circuit resistance.
    Rparallel=(16.0 Ω+13.0 Ω)−1=2.0 Ω,Rtotal=2.0 Ω+2.0 Ω=4.0 ΩR_{\mathrm{parallel}}=\left(\frac{1}{6.0\,\Omega}+\frac{1}{3.0\,\Omega}\right)^{-1}=2.0\,\Omega,\quad R_{\mathrm{total}}=2.0\,\Omega+2.0\,\Omega=4.0\,\Omega
  3. Find total current and series drop
    Use Ohm’s law with the source voltage and total resistance. The same total current passes through the series resistor before reaching the junction.
    Itotal=18 V4.0 Ω=4.5 A,V2.0 Ω=(4.5 A)(2.0 Ω)=9.0 VI_{\mathrm{total}}=\frac{18\,\mathrm{V}}{4.0\,\Omega}=4.5\,\mathrm{A},\quad V_{2.0\,\Omega}=(4.5\,\mathrm{A})(2.0\,\Omega)=9.0\,\mathrm{V}
  4. Find branch values
    The remaining source voltage is across the parallel pair. Both branches share that voltage. Apply Ohm’s law to each branch and check that their currents recombine to the incoming current.
    Vparallel=18 V−9.0 V=9.0 V,I6.0 Ω=1.5 A,I3.0 Ω=3.0 AV_{\mathrm{parallel}}=18\,\mathrm{V}-9.0\,\mathrm{V}=9.0\,\mathrm{V},\quad I_{6.0\,\Omega}=1.5\,\mathrm{A},\quad I_{3.0\,\Omega}=3.0\,\mathrm{A}
Answer: The total current is 4.5 A. The series resistor has a 9.0 V drop, and the parallel pair has 9.0 V across it. The branch currents are 1.5 A and 3.0 A.
Check: The loop drops total 18 V. The branch currents sum to 4.5 A, matching the current entering the junction. Each value has the appropriate SI unit and is consistent with the stated circuit.

Common mistakes and how to avoid them

Adding parallel resistances directly.
Correction: Use the reciprocal rule for parallel groups. The equivalent resistance must be less than the smallest branch resistance.
Assuming parallel branches split voltage equally.
Correction: Parallel branches share the same voltage because they connect to the same two points. Their currents can differ.
Using the total current as the current in every parallel branch.
Correction: At a junction, total current divides. Find each branch current from that branch’s voltage and resistance.
Treating the way components are drawn as proof that they are parallel.
Correction: Trace both ends of each component. Parallel components must connect to the same two points.

Lesson summary

Check your understanding

Question 1

A 10 V source is connected to a 5.0 Ω resistor. What current flows?
  1. 0.50 A
  2. 2.0 A
  3. 5.0 A
  4. 50 A
Show answer and explanation
2.0 A
Ohm’s law gives current as voltage divided by resistance: 10 V ÷ 5.0 Ω = 2.0 A.

Question 2

Two branches carry 1.2 A and 0.8 A into a junction. What current leaves the junction?
  1. 0.4 A
  2. 1.0 A
  3. 2.0 A
  4. 2.4 A
Show answer and explanation
2.0 A
Kirchhoff’s current law says entering current equals leaving current. The total is 1.2 A + 0.8 A = 2.0 A.

Question 3

Two resistors are connected in parallel across a 9.0 V source. What is true of their voltages?
  1. Each has 4.5 V across it.
  2. Each has 9.0 V across it.
  3. The larger resistor has the larger voltage.
  4. Their voltages add to 9.0 V.
Show answer and explanation
Each has 9.0 V across it.
Both resistors connect to the same two points, so each has the full 9.0 V across it.

Question 4

A loop has a 15 V source and one resistor with a 6.0 V drop. What is the drop across the other series resistor?
  1. 6.0 V
  2. 9.0 V
  3. 15 V
  4. 21 V
Show answer and explanation
9.0 V
The voltage drops around the loop must add to the source voltage. The remaining drop is 15 V − 6.0 V = 9.0 V.

Key terms

Conventional current
The circuit-current direction defined as the direction positive charge would move.
Junction
A point where a circuit path splits or joins.
Equivalent resistance
One resistance that has the same overall effect as a group of resistors.
Voltage drop
A decrease in electrical potential across a circuit component in the chosen direction of travel.
Loop
A closed path through a circuit that returns to its starting point.

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Published by DoAssignment. This AI-assisted lesson follows Ontario Grade 11 Physics (SPH3U), expectation F2.2. It is a study resource, not an official curriculum publication.

Before publication, the draft is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability, and then requires administrator approval. Errors can still occur, so corrections are welcomed.

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