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Physics - Electricity: Circuits and Symbols

Grade 7Cambridge (IGCSE)

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

🔑Concepts

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Electric current is the flow of charge (usually electrons) around a circuit. For a current to flow, there must be a complete, closed loop and a source of potential difference (like a cell or battery).

A simple circuit diagram showing a battery connected to a lamp in a closed loop.
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Circuit symbols are used to represent components in a standardized way. A cell provides electrical energy, while a battery consists of multiple cells connected together.

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Ammeters measure current in Amperes (AA) and must always be connected in series with the component being measured. Voltmeters measure potential difference in Volts (VV) and must be connected in parallel across the component.

A circuit showing an ammeter in series and a voltmeter in parallel with a resistor.
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In a series circuit, there is only one path for the current to flow. The current is the same at all points in the circuit: Itotal=I1=I2I_{total} = I_1 = I_2.

A series circuit with two resistors connected one after the other.
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Resistance is a measure of how much a component opposes the flow of current. It is measured in Ohms (Ω\Omega). A higher resistance results in a lower current for a given voltage.

The circuit symbol for a fixed resistor.

📐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.

Problem 4:

A circuit contains a 12V12V battery and two resistors connected in series: R1=2ΩR_1 = 2\Omega and R2=4ΩR_2 = 4\Omega. Calculate the current flowing through the 4Ω4\Omega resistor.

Circuit diagram with a 12V source and two resistors of 2 ohms and 4 ohms in series.

Solution:

Rtotal=R1+R2R_{total} = R_1 + R_2 Rtotal=2Ω+4Ω=6ΩR_{total} = 2\Omega + 4\Omega = 6\Omega I=VRtotalI = \frac{V}{R_{total}} I=12V6Ω=2AI = \frac{12V}{6\Omega} = 2A

Explanation:

In a series circuit, we first find the total resistance by adding the individual resistances. Since the current is the same everywhere in a series circuit, the current through the 4Ω4\Omega resistor is the same as the total current, which is 2A2A.

Problem 5:

A student sets up a circuit with a 15V15V power supply and a single bulb. The ammeter in the circuit reads 3A3A. What is the resistance of the bulb?

Circuit diagram showing a 15V battery, an ammeter reading 3A, and a lamp with unknown resistance R.

Solution:

R=VIR = \frac{V}{I} R=15V3AR = \frac{15V}{3A} R=5ΩR = 5\Omega

Explanation:

To find the resistance, we use Ohm's Law rearranged as R=VIR = \frac{V}{I}. Dividing the voltage (15V15V) by the current (3A3A) gives a resistance of 5Ω5\Omega.