FE Physics
Formulas & Explanation

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

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

-           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