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Physics - Electricity and Simple Circuits

Grade 8Cambridge (IGCSE)

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

🔑Concepts

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Current is the rate of flow of electric charge, measured in Amperes (AA). In a circuit, an ammeter must be connected in series to measure the current flowing through a component.

Circuit diagram showing a battery, an ammeter, and a resistor connected in series.
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Potential Difference (Voltage) is the work done per unit charge moving between two points. A voltmeter must be connected in parallel across a component to measure the potential difference.

Circuit diagram showing a voltmeter connected in parallel across a resistor.
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Resistance is the opposition to the flow of current, measured in Ohms (Ω\Omega). According to Ohm's Law, for a conductor at constant temperature, the current is directly proportional to the potential difference across it.

Standard circuit symbol for a fixed resistor.
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In a parallel circuit, the total current from the source is the sum of the currents through the individual branches, while the potential difference across each branch remains the same.

Circuit diagram showing two resistors connected in parallel.

📐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.

Problem 4:

Find the total current flowing from the battery in a circuit where two resistors, R1=10 ΩR_1 = 10\ \Omega and R2=10 ΩR_2 = 10\ \Omega, are connected in parallel to a 5 V5\ V supply.

Circuit diagram with two 10 Ohm resistors in parallel with a 5V source.

Solution:

  1. Calculate the total resistance (RtotalR_{total}) for the parallel branches: 1Rtotal=1R1+1R2\frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} 1Rtotal=110+110=210\frac{1}{R_{total}} = \frac{1}{10} + \frac{1}{10} = \frac{2}{10} Rtotal=102=5 ΩR_{total} = \frac{10}{2} = 5\ \Omega
  2. Use Ohm's Law to find the total current (II): I=VRtotalI = \frac{V}{R_{total}} I=55=1 AI = \frac{5}{5} = 1\ A

Explanation:

In a parallel circuit, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances. Once the equivalent resistance is found, the total current can be calculated using the supply voltage.

Problem 5:

A circuit consists of a 12 V12\ V battery and three resistors in series: R1=2 ΩR_1 = 2\ \Omega, R2=3 ΩR_2 = 3\ \Omega, and R3=1 ΩR_3 = 1\ \Omega. Determine the potential difference across the 3 Ω3\ \Omega resistor.

Circuit diagram showing three resistors connected in series with a battery.

Solution:

  1. Calculate total resistance (RtotalR_{total}): Rtotal=R1+R2+R3R_{total} = R_1 + R_2 + R_3 Rtotal=2+3+1=6 ΩR_{total} = 2 + 3 + 1 = 6\ \Omega
  2. Calculate the circuit current (II): I=VRtotal=126=2 AI = \frac{V}{R_{total}} = \frac{12}{6} = 2\ A
  3. Calculate the potential difference across R2R_2 (V2V_2): V2=I×R2V_2 = I \times R_2 V2=2×3=6 VV_2 = 2 \times 3 = 6\ V

Explanation:

The current is the same at all points in a series circuit. By finding the total resistance, we find the common current, which is then used to find the specific voltage drop across one resistor.