Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.

Chapter vii direct current circuits new

1,769 views

Published on

Published in: Technology, Business
  • Login to see the comments

Chapter vii direct current circuits new

  1. 1. CHAPTER VIIDIRECT CURRENT CIRCUITS
  2. 2. 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 (i) defined as the amount of charge passing through in every unit of time ( second ). q i = Electric current ( ampere ) i  q = charge ( coulomb ) t t = unit of time ( second ) q n  n = the amount of electron e e = the electron charge/the elementary charge = 1,6 x 10 -19 C
  3. 3. Direction of electric currentDirection of moving electrons Direct current source
  4. 4. B. RESISTANCE OF CONDUCTING WIRE The resistance of a conducting wire depends on :  Length of the wire  Cross section Area  Kind of the wire  Temperature Formula : R = Resistance ( Ohm, Ω )  ρ = Resistivity of the material ( Ω m) Rρ A = Cross-section Area ( m2) Α L = Length (m)
  5. 5.  Resistivities and TemperaturecCoefficients of Resistivity fo various Materials RESISTIVITY TEMPERATURE MATERIALS ρ(Ωm) COEFFICIENT(1/OC) Silver 1,59 x 10-8 3,8 x 10-3 Copper 1,7 x 10-8 3,9 x 10-3 Gold 2,44 x 10-8 3,4 x 10-3 Aluminum 2,82 x 10-8 3,9 x 10-3 Tungsten 5,6 x 10-8 4,5 x 10-3 Iron 10 x 10-8 5,0 x 10-3 Platinum 11 x 10-8 3,92 x 10-3 Lead 22 x 10-8 3,9 x 10-3 Nichrome 1,50 x 10-6 0,4 x 10-3 Carbon 3,5 x 10-5 -0,5 x 10-3 Germanium 0,46 -48 x 10-3 Silicon 640 -75 x 10-3 Glass 1010 - 1014
  6. 6. Temperature Influence for resistivity andresistanceIf temperature of wire is increase, so the resistivity and the resistance of it isincrease Rt  Ro (1   .T ) or R  RO . .Tt   o (1   .T ) or    O . .T ρO = initial of resistivity (Ωm) ρt = final of resistivity (Ωm) Ro = initial of resistance(Ω) Rt = final of resistance (Ω) ΔT = the change of temperature (oC) α = temperature coefficient of resistivity (/ oC) ∆R = The change of resistance ∆ρ = The change of resistivity
  7. 7. C. OHM’S LAWThe ratio of the voltage (V) across a conductor to thecurrent (i) that flows through it is equal to a constant.This constant is called resistance (R) VVi R V  iR A L A = Ammeter V = Voltmeter L = Lamp i = Current (A) V V V = Voltage/the potential difference (V) R = Resistance (Ω) R = tan α  i Graph of V - i
  8. 8. MEASUREMENT OF CURRENT AND VOLTAGE
  9. 9. D. SERIES AND PARALLEL CIRCUIT Kirchhoff’s first rule: The sum of the currents entering the any junction must equal the sum of the currents leaving the junction. Example : i5 i6 I1 i4 I2 i3 i1 + i 2 + i 4 = i3 + i5 + i6
  10. 10. SERIES CIRCUIT (VOLTAGE DIVIDER) R1 R2 R3 Characteristic : I  The current passing V through every resistor is equal. i1 = i2 = i3 = I  The potential difference RS = R1 + R2 + R3 on every resistor is different. V = V1 + V2 + V3 V1 : V2 : V3 = R1 : R2 : R3 V  I  RS R1 R2 R3V1  V V2  V V3  V RS RS RS
  11. 11. PARALLEL CIRCUIT (ELECTRIC CURRENT DIVIDER) R1 i1 i Characteristics : i2 R2  The current passing i3 through the junction is R3 different. V  The potential difference i  i1  i 2  i 3 of every junction is 1 1 1 1 equal.    RP R R R 1 2 3 V1 V 2 V 3 V 1 1 1 i1 : i 2 : i 3  : : R1 R 2 R 3 Rp Rp Rp V i1  I i2  I i3  I I  R1 R2 R3 Rp
