Current, Voltage, and Resistance

I = Q/t; V = IR; measuring with ammeters and voltmeters

# Current, Voltage, and Resistance — GCSE Physics

Electricity powers virtually everything in modern life. Understanding the three fundamental electrical quantities — current, voltage, and resistance — and how they relate through Ohm's Law is essential for GCSE Physics.


1. Electric Current ($I$)

Current is the rate of flow of electric charge.

I=Qt\boxed{I = \frac{Q}{t}}

Where:

  • II = current (amperes, A)
  • QQ = charge (coulombs, C)
  • tt = time (seconds, s)

1 ampere = 1 coulomb per second (1 A = 1 C/s)

Key Points

  • Current is caused by the flow of electrons (in metals) or ions (in solutions)
  • Conventional current flows from positive to negative (historical convention)
  • Electron flow is from negative to positive (actual flow in metals)
  • Current is measured with an ammeter connected in series

2. Potential Difference / Voltage ($V$)

Potential difference (p.d. or voltage) is the energy transferred per unit of charge.

V=EQ\boxed{V = \frac{E}{Q}}

Where:

  • VV = potential difference (volts, V)
  • EE = energy transferred (joules, J)
  • QQ = charge (coulombs, C)

1 volt = 1 joule per coulomb (1 V = 1 J/C)

Key Points

  • Voltage drives the current — it's the "push" that moves charges
  • Measured with a voltmeter connected in parallel across the component
  • The battery provides the voltage (EMF — electromotive force)
  • Voltage is "used up" across components in a circuit

3. Resistance ($R$)

Resistance is the opposition to the flow of current.

V=I×R\boxed{V = I \times R}

(This is Ohm's Law — valid for ohmic conductors at constant temperature)

Where:

  • VV = potential difference (V)
  • II = current (A)
  • RR = resistance (ohms, Ω)

Rearranging

I=VRR=VII = \frac{V}{R} \qquad R = \frac{V}{I}

Defining the Ohm

1 ohm is the resistance when a p.d. of 1 V causes a current of 1 A.

1 Ω=1 V/A1 \text{ Ω} = 1 \text{ V/A}

What Causes Resistance?

  • Electrons collide with positive ions in the metal lattice
  • These collisions transfer energy (heating the wire)
  • More collisions = higher resistance

Factors Affecting Resistance of a Wire

R=ρLAR = \frac{\rho L}{A}

Factor Effect on Resistance
Length (LL) Longer wire → more resistance (directly proportional)
Cross-sectional area (AA) Thinner wire → more resistance (inversely proportional)
Material (ρ\rho) Different materials have different resistivities
Temperature Higher temperature → more resistance (for metals)

4. Ohm's Law and I-V Characteristics

Ohm's Law: For an ohmic conductor (at constant temperature), the current is directly proportional to the potential difference.

I-V Graphs

Ohmic conductor (resistor at constant temp): Straight line through origin. Resistance is constant.

Filament lamp: Curved line. As current increases, the filament heats up, resistance increases. Curve gets shallower.

Diode: Almost no current flows in reverse direction. In forward direction, current flows once threshold voltage (~0.7 V) is reached. Very low resistance in forward bias.

Required Practical: I-V Characteristics

  1. Set up a circuit with the component, ammeter (series), voltmeter (parallel), and variable resistor
  2. Adjust the variable resistor to change the current and voltage
  3. Record I and V readings
  4. Reverse the battery connections and repeat (for negative values)
  5. Plot an I-V graph

Worked Example: Current

Problem

Question: A charge of 600 C passes through a circuit in 2 minutes. Calculate the current.

I=Qt=600120=5 AI = \frac{Q}{t} = \frac{600}{120} = 5 \text{ A}

Solution

Worked Example: Ohm's Law

Problem

Question: A 12 V battery is connected to a 4 Ω resistor. Calculate the current.

I=VR=124=3 AI = \frac{V}{R} = \frac{12}{4} = 3 \text{ A}

Solution

Worked Example: Finding Resistance

Problem

Question: A lamp has a current of 2.5 A when the p.d. is 230 V. Calculate the resistance.

R=VI=2302.5=92 ΩR = \frac{V}{I} = \frac{230}{2.5} = 92 \text{ Ω}

Solution

Worked Example: Charge

Problem

Question: A current of 0.5 A flows for 10 minutes. Calculate the charge.

Q=I×t=0.5×600=300 CQ = I \times t = 0.5 \times 600 = 300 \text{ C}

Solution

Worked Example: Energy

Problem

Question: A 9 V battery delivers 450 C of charge. Calculate the energy transferred.

E=V×Q=9×450=4050 JE = V \times Q = 9 \times 450 = 4050 \text{ J}


Solution

6. Practice Questions

    1. Calculate the current when 1200 C of charge passes a point in 4 minutes. (2 marks)
    1. A resistor has a resistance of 50 Ω. The current through it is 0.4 A. Calculate the voltage. (2 marks)
    1. Describe how you would connect an ammeter and voltmeter to measure the resistance of a lamp. (2 marks)
    1. A filament lamp is connected to a power supply. Explain why its resistance increases as the voltage increases. (3 marks)
    1. A diode is connected in a circuit with a 6 V battery. Explain why reversing the battery stops the current flowing. (2 marks)

    Answers

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Frequently Asked Questions

Is Ohm's Law always true?

No — it only applies to ohmic conductors at constant temperature. Non-ohmic components (filament lamps, diodes, thermistors) don't obey Ohm's Law.

Does current get "used up"?

No! Current is the same everywhere in a series circuit. It's the energy (voltage) that gets transferred across components.

Which way does current really flow?

Electrons flow from negative to positive. Conventional current (used in circuit diagrams) flows from positive to negative. Both conventions give the same results.


Summary

  • Current I=Q/tI = Q/t (amperes) — rate of flow of charge
  • Voltage V=E/QV = E/Q (volts) — energy per unit charge
  • Resistance R=V/IR = V/I (ohms) — opposition to current
  • Ohm's Law: V=IRV = IR (for ohmic conductors at constant temperature)
  • Ammeter in series; voltmeter in parallel
  • I-V graphs: linear for ohmic resistors, curved for lamps, one-way for diodes

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