Design and explain mixed direct-current circuits

Ontario Grade 11 Physics

Electricity and Magnetism

Ontario Grade 11 Physics — F2.3

A mixed direct-current circuit contains at least one series part and one parallel part. You can analyze it by simplifying one part at a time, then using the same rules to find currents and voltage drops. This lesson treats the battery, wires, and resistors as the circuit system. Conventional current is defined as flowing from the battery’s positive terminal through the external circuit toward its negative terminal. This chosen direction is an arrow for keeping track of current; electron flow is in the opposite direction.

What you will learn

Identify series and parallel parts of a direct-current circuit.
Use resistance rules and Ohm’s law to analyze a mixed circuit.
Apply current and voltage rules to explain how a circuit works.
Choose component values for a simple mixed-circuit design.

1. Prerequisite bridge: current, voltage, and resistance

Current describes how much electric charge passes a point each second. Its SI unit is the ampere, written A. Voltage is the energy transferred per unit of charge between two points. Its SI unit is the volt, written V. Resistance describes how strongly a component opposes current. Its SI unit is the ohm, written Ω.

For a resistor, Ohm’s law connects voltage, current, and resistance. Use it when the voltage across that resistor and its resistance are known, or when you need to find one of the three quantities. Rearranging the relationship is ordinary algebra.

Current and voltage are not the same kind of quantity. In these circuit calculations, current is tracked with a chosen direction, while voltage is a difference between two points. Resistors do not have a fixed positive direction in the way a motion vector does; label the current direction you assume and keep it consistent.

V=IR

Use A for current, V for voltage, and Ω for resistance.
Conventional current in the external circuit points from the positive terminal toward the negative terminal.
For a resistor, voltage equals current multiplied by resistance.

2. Read and simplify a mixed circuit

In a series connection, components are joined along one path, so the same current passes through each component. Their resistances add. In a parallel connection, components share the same two connection points, called nodes. The voltage across each parallel branch is the same. The branch currents add to give the total current entering or leaving the branches.

A circuit diagram can be sketched with labels before calculating. For example, a battery can feed resistor

R_1

, followed by a split into two branches containing

R_2

and

R_3

. The branches rejoin and return to the battery. The current through

R_1

is the total current; it divides between the two branches.

For parallel resistors, add the reciprocals of the resistances to find the reciprocal of the equivalent resistance. Equivalent resistance means the single resistance that could replace a group while having the same effect on the rest of the circuit. A parallel combination has less resistance than its smallest branch, because it provides more than one path.

Kirchhoff’s current rule says that current entering a junction equals current leaving it. Kirchhoff’s voltage rule says that the voltage supplied around a complete loop equals the sum of the voltage drops in that loop. These rules help check a circuit calculation: current is conserved at a split, and voltage drops account for the source voltage around a loop.

R_{\mathrm{series}}=R_1+R_2,\quad \frac{1}{R_{\mathrm{parallel}}}=\frac{1}{R_1}+\frac{1}{R_2}

Series: same current; resistances add.
Parallel: same voltage across branches; branch currents add.
Simplify a mixed circuit in stages, then work back to find branch values.

3. Design choices and checks

To design a simple mixed circuit, first decide what the circuit must do. For example, you might want one resistor in series with two parallel branches. Choose component values, draw the connections clearly, and calculate the equivalent resistance and total current for a stated supply voltage. Then calculate the voltage and current in each part.

A design should follow the connection pattern, not just list component values. A resistor is in series only if the same current must pass through it on the single path. Two resistors are in parallel only if both ends of each resistor connect to the same pair of nodes.

Keep units in each calculation. Check that a series equivalent resistance is larger than either component resistance. Check that a parallel equivalent resistance is smaller than the smallest branch resistance. Also check that branch currents add to the total and that the voltage drops around a loop add to the source voltage. These are useful reasonableness checks, not extra measurements.

I_{\mathrm{total}}=I_1+I_2

Draw and label the actual connection pattern before calculating.
Use the circuit rules to predict current and voltage throughout the design.
Check equivalent resistance, junction currents, loop voltages, units, and rounding.

