krit.club logo

Physics - Electricity: Circuits and Symbols

Grade 7IGCSE

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

🔑Concepts

Electric Current (II): The rate of flow of electric charge (electrons) around a circuit. It is measured in Amperes (AA) using an Ammeter which must always be connected in series.

Potential Difference (VV): Also known as voltage, it represents the energy transferred per unit charge between two points in a circuit. It is measured in Volts (VV) using a Voltmeter connected in parallel.

Resistance (RR): The property of a component to oppose or slow down the flow of electric current. It is measured in Ohms (Ω\Omega).

Circuit Symbols: Key symbols include the Cell (single unit), Battery (multiple cells), Switch (open or closed), Lamp (a circle with a cross), Ammeter (circle with 'AA'), Voltmeter (circle with 'VV'), and Resistor (a rectangle).

Series Circuits: A circuit with only one path for current. The current is the same at any point (I1=I2=I3I_1 = I_2 = I_3), but the total voltage is shared between components.

Parallel Circuits: A circuit with multiple branches or paths. The voltage across each branch is the same (V1=V2V_1 = V_2), but the total current is split between the branches (Itotal=I1+I2I_{total} = I_1 + I_2).

📐Formulae

V=I×RV = I \times R

I=VRI = \frac{V}{R}

R=VIR = \frac{V}{I}

Rtotal=R1+R2+R3... (for series circuits)R_{total} = R_1 + R_2 + R_3... \text{ (for series circuits)}

💡Examples

Problem 1:

A 9V9V battery is connected to a resistor. An ammeter shows a current of 0.5A0.5A. Calculate the resistance of the resistor.

Solution:

R=VI=9V0.5A=18ΩR = \frac{V}{I} = \frac{9V}{0.5A} = 18\Omega

Explanation:

By applying the Ohm's Law formula R=VIR = \frac{V}{I}, we divide the potential difference by the current to find the resistance in Ohms.

Problem 2:

Two resistors, R1=4ΩR_1 = 4\Omega and R2=6ΩR_2 = 6\Omega, are connected in a series circuit. What is the total resistance provided to the battery?

Solution:

Rtotal=R1+R2=4Ω+6Ω=10ΩR_{total} = R_1 + R_2 = 4\Omega + 6\Omega = 10\Omega

Explanation:

In a series circuit, the total resistance is the sum of all individual resistances along the path.

Problem 3:

A circuit has a resistance of 20Ω20\Omega and a voltage source of 240V240V. Calculate the current flowing through the circuit.

Solution:

I=VR=240V20Ω=12AI = \frac{V}{R} = \frac{240V}{20\Omega} = 12A

Explanation:

Using the rearranged Ohm's Law formula I=VRI = \frac{V}{R}, the current is calculated as 1212 Amperes.