DIRECT CURRENT CIRCUITS

By

Drs. KISMUNADI, M.PdSMAN 4 SEMARANG

A. DEFINITION OF ELECTRIC CURRENT

Electric current illustrated as a motion of positive charges passing through from the higher potential to the lower potential.

Electric current defined as the amount of charge passing through in every unit of time ( second ).

Formula :I = Electric current

( ampere )q = charge ( coulomb )t = unit of time ( second

)

t

qI

B. RESISTANCE OF WIRE

The resistance of a wire depends on : Length of the wire Width of the wire Kind of the wireFormula :

R = Resistance ( Ohm, ) = Resistivity of the

material ( m) = Length ( m) = Width ( m2)

ΑR

ρ

ρ

A

C. OHM’S LAWSaid : The amount of electric current passing though a conductor is proportionally to the voltage and is reverse to its resistance.

A

V

L

V

V = I . R

R = tg

V

i

Grafik

A = Ammeter

V = Voltmeter

L = Lamp

V = Voltage

Exercise : 1. A wire has a length of 10 m and has a

wide of 2 mm2. The charges passing through the wire is 120 coulomb per minute. If the resistivity of the wire is 4 x 10-7 Ohm meter.Calculate the : a. Electric current

b. Resistance of the wire

c. Potential difference

2. In Ohm’s experiment, the result is :

Determine : a). Resistance of the wire.

b). Coefficient of A.

3. Resistance of a wire is approximately 10 , if the wire folded into the length of 1/4 times from the first, measure the resistance of these wire !

V ( Volt)

4

Ai (ampere)

300

D. SERIES AND PARALLEL CIRCUIT Kirchoff’s first ruleThe sun of the currents entering the function most equal to the sum of the currents leaving it.Example :

II11

ii55ii66

II2 ii33

ii44

II11 + I + I22 = i = i33 + i + i44 + i + i5 5 + i+ i66

Series circuit

VV

RR11 RR22 RR33

ii

R total = RR total = R11 + R + R22 + R + R33

VV11 : V : V2 2 : V: V33 = R = R1 1 : R: R22 : R : R33

V tut = VV tut = V1 1 + V+ V22 + V + V33

Characteristic :Characteristic :

The current The current passing passing through every through every resistor is resistor is equal.equal.

The potential The potential difference on difference on every resistor is every resistor is different. different.

PARALLEL CIRCUIT

II33

RR33

RR11

RR22

II11

II22

R3

I:

R2

I:

R1

II3:I2:I1

V3V2V1

3R

I

2R

I

1R

I

totalR

I

Characteristics : Characteristics :

The current The current passing passing through the through the function is function is different. different.

The potential The potential difference of difference of every function every function is equal.is equal.

Exercise :1. Notice the picture below !

RR11

RR44 RR66

RR22

RR55

RR33

RR77 RR55

I = 2AI = 2A

I = 6AI = 6A I = 10AI = 10A

Calculate the current passing through R1, R3, R4, R5, R6.

2. Look at the picture

V = 75 VoltV = 75 Volt

1010 44 66 55

Measure : a) The current passing Measure : a) The current passing through through every resistor. every resistor.

b) The voltage on every b) The voltage on every resistor.resistor.

3. Look at the picture

RR11 = = 1212

RR22 = 4 = 4

RR33 = 6 = 6

V = 60 V = 60 VoltVolt

II

II33

II44

II11

Calculate :Calculate :

a)a) The total The total current Icurrent I

b)b)The voltage on The voltage on every resistorevery resistor

c)c) The current The current passing passing through every through every resistorresistor

4. Look at the picture

Calculate the total resistance ( a – Calculate the total resistance ( a – b )b )

RR11 = 2 = 2

RR33 = = 1212

RR55 = 6 = 6

RR44 = 2 = 2

RR22 = 4 = 4

aa

bb

5. Look at the picture

AA

BB

CC

DD

22 22 22 22

66 44

22 22 22 22

Calculate :Calculate : a) Total resistance a) Total resistance ABAB

b) Total resistance CDb) Total resistance CD

c) Total resistance ADc) Total resistance AD

d) Total resistance BC d) Total resistance BC

WHEAT STONE’S BRIDGE

RxRxI = 0I = 0

RR11

ll22ll11

rulerruler

If in the If in the galvanometer (G) galvanometer (G) there are no there are no current passed, current passed, then :then :

R1 = Identified resistanceR1 = Identified resistanceRx = Measured resistanceRx = Measured resistance

ll1.1. . . RR1 1

= =

ll22 RxRx

TheThe formula above, also used for the formula above, also used for the circuit belowcircuit below

GG

RR22RR11

RR44RR55

i = 0i = 0RR11 . R . R33 = R = R22 . . RR44

Exercise :1) Look at the picture below !

2) Look at the picture below !

Calculate the total resistance !

