What are the basic electrical quantities covered in the Cambridge IGCSE Coordinated Science syllabus?

The basic electrical quantities covered in the Cambridge IGCSE Coordinated Science syllabus include current, voltage, resistance, and power. Current is the flow of electric charge, measured in amperes (A). Voltage, or potential difference, is the energy per unit charge, measured in volts (V). Resistance is the opposition to the flow of current, measured in ohms (Ω). Power is the rate at which electrical energy is transferred by an electric circuit, measured in watts (W).

How is electrical current measured and what is its significance in circuits?

Electrical current is measured in amperes (A) using an ammeter, which is connected in series with the circuit. Current is significant because it represents the flow of electric charge through a conductor. Understanding current is crucial for analyzing how circuits function, as it affects the operation of electrical devices and the distribution of power within a circuit.

What is Ohm's Law and how does it relate to electrical quantities?

Ohm's Law is a fundamental principle in electricity that relates the electrical quantities of voltage (V), current (I), and resistance (R) in a circuit. It is expressed by the formula V = IR, meaning that the voltage across a conductor is equal to the product of the current flowing through it and its resistance. This law is essential for calculating one of these quantities if the other two are known, and it helps in designing and analyzing electrical circuits.

Why is understanding resistance important in electrical circuits?

Understanding resistance is important because it determines how much current will flow through a circuit for a given voltage. High resistance means less current flow, while low resistance allows more current. Resistance affects the efficiency and safety of electrical devices and circuits. It is crucial for designing circuits that operate correctly and safely, ensuring that components do not overheat or become damaged.

How do you calculate electrical power in a circuit and why is it important?

Electrical power in a circuit is calculated using the formula P = VI, where P is power in watts (W), V is voltage in volts (V), and I is current in amperes (A). Power can also be calculated using P = I²R or P = V²/R, depending on the known quantities. Understanding power is important because it indicates the rate at which energy is consumed or produced by a device. This helps in determining the energy efficiency of electrical appliances and ensuring that circuits are designed to handle the required power levels safely.

Why is "Using time in minutes instead of seconds in $Q = It$" a common mistake in Electrical Quantities?

Why it happens: Students substitute the number as given without checking units. How to avoid it: Always check units. Q = It requires t in seconds. Convert: 1 minute = 60 s.

Why is "Applying Ohm's law without stating the constant-temperature condition" a common mistake in Electrical Quantities?

Why it happens: Students know V = IR but not when it applies. How to avoid it: Always state 'provided temperature (physical conditions) remain constant' when writing Ohm's law. Otherwise resistance changes with temperature (as in a filament lamp).

Why is "Writing $P = VI^{2}$ instead of $P = I^{2}R$ or $P = VI$" a common mistake in Electrical Quantities?

Why it happens: Garbling multiple power formulas together. How to avoid it: The three power formulas: P = IV; P = I^2R (when V not known); P = V^2/R (when I not known). Do not mix terms from different versions.

Cambridge IGCSE Coordinated Science (0654)

Electrical Quantities

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Short Notes - Electrical Quantities

Detailed Study Notes

Detailed notes on Electricity and Magnetism for Cambridge IGCSE Coordinated Science, covering key concepts, explanations, examples, and exam-focused revision points.

Electrical Quantities — Cambridge IGCSE Coordinated Science 0654

Charge, current, voltage, resistance, and power are the fundamental quantities in electricity. Cambridge tests Ohm's law (V = IR), power equations (P = IV, P = I²R), charge (Q = It), and how to measure these quantities in circuits.

At a glance

Electric current I = Q/t (amperes). Conventional current flows from + to −; electrons flow from − to +.
Voltage (potential difference) V: energy transferred per unit charge. V = W/Q. Unit: volts (V).
Resistance R = V/I (ohms, Ω). Ohm's law: V ∝ I at constant temperature.
Power: P = IV = I²R = V²/R (watts). Energy: E = Pt = IVt.
I-V graphs: ohmic conductor → straight line through origin. Filament lamp → S-shaped (resistance increases with temperature). Diode → conducts in one direction only.

What you’ll learn

Mapped to the Cambridge IGCSE 0654 syllabus (2025-2027).

P5.2a — Define current, voltage, and resistance; recall and use V = IR.
P5.2b — Use power equations P = IV, P = I²R.
P5.2c — Describe I-V characteristics of different components.
P5.2d — Use Q = It.

Current, Voltage, and Resistance

The three fundamental quantities; V = IR links them all.

