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Electric current

 

Definition electric current in a wire

 

Electric current : the amount of charge per unit time that crosses the imaginary surface of a wire.

 

I  = Δ Q /Δt

 

I       electric current  in A

ΔQ  the charge passed in C  (1 Coulomb is the unit of charge)

Δt    time in s (!) 

 

Example

The electric current in a wire is 500 mA.

Calculate the charge passed through the imaginary surface of the wire in 2.0 min

 

I  = Δ Q /Δt

 

0.5 = Δ Q/120

 

Δ Q = (0.5) (120) = 60 C

 

Ohm’s Law /Power 

Ohm’s Law                             :    V = I R        

 V electric potential difference in  V     

 I   electric current in A   

 R  electric resistance in  Ω

 

Power      :    P = V I       

P  power in W

 

Example

An electric kettle has a power of  2.0 x 103  W . The voltage is  2.3 x 102 V

Calculate the electric resistance of the heating element.

 

P = V I

2.0 x 103 = (2.3 x 102)(I)

I = (2.0 x 103) /(2.3 x 102) = 8.70 A

 

V = I. R

2.3 x 102 = (8.70 )(R)

R = 26.4 Ω = 2.6 x 101  Ω

 

Electric  resistance of a wire

R  = (ρ.ℓ) / A           length wire in m

                                A cross sectional area in m2

                                                   ρ resistivity of the material in Ωm

Example  

50.0 cm of a constantan wire has a electric resistance of 1.125  Ω

Find the diameter of the wire

ρ of constantan : 0.45 x 10-6 Ωm

 

R = (ρ.)/A

R.A = ρ.ℓ

A = (ρ.ℓ)/R =  (0.45 x 10-6 x 0.50) /1.125 = 2.0 x 10-7 m2 = 2.0 x 10-1 mm2

 

A = π.r2

2.0 x 10-1 = πr2

r2 = (2.0 x10-1)/π = 0.06366

r = 0.25 mm

d = 2r = (2) (0.25) = 0.50 mm

 

Series/Parallel circuit

Example of a circuit partially in series and partially parallel

Voltage difference or electromotive force (emf) of the battery =  9.0 V   

series_parallel_circuit_1

R3 = 3.0 Ω  ;   R1 = 4.0 Ω  ;   R2= 6.0 Ω

R1 en R2  are parallel and this is entirely in serie with R3

 

Calculate the current I1 in resistor   R1

                                                                                                                   

-               At first: find the equivalent resistance of

  R1 en R2

   

    These two resistors are parallel.

    1/R12 = 1/R1 + 1/R2

     

    1/R12  = ¼ + 1/6 = 0.25 + 0.167 = 0.417

 

     R12 = 1/0.417 = 2.40 Ω

series_parallel_circuit_2

    

-           The circuit is equivalent to :

 

RAC  = 2.40 + 3.0 = 5.40 Ω     

                                                                                    

  

   

The main current is  :

        V = IRAC

             

        I = V/ RAC  = (9.0)/ (5.40) = 1.67 A

 

 

 

-            Calculate now  VAB

VAB = IR12 = (1.67) (2.40) = 4.0 V            

 

Or :

                                                                 VBC = I.R3= (1.67)(3.0) = 5.0 V 

                                                                         V = VAB + VBC  

                                                                         9.0 = VAB + 5.0

                                                                         VAB = 4.0 V 

  -     VAB = I1 R1

            4.0 = (I1 )(40)

        I1 = 1.0 A

        

        In the same way

        VAB =I2R2

            4.0 = (I2 )(6.0)

        I2 = 0.67 A

       

    -  Controle : I = I1 + I2              

 

 

Power/Energy

P = E/t        P power in W      E energy in J       t time in s

E = P  t

 

Example

Calculate the electric energy dissipated in a bulb of 50 W during 24 h.

 

E = P t

E = (50) (24 x 3600) = 4.32 x 106 J

Remark  : time  in s(seconds) and the energy in J(Joule)

 

The unit of time can also be : h(our). Then the unity of energy is Wh (Watthour)

 E = P.t

E = (50) (24) = 1200 Wh = 1.2 kWh