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Forces and Energy - Electric Circuits and Conductivity

Grade 5IB

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

🔑Concepts

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Electric current flows in a complete loop called a circuit. For electricity to flow, the circuit must be closed, meaning there are no breaks in the path.

A simple series circuit diagram showing a battery connected to a light bulb in a continuous loop.
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Conductors are materials that allow electricity to flow through them easily, such as Copper (CuCu) or Aluminum (AlAl). Insulators are materials that block the flow of electricity, such as rubber or wood.

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A switch acts as a gate in a circuit. When the switch is 'Open', the circuit is broken and the current stops. When the switch is 'Closed', the path is complete and devices like bulbs will turn on.

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Cells in series increase the total voltage. Adding more cells (Vtotal=V1+V2V_{total} = V_1 + V_2) provides more energy to the circuit, which can make a bulb glow brighter.

Two cells connected in series to a single bulb.

📐Formulae

V=I×RV = I \times R

P=V×IP = V \times I

Total V=V1+V2+V3 (for cells in series)\text{Total } V = V_1 + V_2 + V_3 \text{ (for cells in series)}

💡Examples

Problem 1:

A student builds a circuit with a 9V9V battery and a bulb that has a resistance of 3Ω3\Omega. Calculate the current (II) flowing through the circuit.

Solution:

Using the formula I=VRI = \frac{V}{R}, we get I=9V3Ω=3AI = \frac{9V}{3\Omega} = 3A.

Explanation:

By applying Ohm's Law, we divide the voltage by the resistance to find that the current is 33 Amperes.

Problem 2:

Which material would be the best choice to complete a circuit: a piece of plastic (SiO2SiO_2 based polymer) or a silver (AgAg) paperclip?

Solution:

The silver (AgAg) paperclip.

Explanation:

Silver is a metal and a strong conductor, meaning it has very low resistance (RR), allowing electrons to flow. Plastic is an insulator and will stop the flow of electricity.

Problem 3:

If two 1.5V1.5V cells are placed in series in a flashlight, what is the total voltage (VtotalV_{total}) supplied to the bulb?

Solution:

Vtotal=1.5V+1.5V=3.0VV_{total} = 1.5V + 1.5V = 3.0V

Explanation:

When cells are connected in series, their voltages are added together to provide more energy to the circuit.

Problem 4:

Calculate the total resistance (RtotalR_{total}) in a circuit where two resistors, R1=10ΩR_1 = 10\Omega and R2=5ΩR_2 = 5\Omega, are connected in a single loop (series).

A circuit diagram showing two resistors connected one after another in a series loop.

Solution:

Rtotal=R1+R2R_{total} = R_1 + R_2 Rtotal=10Ω+5ΩR_{total} = 10\Omega + 5\Omega Rtotal=15ΩR_{total} = 15\Omega

Explanation:

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

Problem 5:

A circuit contains three identical light bulbs connected in series to a battery. Each bulb has a resistance of 4Ω4\Omega and the battery provides a total voltage of 12V12V. Calculate the total current (II) flowing through the circuit.

A series circuit diagram showing a 12V battery connected to three bulbs in a single loop, each labeled with 4 Ohms resistance.

Solution:

Rtotal=R1+R2+R3R_{total} = R_1 + R_2 + R_3 Rtotal=4Ω+4Ω+4Ω=12ΩR_{total} = 4\Omega + 4\Omega + 4\Omega = 12\Omega I=VRtotalI = \frac{V}{R_{total}} I=12V12Ω=1AI = \frac{12V}{12\Omega} = 1A

Explanation:

In a series circuit, the total resistance is the sum of all individual resistances. Once the total resistance (12Ω12\Omega) is found, Ohm's Law (V=I×RV = I \times R) is rearranged to I=VRI = \frac{V}{R} to solve for the current. Using the battery voltage of 12V12V and the total resistance of 12Ω12\Omega, we find that 11 Ampere of current flows through the circuit.