Electric charge (Q):

Measured in coulombs (C)
Electrons carry negative charge; protons carry positive charge

Electric current (I):

I = Q / t

I = current (A), Q = charge (C), t = time (s)
Rate of flow of electric charge
Conventional current flows from + (high potential) to − (low potential)
Electrons actually flow in opposite direction (from − to +)
Measured with an ammeter (connected in series)

Potential difference (voltage, V):

V = W / Q

V = potential difference (volts, V), W = energy (J), Q = charge (C)
Energy transferred per unit charge between two points
Measured with a voltmeter (connected in parallel)

Resistance (R):

V = IR (Ohm's Law)

R = resistance (ohms, Ω), V = pd (V), I = current (A)
Resistance = opposition to flow of current
Caused by: collisions between charge carriers (electrons) and positive ions in the conductor
Higher temperature → more collisions → higher resistance (for metals)

Ohm's Law: At constant temperature, V is proportional to I.

I-V characteristics:

Ohmic conductor (e.g., resistor at constant temperature): straight line through origin → constant R
Filament lamp: S-shaped curve (resistance increases as temperature rises at higher currents)
Diode: conducts only in forward bias (I flows); very high resistance in reverse bias
Thermistor: resistance decreases as temperature increases (NTC — negative temperature coefficient)

I = Q/t (A). V = W/Q (V). R = V/I (Ω).
Ammeter in series; voltmeter in parallel.
Ohmic: V ∝ I (straight line). Filament: R increases with T.

Electrical Power and Energy

P = IV and its variants are essential for any calculation involving power or energy in circuits.

Electrical power:

P = IV P = I²R (substitute V = IR) P = V²/R (substitute I = V/R)

P = power (W), I = current (A), V = potential difference (V), R = resistance (Ω)

Electrical energy:

E = Pt = IVt

E = energy (J), t = time (s)

Example 1: A 12 V lamp carries 0.5 A. Find power.

P = IV = 12 × 0.5 = 6 W

Example 2: A 60 W lamp is on for 2 hours. Find energy consumed.

E = Pt = 60 × (2 × 3600) = 432 000 J = 432 kJ

Kilowatt-hour (kWh):

Practical unit of electrical energy used in household billing
1 kWh = 1000 W × 3600 s = 3 600 000 J = 3.6 MJ
Cost = power (kW) × time (h) × price per kWh

Comparing efficiency of electrical devices:

A 10 W LED gives same light output as 60 W incandescent → LED is more efficient (less energy wasted as heat)

Charge flow:

Q = It

E.g., If 2 A flows for 30 s: Q = 2 × 30 = 60 C

P = IV = I²R = V²/R. Choose formula based on given quantities.
E = Pt (J). 1 kWh = 3.6 MJ.
Q = It (coulombs).

Quick recap

I = Q/t (A). V = W/Q (V). R = V/I (Ω). Ohm's law: V = IR at constant temperature.
P = IV = I²R = V²/R. E = Pt = IVt.
I-V graphs: resistor → linear; filament → S-curve; diode → one-way.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

V = IR — Ohm's Law (volts = amps × ohms).
P = IV — power (W) = current (A) × voltage (V).
Q = It — charge (C) = current (A) × time (s).
Ammeter: series. Voltmeter: parallel.
Diode — conducts in one direction only (forward bias).

How it’s examined

Paper 4: 'A resistor has a resistance of 15 Ω and a current of 0.4 A. Calculate the voltage across it' (2 marks — V = 0.4 × 15 = 6 V). 'Calculate the power dissipated' (1 mark — P = IV = 6 × 0.4 = 2.4 W). 'A lamp operates at 230 V and 0.26 A. Calculate the charge flowing in 10 minutes' (2 marks — Q = It = 0.26 × 600 = 156 C). I-V graph: 'Explain why the I-V graph of a filament lamp is not a straight line' (2 marks — as current increases, temperature rises; resistance increases; less current flows than expected for a given V).

Sources: Cambridge IGCSE Coordinated Sciences 0654 syllabus 2025-2027 (P5); 0654 Examiner Reports 2022-2024. Last reviewed 2026-05-14.

Take this whole topic with you

Step-by-step worked examples — Electrical Quantities

Step-by-step solutions to past-paper-style questions on electrical quantities, written exactly the way a tutor would explain them at the board.

