Download Electromotive Force

Electromotive Force
The maximum potential difference between the terminals on a source of electric energy is called the electromotive
force (EMF) and has units of volts
Symbol for EMF in a circuit diagram
Electric Current
The electric current is the amount of charge that passes through a cross-section of a wire per second
Δ𝑞
𝐼=
Δ𝑡
Unit: C/s = A (ampere)
Conventional Current
Conventional current is the hypothetical flow of
positive charges through the circuit that has the
same effect as the flow of negative charges that
actually occurs.
Direct current is a current that only flows in one
direction through the circuit.
Alternating current is a current that varies in
magnitude and reverses direction periodically.
Resistance
Resistance is the ratio of the voltage V applied
across a material to the current running through the
material
𝑅=
𝑉
𝐼
𝑢𝑛𝑖𝑡:
𝑉
𝐴
= Ω (ohm)
This relationship is called Ohm’s Law. The resistance
is a constant. Ohm’s Law is usually written as
V=IR
Resistor
resistor
Any device that offers resistance to the
flow of charges is called a resistor
emf
Circuit diagram
Calculating Resistance
For a wire that has a length, L, and a cross-sectional area, A, the resistance can be calculated
from
𝐿
𝑅=ρ
𝐴
Where ρ is called the resistivity (unit=Ωm) and is a property of the material the wire is made
from. Conductors have a very small resistivity while insulators have a very large resistivity.
Materials that are in between a conductor and insulator are called semiconductors.
Table of Resistivities
Temperature Dependence of Resistivity and Resistance
  o 1   T  To 
temperature coefficient
of resistivity
R  Ro 1   T  To 
Electric Power
In a time of Δt, a charge of Δq will pass through the battery
𝐸𝑛𝑒𝑟𝑔𝑦
Δ𝑞𝑉
𝑃=
=
= 𝐼𝑉
𝑡𝑖𝑚𝑒
Δ𝑡
The power dissipated by the resistor is
2
𝑉
𝑃 = 𝐼𝑉 = 𝐼2 𝑅 =
𝑅
𝑢𝑛𝑖𝑡: 𝑊 (𝑤𝑎𝑡𝑡)
Practice Questions
Question 1:
A steady current of 0.75A flows through a wire. How much charge passes through the wire in 1.5 minutes?
Question 2:
How many electrons flow through a light bulb each second if the current through the light bulb is 0.75A?
Question 3:
A metal rod is 2 m long and 8 mm in diameter. Compute its resistance if the resistivity of the metal is 1.76 X 10-8 Ω m.
Question 4:
An ammeter-voltmeter method is used to measure an unknown resistance R. The ammeter reads 0.3 A, and the voltmeter
reads 1.5 V. Compute the value of R (neglecting the internal resistance of the ammeter and voltmeter).
Question 5:
An aluminum wire has a diameter of 2.59 mm. What is the length of wire required to give a resistance of 1 Ω. ρ for
aluminum is 2.8 X 10-8 Ω m.
Question 6:
The resistance of a coil of copper wire is 3.5 Ω at 0oC. What is the resistance at 50oC. For copper, α = 4.3 X 10-3 oC-1.
Alternating Current
𝑉 = 𝑉0 sin(2π𝑓𝑡)
V0 is the peak voltage and f is the
frequency of the oscillations.
If the circuit only contains resistance
the current is given by
𝑉 𝑉0
𝐼 = = sin 2π𝑓𝑡 = 𝐼0 sin(2π𝑓𝑡)
𝑅 𝑅
Power in Alternating Current
𝑃 = 𝐼𝑉
𝑃 = (𝐼0 sin 2π𝑓𝑡 )(𝑉0 sin(2π𝑓𝑡))
𝑃 = 𝐼0 𝑉0 𝑠𝑖𝑛2 (2π𝑓𝑡)
I oVo  I o  Vo 
P


  I rmsVrms
2
 2  2 
Ohm’s Law for Alternating Current
Vrms = Irms R
P = Irms Vrms
2
P = Irms
R
2
𝑉𝑟𝑚𝑠
𝑃=
𝑅