D.C. Circuits

What you'll learn

How current and potential difference behave in series and parallel circuits.
How to combine resistors and calculate currents and voltages in mixed circuits.
How potential dividers work, including with LDRs and thermistors.
What emf and internal resistance mean, and how to determine internal resistance experimentally.

Prerequisites: the language of circuits

A D.C. circuit is a circuit supplied by direct current, meaning the current flows in one direction around the circuit. In this topic, you will usually treat the current as steady once the switch is closed.

Definition

Current, potential difference and resistance

Current $I$ is the rate of flow of charge: $\frac{\Delta Q}{\Delta t}$ . Its unit is the ampere (A).
Potential difference $V$ is the energy transferred per unit charge between two points: $\frac{W}{Q}$ . Its unit is the volt (V).
Resistance $R$ measures how strongly a component opposes current: $V = I R$ . Its unit is the ohm (Ω).

In circuit calculations, assume connecting wires have negligible resistance unless told otherwise. Ammeters are placed in series; voltmeters are placed in parallel with the component or source being measured.

Series and parallel: the conservation laws

A series circuit has one path for charge. The same current passes through every component.

A parallel circuit has branches. At a junction, charge splits between branches and then recombines.

Series and parallel circuit rules

Key Idea

Why the rules work

In a parallel circuit, the current from the source equals the sum of the currents in the branches because charge is conserved.
In a series circuit, the sum of the potential differences across the components equals the supply potential difference because energy is conserved.
Components in parallel have the same potential difference across them because each branch is connected across the same two points.

So for branches in parallel:

I_1 + I_2 + I_3 + \cdots

For components in series:

Vsupply=V1+V2+V3+⋯V_{\text{supply}} = V_1 + V_2 + V_3 + \cdots

For components in parallel:

V1=V2=VsupplyV_1 = V_2 = V_{\text{supply}}

Combining resistors

For resistors in series, resistances add:

R_1 + R_2 + R_3 + \cdots

For resistors in parallel, reciprocals add:

1R=1R1+1R2+1R3+⋯\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots

Tip

Parallel resistance sanity check

The combined resistance of resistors in parallel is always less than the smallest individual resistance. If your answer is bigger, you have probably added the resistances as if they were in series.

Example

Finding currents in a mixed resistor circuit

A 12.0 V supply is connected to a 4.0 Ω resistor in series with a parallel combination of 6.0 Ω and 12.0 Ω resistors. Find the total current and the current in each parallel branch.

Combine the parallel pair first:
$Ω−1\frac{1}{R_{\text{parallel}}}=\frac{1}{6.0\,\Omega}+\frac{1}{12.0\,\Omega}=0.250\,\Omega^{-1}$
so
$ΩR_{\text{parallel}}=4.0\,\Omega$
Add the series resistor:
$ΩR_{\text{total}}=4.0\,\Omega+4.0\,\Omega=8.0\,\Omega$
Use $V = I R$ to find the current from the supply:
$AI=\frac{V}{R}=\frac{12.0\,\text{V}}{8.0\,\Omega}=1.50\,\text{A}$
The 4.0 Ω series resistor has potential difference:
$VV=IR=1.50\,\text{A}\times 4.0\,\Omega=6.0\,\text{V}$
so the parallel section also has 6.0 V across it.
Find the branch currents:
$AI_{6.0}=\frac{6.0\,\text{V}}{6.0\,\Omega}=1.00\,\text{A}$ $AI_{12.0}=\frac{6.0\,\text{V}}{12.0\,\Omega}=0.500\,\text{A}$
The branch currents add to 1.50 A, matching conservation of charge.

Common Mistake

Using the supply voltage on every component

Only components in parallel with the supply have the full supply potential difference across them. In series sections, the supply potential difference is shared between components.

Potential dividers

A potential divider uses resistors in series to produce a chosen fraction of the supply potential difference. The output is usually taken across one resistor.

Potential divider with sensor examples

For two resistors in series, with the output across $R_2$ :

Vout=R2R1+R2VinV_{\text{out}}=\frac{R_2}{R_1+R_2}V_{\text{in}}

This works because the same current passes through both resistors, so the larger resistance gets the larger share of the potential difference.

An LDR, or light-dependent resistor, has resistance that decreases as light intensity increases. An NTC thermistor has resistance that decreases as temperature increases.

Key Idea

Sensor position matters

If the sensor is the lower resistor and $VoutV_{\text{out}}$ is measured across it, then decreasing the sensor resistance decreases $VoutV_{\text{out}}$ . If the sensor is the upper resistor, the trend reverses.

Example

Calculating a potential divider output

A 9.0 V supply is connected across a potential divider. The upper resistor is 2.2 kΩ and the lower resistor is an LDR. The output is measured across the LDR. The LDR resistance is 1.0 kΩ in bright light and 10 kΩ in the dark. Find the output voltage in each case.

