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Physics - Electricity: Simple series circuits and components

Grade 4IGCSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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A circuit is a closed loop that allows electric current to flow from the positive terminal of a power source to the negative terminal.

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In a series circuit, all components are connected end-to-end in a single path or loop.

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Electric Current (II) is the rate of flow of electric charge (QQ). It is measured in Amperes (AA) using an ammeter connected in series.

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Potential Difference (VV), often called voltage, is the energy transferred per unit charge as it moves between two points. It is measured in Volts (VV) using a voltmeter connected in parallel across a component.

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Resistance (RR) is the opposition to current flow, measured in Ohms (ΩΩ).

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In a series circuit, the current is the same at every point in the loop: Itotal=I1=I2=I3I_{total} = I_1 = I_2 = I_3.

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The total resistance in a series circuit is the sum of the individual resistances: Rtotal=R1+R2+…R_{total} = R_1 + R_2 + \dots.

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The total potential difference provided by the power supply is shared between the components: Vtotal=V1+V2+…V_{total} = V_1 + V_2 + \dots.

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If one component in a series circuit breaks (e.g., a bulb blows), the entire circuit is broken and current stops flowing.

📐Formulae

I=QtI = \frac{Q}{t}

V=I×RV = I \times R

Rtotal=R1+R2+R3+…R_{total} = R_1 + R_2 + R_3 + \dots

Vtotal=V1+V2+V3+…V_{total} = V_1 + V_2 + V_3 + \dots

💡Examples

Problem 1:

A series circuit consists of a 12V12V battery connected to two resistors: R1=4ΩR_1 = 4\Omega and R2=2ΩR_2 = 2\Omega. Calculate the total current flowing through the circuit.

Solution:

First, calculate total resistance: Rtotal=4Ω+2Ω=6ΩR_{total} = 4\Omega + 2\Omega = 6\Omega. Then, use Ohm's Law: I=VR=12V6Ω=2AI = \frac{V}{R} = \frac{12V}{6\Omega} = 2A.

Explanation:

In a series circuit, resistances are additive. Once the total resistance is known, the constant current through the loop is found using the supply voltage.

Problem 2:

If a current of 0.5A0.5A flows through a lamp for 22 minutes, how much charge QQ has passed through the lamp?

Solution:

Convert time to seconds: t=2×60=120st = 2 \times 60 = 120s. Use the formula Q=I×t=0.5A×120s=60CQ = I \times t = 0.5A \times 120s = 60C.

Explanation:

The unit of charge is the Coulomb (CC). Time must always be converted to the SI unit of seconds (ss) before calculation.

Problem 3:

In a series circuit with a 9V9V battery, the potential difference across a bulb is 6V6V. What is the potential difference across the remaining resistor in the circuit?

Solution:

Vresistor=Vtotal−Vbulb=9V−6V=3VV_{resistor} = V_{total} - V_{bulb} = 9V - 6V = 3V.

Explanation:

In a series circuit, the sum of the potential differences across all components must equal the total potential difference provided by the source.

Electricity: Simple series circuits and components Revision - Grade 4 Science IGCSE