Worked example

Find currents in a mixed circuit

A 12 V battery is connected to a 2.0 Ω resistor in series with a parallel pair of 6.0 Ω and 3.0 Ω resistors. Find the total current and the current in each parallel branch.

Set the circuit and direction
The system is the battery and three resistors. Take conventional current from the positive terminal through the 2.0 Ω resistor, then into the parallel branches. The unknowns are total current and the two branch currents.
Replace the parallel pair
The 6.0 Ω and 3.0 Ω resistors share the same two nodes, so use the parallel-resistance rule. Their equivalent resistance is 2.0 Ω.
$R_p=\left(\frac{1}{6.0\,\Omega}+\frac{1}{3.0\,\Omega}\right)^{-1}=2.0\,\Omega$
Find total current
The equivalent parallel part is in series with the 2.0 Ω resistor. Their resistances add, and Ohm’s law gives the current supplied by the battery.
$R_{\mathrm{total}}=2.0\,\Omega+2.0\,\Omega=4.0\,\Omega,\quad I_{\mathrm{total}}=\frac{12\,\mathrm{V}}{4.0\,\Omega}=3.0\,\mathrm{A}$
Find branch currents
The series resistor drops 6.0 V, leaving 6.0 V across each parallel branch. Apply Ohm’s law separately to each branch. Current is directed through each branch toward the battery’s negative terminal.
$I_{6.0\,\Omega}=\frac{6.0\,\mathrm{V}}{6.0\,\Omega}=1.0\,\mathrm{A},\quad I_{3.0\,\Omega}=\frac{6.0\,\mathrm{V}}{3.0\,\Omega}=2.0\,\mathrm{A}$

Answer: The total current is 3.0 A. The currents in the 6.0 Ω and 3.0 Ω branches are 1.0 A and 2.0 A, respectively.

Check: The branch currents add to 3.0 A. The total resistance is 4.0 Ω, so 12 V divided by 4.0 Ω gives 3.0 A. Units and current direction are consistent, and the result is reasonable.

Worked example

Design a circuit for a chosen supply

Design a mixed circuit using a 9.0 V supply, a 3.0 Ω series resistor, and parallel branches of 6.0 Ω and 3.0 Ω. Find the total current and each branch current.

Describe the layout
The system includes the 9.0 V source and all three resistors. Conventional current leaves the positive terminal, passes through the 3.0 Ω series resistor, splits between the two branches, then returns to the negative terminal.
Find the equivalent resistance
The two branches are in parallel. Their equivalent resistance is 2.0 Ω. Add the 3.0 Ω series resistor to find the resistance of the whole circuit.
$R_p=\left(\frac{1}{6.0\,\Omega}+\frac{1}{3.0\,\Omega}\right)^{-1}=2.0\,\Omega,\quad R_{\mathrm{total}}=3.0\,\Omega+2.0\,\Omega=5.0\,\Omega$
Calculate the total current
Use the source voltage and total resistance in Ohm’s law. This is the current through the series resistor and the total current entering the branch junction.
$I_{\mathrm{total}}=\frac{9.0\,\mathrm{V}}{5.0\,\Omega}=1.8\,\mathrm{A}$
Calculate branch currents
The series resistor’s voltage drop is 5.4 V, leaving 3.6 V across each branch. Divide that branch voltage by each branch resistance. The current directions follow the chosen conventional-current path.
$V_p=9.0\,\mathrm{V}-(1.8\,\mathrm{A})(3.0\,\Omega)=3.6\,\mathrm{V},\quad I_{6.0\,\Omega}=0.60\,\mathrm{A},\quad I_{3.0\,\Omega}=1.2\,\mathrm{A}$

Answer: The design has a total current of 1.8 A. The branch currents are 0.60 A through 6.0 Ω and 1.2 A through 3.0 Ω.

Check: The branch currents sum to 1.8 A. The drops are 5.4 V across the series resistor and 3.6 V across the parallel section, which sum to 9.0 V. The units, direction, and two-significant-figure results are consistent.

Worked example

Choose an unknown branch resistor

A 10 V source feeds a 3.0 Ω resistor in series with a parallel pair. One branch is 6.0 Ω. The total current is designed to be 2.0 A. Find the resistance needed in the other branch.