1Calculating charge from current and time
Core• charge, current, Q=It
Question
A current of $2.5\,\text{A}$ flows through a wire for $4.0\,\text{minutes}$ . Calculate the total charge that passes a cross-section of the wire.
Step-by-step solution
1. Step 1
  Convert time to seconds.
  $t = 4.0 \times 60 = 240\,\text{s}$
2. Step 2
  Apply $Q = It$ .
  $Q = 2.5 \times 240 = 600\,\text{C}$
Answer
$Q = 600\,\text{C}$
2Ohm's law and resistance
Extended• Ohm's law, resistance, R=V/I
Question
A potential difference of $12\,\text{V}$ is applied across a resistor. A current of $0.40\,\text{A}$ flows. (a) Calculate the resistance. (b) If the potential difference is doubled (temperature unchanged), predict the new current.
Step-by-step solution
1. Step 1
  Calculate resistance using $R = V/I$ .
  $R = \dfrac{12}{0.40} = 30\,\Omega$
2. Step 2
  If temperature is unchanged, the resistor is ohmic — $R$ is constant at $30\,\Omega$ . New PD $= 24\,\text{V}$ .
  $I_{\text{new}} = \dfrac{V}{R} = \dfrac{24}{30} = 0.80\,\text{A}$
Answer
$R = 30\,\Omega$ ; new current $= 0.80\,\text{A}$
3Interpreting I-V characteristics
Challenge• I-V characteristics, filament lamp, diode, Ohm's law
Question
Sketch I-V characteristic curves for (a) a fixed resistor, (b) a filament lamp, and (c) a diode. Explain the shape of each curve.
Step-by-step solution
1. Step 1
  (a) Fixed resistor: straight line through the origin (symmetric for positive and negative V). Constant gradient = constant resistance = constant $V/I$ → ohmic conductor.
2. Step 2
  (b) Filament lamp: S-shaped curve, symmetric. As current increases, filament heats up → resistance increases → gradient decreases (current increases less steeply). Not ohmic.
3. Step 3
  (c) Diode: for forward bias (V > ~0.6 V for silicon), current increases steeply. For reverse bias, current $\approx 0$ (blocked). The I-V graph is highly asymmetric. Not ohmic.
Answer
Resistor: straight line (ohmic). Filament lamp: S-curve (resistance increases with temperature). Diode: asymmetric — conducts in forward direction only.

Key Formulae — Electrical Quantities

The formulae you need to memorise for electrical quantities on the Cambridge IGCSE 0654 paper, with every variable defined in plain English and a note on when to use it.

Electric charge
$Q = It$
$Q$
charge (C)
$I$
current (A)
$t$
time (s)
When to use
Finding total charge from current and time.
Potential difference
$V = \dfrac{W}{Q}$
$V$
potential difference (V)
$W$
work done / energy transferred (J)
$Q$
charge (C)
When to use
Relating potential difference to energy transferred per unit charge.
Resistance (Ohm's law)
$R = \dfrac{V}{I}$
$R$
resistance (Ω)
$V$
potential difference (V)
$I$
current (A)
When to use
Finding resistance, or checking whether a component is ohmic.
Electrical power
$P = IV = I^{2}R = \dfrac{V^{2}}{R}$
$P$
power (W)
$I$
current (A)
$V$
potential difference (V)
$R$
resistance (Ω)
When to use
Calculating power dissipated in a component when any two of $I$ , $V$ , $R$ are known.

Key Definitions and Keywords — Electrical Quantities

Definitions to memorise and the exact keywords mark schemes credit for electrical quantities answers — sharpened from recent examiner reports for the 2026 0654 sitting.

Electric current
Examiner keyword
The rate of flow of charge. $I = Q/t$ ; SI unit: ampere (A), where $1\,\text{A} = 1\,\text{C/s}$ .
Potential difference (voltage)
Examiner keyword
The energy transferred per unit charge between two points in a circuit. $V = W/Q$ ; SI unit: volt (V), where $1\,\text{V} = 1\,\text{J/C}$ .
Resistance
Examiner keyword
The opposition to the flow of current. $R = V/I$ ; SI unit: ohm ( $\Omega$ ).
Ohm's law
Examiner keyword
The current through a conductor is directly proportional to the potential difference across it, provided temperature (and other physical conditions) remain constant.
Ohmic conductor
Examiner keyword
A component that obeys Ohm's law — its resistance is constant at constant temperature. A straight-line $I$ - $V$ graph through the origin.
Related: Ohm's law, resistance

Common Mistakes and Misconceptions — Electrical Quantities

The traps other students keep falling into on electrical quantities questions — taken from recent Cambridge IGCSE 0654 examiner reports and mark schemes — and how to avoid them.