Use the divider equation with the LDR as $R_2$ :
$Vout=R2R1+R2VinV_{\text{out}}=\frac{R_2}{R_1+R_2}V_{\text{in}}$
In bright light:
$VV_{\text{out}}=\frac{1.0\,\text{k}\Omega}{2.2\,\text{k}\Omega+1.0\,\text{k}\Omega}\times 9.0\,\text{V}=2.8\,\text{V}$
In the dark:
$VV_{\text{out}}=\frac{10\,\text{k}\Omega}{2.2\,\text{k}\Omega+10\,\text{k}\Omega}\times 9.0\,\text{V}=7.4\,\text{V}$
The output is larger in the dark because the LDR resistance is larger, so it takes a larger fraction of the supply potential difference.

Common Mistake

Loaded potential dividers

The simple divider formula assumes the output is measured by a very high-resistance voltmeter or device, so almost no current is drawn from the output. If a low-resistance load is connected, it is effectively in parallel with $R_2$ and changes the divider ratio.

Emf and internal resistance

The emf $E$ of a source is the energy transferred by the source per unit charge. Its unit is the volt (V), the same as potential difference.

Definition

Emf

The emf $E$ of a source is the energy supplied by the source to each coulomb of charge passing through it. When no current is drawn, the terminal potential difference is equal to the emf.

Real cells are not ideal. They have internal resistance $r$ , meaning some energy is transferred inside the cell itself, usually as thermal energy. The potential difference available across the external terminals is called the terminal potential difference $V$ .

The Eduqas equation is:

V = E - I r

Here, $I r$ is called the lost volts: the potential difference across the internal resistance.

Internal resistance practical circuit and graph

Example

Using internal resistance with cells in series

Two identical cells are connected in series, each with emf 1.50 V and internal resistance 0.30 Ω. They supply two external resistors, 5.0 Ω and 10.0 Ω, connected in series. Find the current and the terminal potential difference of the battery.

Add the emfs and internal resistances because the cells are in series and facing the same direction:
$VE_{\text{total}}=1.50\,\text{V}+1.50\,\text{V}=3.00\,\text{V}$ $Ωr_{\text{total}}=0.30\,\Omega+0.30\,\Omega=0.60\,\Omega$
Add the external series resistance:
$ΩR_{\text{external}}=5.0\,\Omega+10.0\,\Omega=15.0\,\Omega$
Use the total circuit resistance, including internal resistance:
$AI=\frac{E_{\text{total}}}{R_{\text{external}}+r_{\text{total}}} =\frac{3.00\,\text{V}}{15.0\,\Omega+0.60\,\Omega} =0.192\,\text{A}$
Find the terminal potential difference:
$VV=E-Ir=3.00\,\text{V}-(0.192\,\text{A}\times 0.60\,\Omega)=2.88\,\text{V}$
Check against the external circuit:
$VV=IR_{\text{external}}=0.192\,\text{A}\times 15.0\,\Omega=2.88\,\text{V}$

Specified practical: determining internal resistance

Set up a cell connected to a switch, ammeter and variable resistor in series. Place a voltmeter across the cell terminals. Change the resistance to get different currents, and record the current $I$ and terminal potential difference $V$ each time.

Plot $V$ on the vertical axis against $I$ on the horizontal axis. Compare:

V = E - I r

with the straight-line form:

y = m x + c

So the y-intercept gives $E$ , and the gradient gives $- r$ .

Example

Finding internal resistance from a graph

A graph of terminal potential difference against current has a y-intercept of 1.54 V. Two points on the best-fit line are (0.20 A, 1.42 V) and (1.00 A, 0.94 V). Find the emf and internal resistance.

Read the emf from the y-intercept:
$VE=1.54\,\text{V}$
Calculate the gradient of the graph:
$Ω\text{gradient}=\frac{0.94\,\text{V}-1.42\,\text{V}}{1.00\,\text{A}-0.20\,\text{A}} =\frac{-0.48\,\text{V}}{0.80\,\text{A}} =-0.60\,\Omega$
Since the gradient is $- r$ :
$Ωr=0.60\,\Omega$

For good practical technique, switch off between readings to reduce heating, avoid very low load resistances that could cause large currents, and choose meter ranges that give sensible precision. Record meter resolution, repeat readings where possible, and use error bars or steepest and shallowest acceptable gradients to estimate uncertainty in $r$ .

Exam technique

In the exam

Mark junctions and loops first: use current conservation at junctions and energy conservation around loops.
Reduce resistor networks in stages, starting with obvious series or parallel groups.
In potential dividers, identify exactly which resistor $VoutV_{\text{out}}$ is measured across before substituting.
For internal resistance graphs, plot $V$ against $I$ : intercept is $E$ , gradient is $- r$ .

