**Definition of Electric Currents**

Electric current occurs because of the flow of electrons in which each electron has a charge of the same amount. If we have a negatively charged object, it means that the object has an excess of electrons. The degree to which the object is filled is measured by the number of excess electrons present. The charge of an electron, often represented by the symbol q or e, is expressed in coulombs, which is

* q *≈ 1.6 ⋅ 10^{-19} coulombs.

Suppose we have a piece of copper wire which is usually used as an electricity conductor on the grounds that the price is relatively cheap, strong and resistant against corrosion. The amount of conduction on the wire depends only on the presence of free electrons (from the valence electrons), because the nuclear charge and the electrons in the inner path are tightly bound to the crystal structure.

Basically, in the conducting wire there is a very large flow of electrons, if the number of electrons moving to the right and to the left is the same it is as if nothing happened. However, if the right end of the wire attracts electrons while the left end releases it, there will be a flow of electrons to the right (but remember, in this case it is agreed that the direction of the current is left). This flow of electrons is hereinafter referred to as electric current.

The amount of electric current is measured in units of the number of electrons per second, however this is not a practical unit because the price is too small. The unit used is the ampere, where the

Current and Voltage of Electricity 1

*i = dq / s *

1 ampere = 1 coulomb / sec.

The example below illustrates the amount of electric current for some equipment:

Power station ………………. 1000 A

Car starter …………. …… 100 A

Light bulb ………………. 1 A

Small radio …………….. .. 10 mA

Watches ………………. 1 ∝A

**Definition of Voltage**

It is easy to analogize the flow of electricity with the flow of water. Suppose we have 2 tubes that are connected by pipes as shown in Figure 1.1. If both tubes are placed on the table then the water level in both tubes will be the same and in this case there is no water flow in the pipe. If one of the tubes is lifted, water will automatically flow from the tube to the lower tube. The higher the tube is lifted, the faster the water will flow through the pipe.

Figure 1.1 The flow of water in the related vessel .

The occurrence of this flow can be understood with the concept of potential energy. The height of the tube shows the amount of potential energy it has. The most important of this case is the difference in height of the two tubes which simultaneously determines the magnitude of the potential difference. So the greater the potential difference, the faster the water flow in the pipe.

The same concept applies to the flow of electrons in a conductor. What determines how much current flows is the amount of potential difference (expressed in volts). So for a conductor the greater the potential difference, the greater the current flowing.

It should be noted that the potential difference is measured between the ends of a conductor. But sometimes we talk about potential at some point. In this case we are actually measuring the potential difference at that point against a certain reference point. As a point of reference standards usually been a point of land(ground).

Furthermore, we can analogize a battery or battery as a water tube that is lifted. This battery has chemical energy that is ready to be converted into electrical energy. If the battery is not used, no energy is released, but keep in mind that the potential of the battery is there. Almost all batteries providepotential ( *electromotive force *almost the same- emf) even though current is drawn from the battery.

**Ohm’s Law **

In most metal conductors, the relationship between current and potential is governed by Ohm’s Law. Ohm uses a simple experimental circuit as shown in Figure 1.2. He used a series of potential sources in series, measured the amount of current flowing and found a simple linear relationship, written as

*V = IR *

where *R = V / I is *called the resistance of the load. This name fits perfectly because *R is *a measure of how much the conductor holds the flow rate of electrons. Beware, the application of the ohm law is very limited under certain conditions, even this law does not apply if the temperature of the conductor changes. For certain materials or electronic devices such as diodes and transistors, therelationship is *I *and *V *not linear.

Figure 2 Experiments of Ohm’s law series

**Power**

Suppose a potential *v is *applied to a load and current i flows as shown in Figure 1.3. The energy supplied to each electron that produces an electric current is proportional to *v *(potential difference). Thus the total energy applied to the number of electrons for a total charge of *dq *is proportional to *v *⋅ *dq*.

The energy given to electrons per unit time is defined as power (*power*) p of

*p = v dq / dt = vi *(1.2) in watts

where 1 watt = 1 volt ⋅ 1 ampere

Figure 1.3 Current flow in the load due to potential *v *

**Power on resistance (resistor) **

If a voltage *V is *applied to a resistance *R *then the amount of current flowing is

*I = V / R *(Ohm’s law)

and the power given is

*P = V*⋅ *I *

* = V ^{2}*

*/ R*

* = I ^{2}*

*R*

For certain cases the problem becomes different if the potential given is not constant, for example in the form of a sine function with time (as in alternating current)

*v = V *sin ω *t *

, thus

*i = v / R *

* = (V / R*) sin ω *t *

Electric Current and Voltage 5

and

*p = v *⋅ *i *

* = (V*^{2}*/ R) *sin^{2} ω *t *(1.4)

*p is *always positive so that power will always be lost at any time, changing to heat on resistance. This power changes over time, has a value of zero when sin ω*t *= 0, and a maximum of *V**2**/ R *when sin ω *t *= 1.

To determine the heating effect of the above signal, the power equation above can be written as

*p= 1/2 *(*V ^{2} */

*R*)(1 – cos2ω

*t*)

*cos 2*ω*t *will be positive or negative as often, so the average is zero. Thus the average power loss is

*P = 1/2 *(*V ^{2} / R*) = (V /√2 )

^{2}

*/ R*

This is the power dissipated at R if the voltage is constant Vp /√2 imposed on him. The value of Vp /√2 = 0.707 *V is *often used as a measure if the sine voltage is used in a circuit and the value of this voltage is often referred to as the *root-mean-square (*RMS) value. In this case we have to be careful to determine the 3 measurements used, namely

Price RMS = Vp /√2

Peak amplitude = *V**p** *

Peak-to-peak price = *2V**p*

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