✕Using time in minutes instead of seconds in $Q = It$
Why it happens
Students substitute the number as given without checking units.
How to avoid it
Always check units. $Q = It$ requires $t$ in seconds. Convert: 1 minute = 60 s.
✕Applying Ohm's law without stating the constant-temperature condition
Why it happens
Students know $V = IR$ but not when it applies.
How to avoid it
Always state 'provided temperature (physical conditions) remain constant' when writing Ohm's law. Otherwise resistance changes with temperature (as in a filament lamp).
✕Writing $P = VI^{2}$ instead of $P = I^{2}R$ or $P = VI$
Why it happens
Garbling multiple power formulas together.
How to avoid it
The three power formulas: $P = IV$ ; $P = I^{2}R$ (when $V$ not known); $P = V^{2}/R$ (when $I$ not known). Do not mix terms from different versions.

Electrical Quantities — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge IGCSE Coordinated Science (0654)

Electrical Quantities

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3 tasks remaining

Short Notes - Electrical Quantities

Detailed Study Notes

Detailed notes on Electricity and Magnetism for Cambridge IGCSE Coordinated Science, covering key concepts, explanations, examples, and exam-focused revision points.

Electrical Quantities — Cambridge IGCSE Coordinated Science 0654

At a glance

Electric current I = Q/t (amperes). Conventional current flows from + to −; electrons flow from − to +.
Voltage (potential difference) V: energy transferred per unit charge. V = W/Q. Unit: volts (V).
Resistance R = V/I (ohms, Ω). Ohm's law: V ∝ I at constant temperature.
Power: P = IV = I²R = V²/R (watts). Energy: E = Pt = IVt.
I-V graphs: ohmic conductor → straight line through origin. Filament lamp → S-shaped (resistance increases with temperature). Diode → conducts in one direction only.

What you’ll learn

Mapped to the Cambridge IGCSE 0654 syllabus (2025-2027).

P5.2a — Define current, voltage, and resistance; recall and use V = IR.
P5.2b — Use power equations P = IV, P = I²R.
P5.2c — Describe I-V characteristics of different components.
P5.2d — Use Q = It.

Current, Voltage, and Resistance

The three fundamental quantities; V = IR links them all.

Electric charge (Q):

Measured in coulombs (C)
Electrons carry negative charge; protons carry positive charge

Electric current (I):

I = Q / t

I = current (A), Q = charge (C), t = time (s)
Rate of flow of electric charge
Conventional current flows from + (high potential) to − (low potential)
Electrons actually flow in opposite direction (from − to +)
Measured with an ammeter (connected in series)

Potential difference (voltage, V):

V = W / Q

V = potential difference (volts, V), W = energy (J), Q = charge (C)
Energy transferred per unit charge between two points
Measured with a voltmeter (connected in parallel)

Resistance (R):

V = IR (Ohm's Law)

R = resistance (ohms, Ω), V = pd (V), I = current (A)
Resistance = opposition to flow of current
Caused by: collisions between charge carriers (electrons) and positive ions in the conductor
Higher temperature → more collisions → higher resistance (for metals)

Ohm's Law: At constant temperature, V is proportional to I.

I-V characteristics:

Ohmic conductor (e.g., resistor at constant temperature): straight line through origin → constant R
Filament lamp: S-shaped curve (resistance increases as temperature rises at higher currents)
Diode: conducts only in forward bias (I flows); very high resistance in reverse bias
Thermistor: resistance decreases as temperature increases (NTC — negative temperature coefficient)

I = Q/t (A). V = W/Q (V). R = V/I (Ω).
Ammeter in series; voltmeter in parallel.
Ohmic: V ∝ I (straight line). Filament: R increases with T.

Electrical Power and Energy

P = IV and its variants are essential for any calculation involving power or energy in circuits.

Electrical power:

P = IV P = I²R (substitute V = IR) P = V²/R (substitute I = V/R)

P = power (W), I = current (A), V = potential difference (V), R = resistance (Ω)

Electrical energy:

E = Pt = IVt

E = energy (J), t = time (s)

Example 1: A 12 V lamp carries 0.5 A. Find power.

P = IV = 12 × 0.5 = 6 W

Example 2: A 60 W lamp is on for 2 hours. Find energy consumed.

