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Physics - Electricity and Simple Circuits

Grade 8IGCSE

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

🔑Concepts

Electric charge (QQ) is measured in Coulombs (CC). It can be positive or negative; like charges repel and opposite charges attract.

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

Potential Difference (VV), or voltage, is the work done (energy transferred) per unit charge. It is measured in Volts (VV) using a voltmeter connected in parallel.

Resistance (RR) is the property of a component that opposes the flow of current, measured in Ohms (Ω\Omega).

In a series circuit, there is only one path for the current. The current II is the same at all points, but the total voltage VtotalV_{total} is shared between components.

In a parallel circuit, there are multiple branches. The voltage VV is the same across each branch, but the total current ItotalI_{total} is the sum of the currents in each branch.

Conductors (like copper) allow electrons to flow easily, whereas insulators (like plastic) have a very high resistance and prevent the flow of charge.

📐Formulae

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

V=WQV = \frac{W}{Q}

V=I×RV = I \times R

Rtotal=R1+R2+...R_{total} = R_1 + R_2 + ...

1Rtotal=1R1+1R2+...\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} + ...

💡Examples

Problem 1:

A charge of 30 C30\ C passes through a light bulb in 15 seconds15\ seconds. Calculate the current flowing through the bulb.

Solution:

I=Qt=30 C15 s=2 AI = \frac{Q}{t} = \frac{30\ C}{15\ s} = 2\ A

Explanation:

To find current, divide the total charge by the time in seconds. The unit is Amperes (AA).

Problem 2:

Calculate the resistance of a component if a potential difference of 12 V12\ V causes a current of 0.5 A0.5\ A to flow through it.

Solution:

R=VI=12 V0.5 A=24 ΩR = \frac{V}{I} = \frac{12\ V}{0.5\ A} = 24\ \Omega

Explanation:

Using Ohm's Law (V=IRV = IR), we rearrange the formula to R=VIR = \frac{V}{I} to find the resistance in Ohms.

Problem 3:

Two resistors, R1=4 ΩR_1 = 4\ \Omega and R2=6 ΩR_2 = 6\ \Omega, are connected in series to a 20 V20\ V power supply. Calculate the total resistance and the current in the circuit.

Solution:

Rtotal=4 Ω+6 Ω=10 ΩR_{total} = 4\ \Omega + 6\ \Omega = 10\ \Omega; I=VRtotal=20 V10 Ω=2 AI = \frac{V}{R_{total}} = \frac{20\ V}{10\ \Omega} = 2\ A

Explanation:

In a series circuit, total resistance is the sum of individual resistances. Then, Ohm's Law is used to find the total current.

Electricity and Simple Circuits - Revision Notes & Key Formulas | IGCSE Grade 8 Science