PHYSICSDYNAMIC ELECTRICITY

Class IX A

Femi Tania

Acknowledgement

Puji syukur kita panjatkan ke hadirat Tuhan Yang Maha Esa karena atas rahmat-Nya saya dapat menyelesaikan makalah ini tentang Dynamic Electricity. Saya juga mengucapkan terima kasih kepada Ibu Rismawati karena telah memberi saya kesempatan untuk mempebaiki nilai saya dengan memberikan tugas ini.

Saya menyadari pada makalah ini terdapat kekurangan, akan tetapi saya akan terus memperbaiki kekurangan itu. Oleh karena itu, saya harap dimaklumi dan saya senantiasa mengharapkan masukan demi penyempurnaan makalah ini. Semoga makalah ini bermanfaat.

Femi Tania (IX A)

DYNAMIC ELECTRICITYA. Electric Current

• Electric current ( I ) has a charge flow from higher to lower potentials.

• Electron ( - ) flows from lower potential to higher potential.

1. Electric Current

Formula :

WHERE:

I = Electric Current (A)

Q = Electric Charge (C)

t = time (s)

I = Qt

= Coulombsecond

= Cs

t = QI

= CoulombAmpere

= CA

Q = I . t

= Ampere . second

= A . s

= Coulomb (C)

1 A = 103 mA 1 mA = 103 MA 1 MA = 10-6 A

1 A = 106 MA 1 mA = 10-3 A 1 MA = 10-3 mA

Close circuit

Electric current is measured in coulombs per second and measure by using tool named amperemeter, setting in series in a circuit, it’s unit is ampere (A).

2. Switch and Fuse

Switch is to disconnect and connect the electric current.

Fuse or protective tool is always set in each electric circuit like the electric current in house, television or car. As an automatic current breaker.

FIGURE :

Switch Fuse

The Formula:

= scale that pointedby the pointer

maximal scalex measure limit

B. Electric Resistance

1. Ohm’s Law

A German physicist, Georg Simon Ohm, in 1827 found the relationship between electric current, resistance, and voltage source. The finding, then, is known as Ohm’s Law. In Ohm’s law, concluded that the potential difference is proportional to the current ( V I )

FORMULA:

I ~ V Electric current is directly proportional to the current.

I ~ 1R

Electric current is inversely proportional to the resistance.

The tool to measure resistance is ohmmeter.

V = I . R

= Ampere . Ohm

= A .

= volt (V)

R = VI

= VoltAmpere

= Ohm ()

I = VR

= VoltOhm

=Ampere (A)

A V O meter

Amperemeter Voltmeter Ohmmeter

Electric current (I) potential different (V) resistance (R)

Ampere (A) Volt (V) Ohm ()

2. The Factors Influencing Resistance of a Conductor

The resistance of a conductor is smaller if the cross area is enlarged. If the wire length is expressed with (l), the resistance in each wire length wire with resistivity (ρ), and the wire cross area with (A). The formula:

3. Resistance in Daily Life

A resistor functions to manage the electric voltage and current so that the components in an electric circuit might function well. Constant resistance are commonly used as components in radio, television, and other electronic appliances.

R=ρ lA

= ohmm xm

m2

= ohmm2

m2

= ohm ()

A=ρ lR

= ohmm xmohm

= m2

l=R Aρ

= ohmxm2

ohmm

= m

Carbon-type Resistor

R . A=ρ . l

C. Resistance Circuit

1. In Series:

Problem Solving :

Given: R1 = 50, R2 = 75, R3 = 100, R4 = 200, R5 = 400, R6 = 125

Asked: Rt = ?

Answer : Rt = R1 + R2 + R3 + R4 +R5 + R6

= 50 + 75 + 100 + 200 + 400 + 125

= 950 ohms ()

2. In Parallel :

If the parallel circuit just have two branches, we can use:

Rp=R1 x R2R1+R2

Problem Solving :

D. Kirchoff’s Rules

1. Current in Non-Branched Circuit

The current flows from positive pole of the battery to the bulb and them to the negative pole. As the source of electric current voltage, the battery returns the electron to the positive pole. The electric current flows without starting do not point nor end that it is said that the current flows in a closed circuit. The current in a non-branched circuit is equal any where.

¿ 20010

=¿

20

Given : R1= 50, R2= 40, R3= 200

Asked : Rp = ?

Answer: 1R p

= 1R1

+ 1R2

+ 1R3

¿150

+ 140

+ 1200

¿4200

+ 5200

+ 1200

= 10200

¿20010

=¿20

2. Current in Branched Circuit

The amount of ingoing current in a branching point is equal to that of outgoing current from the branching.

This satement is known as the first of Khircoff’s rule, formulated as follows :

Problem Solving:

I=I 1+ I 2+ I 3=I'

Given : I2 = 5A, I3 = 4A, I4 = 6A, I5 = 3A

Asked : I1 = ?

Answer : Σ I ¿=Σ I out

I1 + I3 + I5 = I2 + I4

I1 + 4 + 3 = 5 + 6

I1 + 7 = 11

I1 = 11 – 7

= 4 A