E = Pt = 60 × (2 × 3600) = 432 000 J = 432 kJ

Kilowatt-hour (kWh):

Practical unit of electrical energy used in household billing
1 kWh = 1000 W × 3600 s = 3 600 000 J = 3.6 MJ
Cost = power (kW) × time (h) × price per kWh

Comparing efficiency of electrical devices:

A 10 W LED gives same light output as 60 W incandescent → LED is more efficient (less energy wasted as heat)

Charge flow:

Q = It

E.g., If 2 A flows for 30 s: Q = 2 × 30 = 60 C

P = IV = I²R = V²/R. Choose formula based on given quantities.
E = Pt (J). 1 kWh = 3.6 MJ.
Q = It (coulombs).

Quick recap

I = Q/t (A). V = W/Q (V). R = V/I (Ω). Ohm's law: V = IR at constant temperature.
P = IV = I²R = V²/R. E = Pt = IVt.
I-V graphs: resistor → linear; filament → S-curve; diode → one-way.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

V = IR — Ohm's Law (volts = amps × ohms).
P = IV — power (W) = current (A) × voltage (V).
Q = It — charge (C) = current (A) × time (s).
Ammeter: series. Voltmeter: parallel.
Diode — conducts in one direction only (forward bias).

How it’s examined

Sources: Cambridge IGCSE Coordinated Sciences 0654 syllabus 2025-2027 (P5); 0654 Examiner Reports 2022-2024. Last reviewed 2026-05-14.

Take this whole topic with you

Step-by-step worked examples — Electrical Quantities

Step-by-step solutions to past-paper-style questions on electrical quantities, written exactly the way a tutor would explain them at the board.

1Calculating charge from current and time
Core• charge, current, Q=It
Question
A current of $2.5\,\text{A}$ flows through a wire for $4.0\,\text{minutes}$ . Calculate the total charge that passes a cross-section of the wire.
Step-by-step solution
1. Step 1
  Convert time to seconds.
  $t = 4.0 \times 60 = 240\,\text{s}$
2. Step 2
  Apply $Q = It$ .
  $Q = 2.5 \times 240 = 600\,\text{C}$
Answer
$Q = 600\,\text{C}$
2Ohm's law and resistance
Extended• Ohm's law, resistance, R=V/I
Question
A potential difference of $12\,\text{V}$ is applied across a resistor. A current of $0.40\,\text{A}$ flows. (a) Calculate the resistance. (b) If the potential difference is doubled (temperature unchanged), predict the new current.
Step-by-step solution
1. Step 1
  Calculate resistance using $R = V/I$ .
  $R = \dfrac{12}{0.40} = 30\,\Omega$
2. Step 2
  If temperature is unchanged, the resistor is ohmic — $R$ is constant at $30\,\Omega$ . New PD $= 24\,\text{V}$ .
  $I_{\text{new}} = \dfrac{V}{R} = \dfrac{24}{30} = 0.80\,\text{A}$
Answer
$R = 30\,\Omega$ ; new current $= 0.80\,\text{A}$
3Interpreting I-V characteristics
Challenge• I-V characteristics, filament lamp, diode, Ohm's law
Question
Sketch I-V characteristic curves for (a) a fixed resistor, (b) a filament lamp, and (c) a diode. Explain the shape of each curve.
Step-by-step solution
1. Step 1
  (a) Fixed resistor: straight line through the origin (symmetric for positive and negative V). Constant gradient = constant resistance = constant $V/I$ → ohmic conductor.
2. Step 2
  (b) Filament lamp: S-shaped curve, symmetric. As current increases, filament heats up → resistance increases → gradient decreases (current increases less steeply). Not ohmic.
3. Step 3
  (c) Diode: for forward bias (V > ~0.6 V for silicon), current increases steeply. For reverse bias, current $\approx 0$ (blocked). The I-V graph is highly asymmetric. Not ohmic.
Answer
Resistor: straight line (ohmic). Filament lamp: S-curve (resistance increases with temperature). Diode: asymmetric — conducts in forward direction only.

Key Formulae — Electrical Quantities

The formulae you need to memorise for electrical quantities on the Cambridge IGCSE 0654 paper, with every variable defined in plain English and a note on when to use it.

