krit.club logo

Physics - Electricity: Simple series circuits and components

Grade 4Cambridge (IGCSE)

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

🔑Concepts

•

In a series circuit, there is only one path for the electric current (II) to flow; if the circuit is broken at any point, the current stops flowing everywhere.

A simple series circuit diagram showing a battery connected to two lamps in a single loop.
•

The current (II) remains the same at every point in a series circuit. This means Itotal=I1=I2=I3I_{total} = I_1 = I_2 = I_3.

•

The total resistance (RtotalR_{total}) of the circuit is the sum of the individual resistances of all components connected in series.

•

The total potential difference (VtotalV_{total}) supplied by the battery is shared between the components. The sum of the voltages across each component equals the battery voltage.

Series circuit with a battery and two lamps showing how the total voltage is divided.

📐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.

Problem 4:

A circuit contains a battery and two identical bulbs. An ammeter measures a current of 0.4A0.4A leaving the battery. If the voltage across one bulb is 4.5V4.5V, what is the total voltage of the battery?

Circuit with an ammeter showing 0.4A and two lamps each labeled with 4.5V.

Solution:

  1. Identify the voltage across the second bulb: Since the bulbs are identical and the current is the same, the voltage across the second bulb is also V2=4.5VV_2 = 4.5V.
  2. Calculate total voltage: Vtotal=V1+V2=4.5V+4.5V=9VV_{total} = V_1 + V_2 = 4.5V + 4.5V = 9V

Explanation:

In a series circuit, identical components will share the total voltage equally. The sum of the potential differences across the lamps equals the electromotive force (EMF) of the battery.

Problem 5:

A series circuit is constructed with a 20V20V DC power supply, a 5Ω5Ω resistor, and a 15Ω15Ω heating element. Calculate the total resistance of the circuit and the current flowing through the 15Ω15Ω heating element.

A series circuit diagram showing a 20V battery connected in a single loop to a 5 ohm resistor and a 15 ohm resistor.

Solution:

Rtotal=R1+R2R_{total} = R_1 + R_2 Rtotal=5Ω+15Ω=20ΩR_{total} = 5\Omega + 15\Omega = 20\Omega I=VRtotalI = \frac{V}{R_{total}} I=20V20Ω=1AI = \frac{20V}{20\Omega} = 1A

Explanation:

In a series circuit, the total resistance is the sum of the individual resistances. Since there is only one path for the current, the current II is the same at all points in the circuit. Therefore, the current through the heating element is the same as the total current calculated using Ohm's Law.