GG

RR11=1=100

R=12R=12

RxRx

2 cm2 cm 5 cm5 cm

1010

88

44

22 66bbaa

Find RxFind Rx

E. THE SERIES AND PARALLEL OF ...........Ohm’s second lowThe current flows on a resistor is proportional to the potential difference and inversely proportional to the resistance.

E = Potential difference

r = The resistor of the battery

EE

rr

II

RR

E = I ( R + E = I ( R + r )r )E = I R + I E = I R + I rr

E = V + E = V + VVll

V = V = V = V = IRIR

VVll = Ir = Ir

Is called the lost of potential difference

Battery in series

n = Amount

Battery in parallel

Etotal = n E

rtotal = n rRR

EE11 EE22 EE33

rr11 rr22 rr33

rr

rr

rrEE

EE

EE EEtotaltotal = E = E

rrrrllll

total

Exercise1. Look the picture :

2. Look the picture :

E = 27 voltE = 27 volt

r r = 2/3= 2/3

II II

11 II

22

II

33

RR

11RR

22RR

33

RR11 = 12 = 12

RR22 = 16 = 16

RR33 = 24 = 24

Hitung : a) I, IHitung : a) I, I11, I, I22 dan I dan I33

b) Tegangan jepit (klem b) Tegangan jepit (klem sparing) sparing) rangkaian rangkaian

RR11

RR22

RR33

II

11 II

22

II

33

II

rr rr rr rr

EE EE EE EE E = 15 E = 15 VoltVolt

r =r = 0,25 0,25

RR1 1 = 8= 8 R R33 = = 2424

RR22 = 12 = 12

Hitung : a) I, IHitung : a) I, I11, I, I22 dan I dan I33

b) Tegangan jepit b) Tegangan jepit rangkaianrangkaian

3. Look the picture :

4. Look the picture :

4Volt4Volt

0,20,2

33

22

2 2 VoltVolt 0,10,1

11

0,40,4

8 8 VoltVolt

0,0,33

6 6 VoltVolt

33

Find the current Find the current flow throungh the flow throungh the circuitcircuit

IIII

11

II

22

66

3030 0,0,88

EE

EE

EErr

rr

rr

E = 18 VoltE = 18 Volt

rr = 0,6 = 0,6 Hitung : a) I, IHitung : a) I, I11 end I end I22

b) Tegangan jepit b) Tegangan jepit

c) The lost of pot c) The lost of pot difference difference

5. Look the picture :

Each branch cousists of 6 battery Each branch cousists of 6 battery

E = 4 VoltE = 4 Volt

R = 0,5R = 0,5

Hitung : a) I, I1, I2Hitung : a) I, I1, I2

b) Tegangan jepitb) Tegangan jepit

c) The lost of pot differencec) The lost of pot difference

EE

EE

EE

II 77 55II11

44II22

F. KIRCHOFF SECOND LOWThe sum of the drops in potential difference in a close circuit is equal to zero.

oiv R εε oiRE εεOr Or

V = Potential differenceV = Potential difference

E = Elf ( electromotive force )E = Elf ( electromotive force )

Exercise :1. Look the picture :

2. Look the picture :

EE11 = = 9V9Vrr11 = = 0,50,5

EE11 = = 4V4Vrr11 = = 0,50,5

AA

3

Arde

Find :

a)The current how the circuit

b) The potential difference in A

EE11 = = 9V9Vrr11 = = 0,50,5

EE11 = = 4V4Vrr11 = = 0,50,5

AA

3

Arde

Find :

a)The current how the circuit

b) The potential difference in A

EE11 = = 12V12V

4R1 R2 = 3

V2 = 6V

Find :

a) The currents on each prances.

b)The potential difference A.B

3. Look the picture R1 = 6

E1 = 10Vr1 = 0,5

E2 = 4VoltR2 = 0,5R2 = 2

R3 = 5

E3 = 12 Vr3 = 0,5

B

A

Find :

a)The currents on each prances

b) The potential di ference A. B

G. ENERGY AND ELECTRICAL......On a conductor of electrically charged in a moving or steady charge is called electrical energy.Rumus W = V i t

W = Electrical energy V = Electrical of potential

difference i = Electric current t = Time ( s )

The rate of electrical energy per second is called electrical power.t

W p P = daya listrik

(Walt) P = V . i

Exercise :1. An electrical iron is con summed 500

watt of power. The iron is used for ½ hour.Find : a) The electric current

b) The energy consumed2. three lamps each have

100W/110V,25W/110V and 20 W/110V is arranged in series and connected to potential difference of 110V. Find : a) The power consumed by the lamps

b) The electric current through each lamp.

3. Look at the picture

Hitung :a) The power consumed.b) The energy used for 1 hour.

4. A refrigerator of 90W/220V is used for hours energy lay. If the cost of I Kwh is Rp. 700,- now much is the cost in one month.

R1 = 20

R2 = 30

R3 = 60

A

B

C

D

R1 R2 R3