Identify what is known
The system is the source and the series and parallel resistors. Conventional current leaves the positive terminal and splits at the parallel junction. The unknown is the resistance of the second branch.
Find the required total and parallel resistances
Use Ohm’s law to find the total resistance that gives 2.0 A from a 10 V source. Subtract the known series resistance to get the required equivalent resistance of the parallel pair.
$R_{\mathrm{total}}=\frac{10\,\mathrm{V}}{2.0\,\mathrm{A}}=5.0\,\Omega,\quad R_p=5.0\,\Omega-3.0\,\Omega=2.0\,\Omega$
Solve for the branch resistance
The parallel rule relates the 2.0 Ω equivalent resistance to the known 6.0 Ω branch and the unknown branch. Rearrange the relationship to find the unknown.
$\frac{1}{R_p}=\frac{1}{6.0\,\Omega}+\frac{1}{R},\quad R=3.0\,\Omega$
Check the circuit
The parallel equivalent of 6.0 Ω and 3.0 Ω is 2.0 Ω. Adding the 3.0 Ω series resistor gives 5.0 Ω, which draws 2.0 A from 10 V. The branch current directions are away from the junction along each branch.
$R_p=\left(\frac{1}{6.0\,\Omega}+\frac{1}{3.0\,\Omega}\right)^{-1}=2.0\,\Omega$

Answer: Use a 3.0 Ω resistor in the second parallel branch.

Check: The parallel equivalent is less than the smallest branch resistance, as expected. Total resistance and current reproduce the design targets, with units and significant figures consistent.

Common mistakes and how to avoid them

Adding the resistances of parallel branches.

Correction: Use the reciprocal rule for parallel resistors. The equivalent resistance must be smaller than either branch.

Assuming current is the same in every part of a mixed circuit.

Correction: Current is the same along a single series path. At a parallel junction it divides, and the branch currents add back to the total.

Assuming parallel branches have the same current.

Correction: Parallel branches have the same voltage. Their currents depend on their resistances.

Reporting a numerical answer without units or a circuit check.

Correction: Include SI units, use sensible significant figures, and verify junction currents and loop voltage drops.

Lesson summary

A mixed circuit contains series and parallel connections.
Simplify one connection group at a time, then use Ohm’s law to find currents and voltage drops.
Series parts share current; parallel branches share voltage.
Check that branch currents add to the total and loop voltage drops add to the source voltage.

Check your understanding

Question 1

Two 4.0 Ω resistors are connected in parallel. What is their equivalent resistance?

2.0 Ω
4.0 Ω
8.0 Ω
correctIndex":0,"explanation":"The reciprocal rule gives an equivalent resistance of 2.0 Ω, which is less than either 4.0 Ω branch."}

Show answer and explanation

2.0 Ω

The reciprocal rule gives an equivalent resistance of 2.0 Ω, which is less than either 4.0 Ω branch.

Question 2

A total current of 2.5 A reaches a junction. One branch carries 1.0 A. What current flows in the other branch?

1.5 A
2.5 A
3.5 A
correctIndex":0,"explanation":"Current entering a junction equals current leaving it, so the other branch carries 2.5 A minus 1.0 A, or 1.5 A."}

Show answer and explanation

1.5 A

Current entering a junction equals current leaving it, so the other branch carries 1.5 A.

Key terms

Direct current: Electric current that flows in one direction through a circuit.
Series connection: A connection with one path through the components, so the same current passes through each.
Parallel connection: A connection in which components share the same two nodes and have the same voltage across them.
Equivalent resistance: The resistance of a single replacement resistor that has the same effect as a group of resistors.
Conventional current: The chosen direction of current from the positive terminal toward the negative terminal in the external circuit.

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About this lesson and its review

Published by DoAssignment. This reviewed lesson follows Ontario Grade 11 Physics (SPH3U), expectation F2.3. It is a study resource, not an official curriculum publication.

Before publication, content is checked for structure, mathematical or chemical notation, calculations, course boundaries, and readability. Errors can still occur, so corrections are welcomed.

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