  12. 12. E. WHEAT STONE’S BRIDGEIf in the galvanometer R1 R2(G) there are no electric Gcurrent passed, called a LA LBgalvanometer in Conducting wireequilibrium condition R1 . R B = R2 . R A because  R ρ Α so; RA= wire resistance of part A R1 . L B = R2 . L A RB= wire resistance of part B LA= wire length of part A LB= wire length of part B
  13. 13. The forms of Wheat stone bridge: R1 R2 If:@ R5 R 1 . R3 = R 2 . R 4 R4 R3 so, R5 can be ignored and then the wheat R1 R2 stone bridge circuit can be simplified to:@ R5 R4 R3 R1 R2 R1 R4 R3 R4 R5 R2@ R3
  14. 14. If R1 . R3 ≠ R2 . R4so, the circuits can be transforms to form Y (transformation of ∆ toY) R1 R2 R2 Rb Rb R5 Ra Ra Rc Rc R3 R3 R4 R1  R4 R4  R5 Ra  Rc  R1  R4  R5 R1  R4  R5 R1  R5 Rb  R1  R4  R5
  15. 15. F. SOURCE OF ELECTROMOTIVE FORCE (EMF) Current in conductor is produced by an electric field, and electric field is formed by the potential difference, devices such as batteries and dynamos should be connected to the circuit. These sources of electric energy are called source of electromotive force (ε) R K ● ● i ε r V • When the switch K is open, the voltmeter reads is EMF (ε) • When the switch K is closed, the voltmeter reads is clamping voltage (V) V= i R V= clamping voltage = potential difference on the external resistance
  16. 16. ε=iR+ir ε=i(R+r) ε = EMF (volt) r = internal resistance (Ω ) R = external resistance (Ω )Series Connection of Batteries ε1 ε2 ε3 Σε = ε1 + ε2 +ε3 r1 r2 r3 i Σr = r1 + r2 + r3 R If the batteries are identical, and each has an EMF ε, and an Σε = n ε internal resistance r Σr = n r
  17. 17.  Parallel Connection of Batteries ε1 r1 ε2 r2 ε3 i r3 R For identical batteries: Σε = ε
  18. 18.  Compound Connection of Batteries E1 E2 E3 E4 E5 r1 r2 r3 r4 r5 E6 E7 E8 E9 E10 r1 r2 r3 r4 r5 E11 E12 E13 E14 E15 r11 r12 r13 r14 r15
  19. 19. G. KIRCHHOFF’S SECOND RULESThe sum of the drops in potential difference in a close circuit isequal to zero. Σε + Σ (i . R) = 0 or Σε = Σ (i. R) Σ(i.R) = Dropping Potential difference ε = EMF ( electromotive force ) In applying Kirchhoff’s rules, the following rules should be noted:1. Assign a symbol and direction to the currents in each part of the circuit2. Loops are chosen and the direction around each loop is designated3. The sign of the current are taken “+” when they are in the same direction of loops, and taken “-” when they are in the opposite direction of loops4. The sign of the EMF are taken “+” when loops inside polar (+) of elements, and taken “-” when loops inside polar (-) of elements
  20. 20. G. WORK DONE BY THE ELECTRIC CURRENT ( JOULE’S LAW) The amount of heat dissipated from a current carrying conductor is proportional to the resistance of the conductor, the square of current and the time needed for the current to pass trough the conductor W=qV V2 Since q = i t, W=Vit i V W R R And V= I R W = i2 R t W = electrical energy (J) V = potential difference (volt) q = charge (C) i = electric current (A) t = time ( s )
  21. 21. The electrical energy dissipated per unit time (second) is calledelectrical power. Vi t P PVi t WP P i 2 Rt Pi R 2 t V2 t t V2 P R P t R P = Electric Power (Watt)

×