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The three concepts of charge, current and voltage are all very closely linked. Here we will explore the way in which they relate to each other and distribute energy.
Charge
Current is a flow of electrical charge. The charge is carried by the electrons which flow around the circuit. The amount of electrical charge that moves around a circuit is dependent upon the size of the current and the length of time the current flows for. The unit for electrical charge is the coulomb (C). One coulomb is the amount of charge carried by one ampere of current in one second. The equation used to calculate charge is:
![Rendered by QuickLaTeX.com \[ Q = I \times t \]](https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-cd6cec628b3cf8fad23bee31aa5322a0_l3.png)
You must know and be able to use this equation as it will not be provided in the exam.
Example
Calculate the charge if a current of 8A flows through a circuit for 10 seconds.
Charge = 8A x 10 seconds
Charge = 80C
Current
Current () is a measure of the amount of charge () carried past a point over a certain time period (t). This is known as the rate of flow of charge and it is measured in amperes or amps (A). The current of any circuit can be determined by measuring the amount of charge (Q) that passes a point in a certain period of time, and using the values in the rearranged equation for charge as shown below:
![Rendered by QuickLaTeX.com \[ I = \frac{Q}{t} \]](https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-4136025b8065e8ee13bffff7a19d19e4_l3.png)
Example
Calculate the current of the circuit if 200C of charge is passed through the circuit over 20 seconds.
Current = 200 ÷ 20 seconds
Current = 10A
Energy transferred
The amount of energy transferred to a component in a circuit can be measured using the values for charge and voltage as shown in the equation below:
![Rendered by QuickLaTeX.com \[ \text{Energy transferred} = \text{charge} \times \text{voltage} \]](https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-d296f0877302c07a44a326edb8898221_l3.png)
![Rendered by QuickLaTeX.com \[ E = Q \times V \]](https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-80daf39e9ea716195d8505c01b080865_l3.png)
Energy transferred is measured in joules (J). The charge is measured in coulomb (C) and voltage in volts (V). You must know and be able to use this equation as it will not be provided in the exam.
Example
Calculate the energy transferred if the charge is 75C and the voltage is 2.5V.
Energy transferred = 75C x 2.5V
Energy transferred = 187.5J
Voltage
Voltage is also sometimes known as potential difference and is a measure of the energy transferred per unit of charge passed. Voltage can be calculated using the equation below:
![Rendered by QuickLaTeX.com \[ \text{Voltage} = \text{Energy transferred} \div \text{Charge} \]](https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-782c1767f208032e7879416641a6b236_l3.png)
![Rendered by QuickLaTeX.com \[ V = \frac{E}{Q} \]](https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-68bea758c08e132d9588869ed5bf19f1_l3.png)
Voltage is measured in volts (V) where 1 volt is 1 joule of energy per coulomb. Therefore, if something has a voltage of 10 volts it must have 10 joules of energy transferred per unit of charge passed. (This is the same as saying 10 joules per coulomb, since 1 coulomb is 1 unit of charge passed.)
Example
Calculate the voltage needed to transfer 200J of energy with a charge of 50C.
![Rendered by QuickLaTeX.com \[ \text{Voltage} = 200J \div 50C \]](https://online-learning-college.com/wp-content/ql-cache/quicklatex.com-bb9f1645641f5d8faccad010e3dc2022_l3.png)
Voltage = 4V
Understanding the link between current, voltage and resistance
It can be a little tricky to really get to grips with the links between current, voltage and resistance. One common analogy used to help you to understand the terms uses the example of a water tank. The amount of water represents the charge, the pressure of the water represents voltage and the rate of flow of the water represents current. At the end of this water tank is a hose which the water passes through. The diameter of the hose can be altered which represents the way in which the size of wire used in a circuit can vary.
The size or thickness of the wire affects the resistance of the circuit. Resistance is a measure of how easily the electrical charge can flow along the wire. A thinner wire has a much higher resistance and slows down the rate of flow of electrical charge. A thicker wire has a lower resistance, which means that the rate of flow of charge can increase.

Current is represented by the rate of flow of water. To measure this, we look at the volume of water that flows out of the hose over a period of time. In an electrical circuit, this is equivalent to the amount of charge flowing through the circuit every second. Measurement of the flow tells us how much water has passed a point – much like the current telling us how much charge passes a point in a circuit.
By comparing two different water tanks with hoses of different diameters, we can evaluate the effect of changing the diameter on the rate of the flow of water. If the two tanks contain the same amount of water but have different hose diameters, the one with the larger diameter hose will allow more water to pass at once.
This is analogous to the effects of changing the thickness of the wires in an electrical circuit. A circuit with a thinner wire has a higher resistance. If a circuit has a high resistance, then it slows down the rate of flow of electrical charge in the same way that use of the hose with the smaller diameter slows down the rate of flow of the water.

To reduce the effects of the resistance from the smaller hose, we can add more water to the tank. This increases the pressure in the hose causing the water to squirt out faster, as shown in the diagram below:

The amount of water in the two tanks then decreases at the same rate meaning that the flow of water is equal. This is the same as having a circuit with a high resistance (a small hose). If we increase the voltage, the current also increases.