Self review

Check yourself

Why does the current split in a parallel circuit but stay the same in a series circuit?
If an LDR is the lower resistor in a potential divider, what happens to $VoutV_{\text{out}}$ when light intensity increases?
On a graph of terminal potential difference against current, how would you find the internal resistance?

What you'll learn

How current and potential difference behave in series and parallel circuits.
How to combine resistors and calculate currents and voltages in mixed circuits.
How potential dividers work, including with LDRs and thermistors.
What emf and internal resistance mean, and how to determine internal resistance experimentally.

Prerequisites: the language of circuits

Definition

Current, potential difference and resistance

Current $I$ is the rate of flow of charge: $\frac{\Delta Q}{\Delta t}$ . Its unit is the ampere (A).
Potential difference $V$ is the energy transferred per unit charge between two points: $\frac{W}{Q}$ . Its unit is the volt (V).
Resistance $R$ measures how strongly a component opposes current: $V = I R$ . Its unit is the ohm (Ω).

Series and parallel: the conservation laws

A series circuit has one path for charge. The same current passes through every component.

A parallel circuit has branches. At a junction, charge splits between branches and then recombines.

Series and parallel circuit rules

Key Idea

Why the rules work

In a parallel circuit, the current from the source equals the sum of the currents in the branches because charge is conserved.
In a series circuit, the sum of the potential differences across the components equals the supply potential difference because energy is conserved.
Components in parallel have the same potential difference across them because each branch is connected across the same two points.

So for branches in parallel:

I_1 + I_2 + I_3 + \cdots

For components in series:

Vsupply=V1+V2+V3+⋯V_{\text{supply}} = V_1 + V_2 + V_3 + \cdots

For components in parallel:

V1=V2=VsupplyV_1 = V_2 = V_{\text{supply}}

Combining resistors

For resistors in series, resistances add:

R_1 + R_2 + R_3 + \cdots

For resistors in parallel, reciprocals add:

1R=1R1+1R2+1R3+⋯\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \cdots

Tip

Parallel resistance sanity check

The combined resistance of resistors in parallel is always less than the smallest individual resistance. If your answer is bigger, you have probably added the resistances as if they were in series.

Example

Finding currents in a mixed resistor circuit

A 12.0 V supply is connected to a 4.0 Ω resistor in series with a parallel combination of 6.0 Ω and 12.0 Ω resistors. Find the total current and the current in each parallel branch.

Combine the parallel pair first:
$Ω−1\frac{1}{R_{\text{parallel}}}=\frac{1}{6.0\,\Omega}+\frac{1}{12.0\,\Omega}=0.250\,\Omega^{-1}$
so
$ΩR_{\text{parallel}}=4.0\,\Omega$
Add the series resistor:
$ΩR_{\text{total}}=4.0\,\Omega+4.0\,\Omega=8.0\,\Omega$
Use $V = I R$ to find the current from the supply:
$AI=\frac{V}{R}=\frac{12.0\,\text{V}}{8.0\,\Omega}=1.50\,\text{A}$
The 4.0 Ω series resistor has potential difference:
$VV=IR=1.50\,\text{A}\times 4.0\,\Omega=6.0\,\text{V}$
so the parallel section also has 6.0 V across it.
Find the branch currents:
$AI_{6.0}=\frac{6.0\,\text{V}}{6.0\,\Omega}=1.00\,\text{A}$ $AI_{12.0}=\frac{6.0\,\text{V}}{12.0\,\Omega}=0.500\,\text{A}$
The branch currents add to 1.50 A, matching conservation of charge.

Common Mistake

Using the supply voltage on every component

Only components in parallel with the supply have the full supply potential difference across them. In series sections, the supply potential difference is shared between components.

Potential dividers

A potential divider uses resistors in series to produce a chosen fraction of the supply potential difference. The output is usually taken across one resistor.

Potential divider with sensor examples

For two resistors in series, with the output across $R_2$ :

Vout=R2R1+R2VinV_{\text{out}}=\frac{R_2}{R_1+R_2}V_{\text{in}}

This works because the same current passes through both resistors, so the larger resistance gets the larger share of the potential difference.

An LDR, or light-dependent resistor, has resistance that decreases as light intensity increases. An NTC thermistor has resistance that decreases as temperature increases.

Key Idea

Sensor position matters

Example

Calculating a potential divider output

Use the divider equation with the LDR as $R_2$ :
$Vout=R2R1+R2VinV_{\text{out}}=\frac{R_2}{R_1+R_2}V_{\text{in}}$
In bright light:
$VV_{\text{out}}=\frac{1.0\,\text{k}\Omega}{2.2\,\text{k}\Omega+1.0\,\text{k}\Omega}\times 9.0\,\text{V}=2.8\,\text{V}$
In the dark:
$VV_{\text{out}}=\frac{10\,\text{k}\Omega}{2.2\,\text{k}\Omega+10\,\text{k}\Omega}\times 9.0\,\text{V}=7.4\,\text{V}$
The output is larger in the dark because the LDR resistance is larger, so it takes a larger fraction of the supply potential difference.

