The Ohm s Law. 180 years old battery

Transcription

1 The Ohm s aw George Ohm has established experimentally in 827 the following law V is the voltage across a conductor I= V I is the current thought a conductor / is a proportionality factor; is called resistance + V A I Electric current 80 years old battery Voltage

2 Ohm s aw General type electric circuit: Wires V Electri c power supply Wires Curren t oad (motor, electronic device, appliances etc.) Given the power supply voltage V, how much current flows through the load? 2

3 Ohm s aw and electron charge transfer rate comparison I = e N v I = V Electric current Electric current Electric field (charge velocity) Voltage 3

4 Experimental observations of Ohm s aw I = V Electric current Voltage Conductor length decreases Current increases, i.e. decreases 4

5 Experimental observations of Ohm s aw I = V Electric current Voltage Conductor area increases Current increases, i.e. decreases 5

6 esistance and resistivity I V = = ρ A is the conductor length (along the current direction) A is the cross section area (with respect to the current direction) ρ is called resistivity of the material; ρ does not depend on the conductor shape; it depends on mobile charge concentration and mobility in the material. 6

7 Ohm s law, resistance and resistivity summary I = V V= I = = ρ A ρ A Ohm s law in the form of I(V): the current through the conductor = applied voltage divided by the resistance Ohm s law in the form of I(V): the voltage needed to maintain the current I = current times resistance. A and describe the geometry of the sample or wire. The larger is the cross section A the smaller is, the higher is the current. The longer is the wire length, the higher is, the smaller is the current; ρ describes the material ability to conduct the current. The higher is ρ, the higher is and the lower is the current 7

8 The units for resistance and resistivity V = I = esistance is measured in Ohms (Ohm, Ω) Ohm is the resistance of the sample that passes the current of A when the voltage of V is applied across it. V I = ρ A 2 A m ρ = Ohm = Ohm m m esistivity is measured in Ohm meters (Ohm m) 8

9 Ohms law using conductance and conductivity Conductance G = / Using the conductance, the Ohm s law can be written as Also, from G= σ ρ / I= V = G V and = ρ A, G= / = A ρ = σ is called conductivity of the material (does not depend on the conductor shape) Using conductivity, the conductance of the sample is given by G A = σ 9

10 Ohm s aw using conductance summary The expression that relates the electric current to the applied voltage is called the Ohm s aw (established experimentally in 827) I = G V, where A G = σ is the conductance of the sample (wire), σ is the conductivity of the material A is the conductor cross section area; is the conductor length along the current direction 0

11 The units for conductance and conductivity I = G V G = Conductance is measured in Siemens (S) S is the conductance of the sample that passes the current of A when the voltage of V is applied across it. I V G A = σ m σ = G S = S / m 2 A m Conductivity is measured in Siemens per meter (S/m)

12 Units and Dimensions Charge: Coulomb [C] abs (electron charge) e = C Electric current: Ampere [A] Electric voltage: Volt esistance: Ohm esistivity Conductance: Siemens [V] [Ohm] [Ohm m] [S] Conductivity: [S/m, (Ohm m) ] 2

13 Example : the Ohm s aw V = I I = V / I = 2V / 3 Ω = 4 A Note: the notation for the battery voltage is usually V or E 3

14 esistivity of different materials Material Wires are used to connect different components in the network; Wires have very low resistance esistors are used to dissipate the power and to change the voltage (potential). Aluminum [Al] 27 Aluminum Alloy 50 Brass 20 6 Carbon [C] Copper [Cu] 7 Copper Alloy Gold [Au] 24 Iron [Fe] 97 Electric esistivity ( 0 9 Ohm m) 4

15 Example 2 Find the resistance of a wire made of copper [Cu] (ρ = 7*0 9 Ohm m). The wire is m long and is mm in diameter. = ρ */A = m; D = mm = 0 3 m; The area, A = π*d 2 /4 = 3.4*(0 3 ) 2 /4 = 7.85*0 7 m 2 The resistance = 7*0 9 Ohm*m*m/ 7.85*0 7 m 2 = 2.7*0 2 Ohm = 2.7 mohm What voltage across this wire is required to have the 00 ma current through it? V = I = 00 e-3 2.7e-3 = V 5

16 Example 3 Find the resistance of a carbon resistor, which is cm long and 0. mm in diameter = ρ */A; ρ = * 0 9 Ohm*m = Ohm*m ; = cm = 0 2 m; A = 7.85*0 9 m 2 =.4*0 5 Ohm*m*0 2 m/ 7.85*0 9 m 2 = 7.8 Ohm Compare: Cu wire, = 2.7 mohm 0.02 Ohm; Carbon resistor, 2 = 7.8 Ohm 8 Ohm; To have the current through the Cu wire of 0. A, the required voltage across the wire, V = I* = V; To have the same current through the Carbon resistor, the required voltage across the resistor is V 2 = I* 2 =.8 V; 6