Electric charge
$Q = It$
$Q$
charge (C)
$I$
current (A)
$t$
time (s)
When to use
Finding total charge from current and time.
Potential difference
$V = \dfrac{W}{Q}$
$V$
potential difference (V)
$W$
work done / energy transferred (J)
$Q$
charge (C)
When to use
Relating potential difference to energy transferred per unit charge.
Resistance (Ohm's law)
$R = \dfrac{V}{I}$
$R$
resistance (Ω)
$V$
potential difference (V)
$I$
current (A)
When to use
Finding resistance, or checking whether a component is ohmic.
Electrical power
$P = IV = I^{2}R = \dfrac{V^{2}}{R}$
$P$
power (W)
$I$
current (A)
$V$
potential difference (V)
$R$
resistance (Ω)
When to use
Calculating power dissipated in a component when any two of $I$ , $V$ , $R$ are known.

Key Definitions and Keywords — Electrical Quantities

Definitions to memorise and the exact keywords mark schemes credit for electrical quantities answers — sharpened from recent examiner reports for the 2026 0654 sitting.

Electric current
Examiner keyword
The rate of flow of charge. $I = Q/t$ ; SI unit: ampere (A), where $1\,\text{A} = 1\,\text{C/s}$ .
Potential difference (voltage)
Examiner keyword
The energy transferred per unit charge between two points in a circuit. $V = W/Q$ ; SI unit: volt (V), where $1\,\text{V} = 1\,\text{J/C}$ .
Resistance
Examiner keyword
The opposition to the flow of current. $R = V/I$ ; SI unit: ohm ( $\Omega$ ).
Ohm's law
Examiner keyword
The current through a conductor is directly proportional to the potential difference across it, provided temperature (and other physical conditions) remain constant.
Ohmic conductor
Examiner keyword
A component that obeys Ohm's law — its resistance is constant at constant temperature. A straight-line $I$ - $V$ graph through the origin.
Related: Ohm's law, resistance

Common Mistakes and Misconceptions — Electrical Quantities

The traps other students keep falling into on electrical quantities questions — taken from recent Cambridge IGCSE 0654 examiner reports and mark schemes — and how to avoid them.

✕Using time in minutes instead of seconds in $Q = It$
Why it happens
Students substitute the number as given without checking units.
How to avoid it
Always check units. $Q = It$ requires $t$ in seconds. Convert: 1 minute = 60 s.
✕Applying Ohm's law without stating the constant-temperature condition
Why it happens
Students know $V = IR$ but not when it applies.
How to avoid it
Always state 'provided temperature (physical conditions) remain constant' when writing Ohm's law. Otherwise resistance changes with temperature (as in a filament lamp).
✕Writing $P = VI^{2}$ instead of $P = I^{2}R$ or $P = VI$
Why it happens
Garbling multiple power formulas together.
How to avoid it
The three power formulas: $P = IV$ ; $P = I^{2}R$ (when $V$ not known); $P = V^{2}/R$ (when $I$ not known). Do not mix terms from different versions.

Electrical Quantities — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Electrical Quantities

Short Notes - Electrical Quantities

Detailed Study Notes

At a glance

What you’ll learn

Current, Voltage, and Resistance

Electrical Power and Energy

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Calculating charge from current and time

2Ohm's law and resistance

3Interpreting I-V characteristics

Electric charge

Potential difference

Resistance (Ohm's law)

Electrical power

Electric current

Potential difference (voltage)

Resistance

Ohm's law

Ohmic conductor

✕Using time in minutes instead of seconds in Q=ItQ = ItQ=It

✕Applying Ohm's law without stating the constant-temperature condition

✕Writing P=VI2P = VI^{2}P=VI2 instead of P=I2RP = I^{2}RP=I2R or P=VIP = VIP=VI

Electrical Quantities — frequently asked questions

Electrical Quantities

Short Notes - Electrical Quantities

Detailed Study Notes

At a glance

What you’ll learn

Current, Voltage, and Resistance

Electrical Power and Energy

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Calculating charge from current and time

2Ohm's law and resistance

3Interpreting I-V characteristics

Electric charge

Potential difference

Resistance (Ohm's law)

Electrical power

Electric current

Potential difference (voltage)

Resistance

Ohm's law

Ohmic conductor

✕Using time in minutes instead of seconds in Q=ItQ = ItQ=It

✕Applying Ohm's law without stating the constant-temperature condition

✕Writing P=VI2P = VI^{2}P=VI2 instead of P=I2RP = I^{2}RP=I2R or P=VIP = VIP=VI

Electrical Quantities — frequently asked questions

✕Using time in minutes instead of seconds in $Q = It$

✕Writing $P = VI^{2}$ instead of $P = I^{2}R$ or $P = VI$

✕Using time in minutes instead of seconds in $Q = It$

✕Writing $P = VI^{2}$ instead of $P = I^{2}R$ or $P = VI$