Common Mistake

Loaded potential dividers

Emf and internal resistance

The emf $E$ of a source is the energy transferred by the source per unit charge. Its unit is the volt (V), the same as potential difference.

Definition

Emf

The emf $E$ of a source is the energy supplied by the source to each coulomb of charge passing through it. When no current is drawn, the terminal potential difference is equal to the emf.

The Eduqas equation is:

V = E - I r

Here, $I r$ is called the lost volts: the potential difference across the internal resistance.

Internal resistance practical circuit and graph

Example

Using internal resistance with cells in series

Add the emfs and internal resistances because the cells are in series and facing the same direction:
$VE_{\text{total}}=1.50\,\text{V}+1.50\,\text{V}=3.00\,\text{V}$ $Ωr_{\text{total}}=0.30\,\Omega+0.30\,\Omega=0.60\,\Omega$
Add the external series resistance:
$ΩR_{\text{external}}=5.0\,\Omega+10.0\,\Omega=15.0\,\Omega$
Use the total circuit resistance, including internal resistance:
$AI=\frac{E_{\text{total}}}{R_{\text{external}}+r_{\text{total}}} =\frac{3.00\,\text{V}}{15.0\,\Omega+0.60\,\Omega} =0.192\,\text{A}$
Find the terminal potential difference:
$VV=E-Ir=3.00\,\text{V}-(0.192\,\text{A}\times 0.60\,\Omega)=2.88\,\text{V}$
Check against the external circuit:
$VV=IR_{\text{external}}=0.192\,\text{A}\times 15.0\,\Omega=2.88\,\text{V}$

Specified practical: determining internal resistance

Plot $V$ on the vertical axis against $I$ on the horizontal axis. Compare:

V = E - I r

with the straight-line form:

y = m x + c

So the y-intercept gives $E$ , and the gradient gives $- r$ .

Example

Finding internal resistance from a graph

A graph of terminal potential difference against current has a y-intercept of 1.54 V. Two points on the best-fit line are (0.20 A, 1.42 V) and (1.00 A, 0.94 V). Find the emf and internal resistance.

Read the emf from the y-intercept:
$VE=1.54\,\text{V}$
Calculate the gradient of the graph:
$Ω\text{gradient}=\frac{0.94\,\text{V}-1.42\,\text{V}}{1.00\,\text{A}-0.20\,\text{A}} =\frac{-0.48\,\text{V}}{0.80\,\text{A}} =-0.60\,\Omega$
Since the gradient is $- r$ :
$Ωr=0.60\,\Omega$

Exam technique

In the exam

Mark junctions and loops first: use current conservation at junctions and energy conservation around loops.
Reduce resistor networks in stages, starting with obvious series or parallel groups.
In potential dividers, identify exactly which resistor $VoutV_{\text{out}}$ is measured across before substituting.
For internal resistance graphs, plot $V$ against $I$ : intercept is $E$ , gradient is $- r$ .

Self review

Check yourself

Why does the current split in a parallel circuit but stay the same in a series circuit?
If an LDR is the lower resistor in a potential divider, what happens to $VoutV_{\text{out}}$ when light intensity increases?
On a graph of terminal potential difference against current, how would you find the internal resistance?

What you'll learn

Prerequisites: the language of circuits

Series and parallel: the conservation laws

Combining resistors

Potential dividers

Emf and internal resistance

Specified practical: determining internal resistance

Review D.C. Circuits by teaching Genie

Remember key concepts with flashcards

D.C. Circuits

What you'll learn

Prerequisites: the language of circuits

Series and parallel: the conservation laws

Combining resistors

Potential dividers

Emf and internal resistance

Specified practical: determining internal resistance

Review D.C. Circuits by teaching Genie

Remember key concepts with flashcards

D.C. Circuits

What you'll learn

Prerequisites: the language of circuits

Series and parallel: the conservation laws

Combining resistors

Potential dividers

Emf and internal resistance

Specified practical: determining internal resistance

Review D.C. Circuits by teaching Genie

Remember key concepts with flashcards

D.C. Circuits Revision Guide

D.C. Circuits

What you'll learn

Prerequisites: the language of circuits

Series and parallel: the conservation laws

Combining resistors

Potential dividers

Emf and internal resistance

Specified practical: determining internal resistance

Review D.C. Circuits by teaching Genie

Remember key concepts with flashcards

D.C. Circuits Revision Guide