CHAPTER VII Direct Current Circuits
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Transcript of CHAPTER VII Direct Current Circuits
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