1
ELECTRICITY
 Charge
atom
 comes in two forms
 quantized (there is a smallest amount – quant)
• cannot divide up charge into smaller units than that of electron (or proton) i.e. all objects have a charge
that is
a whole-number multiple of charge of the smallest amount (a single ±e).
• the net charge is the algebraic sum of the individual charges.
law of conservation for charges – charge is conserved, can not be created or destroyed, just transferred
 Force between TWO POINT charges q1 and q2 at distance r :
F=k
q1q2
r2
(vector algebra)
 Electric field (vector algebra)
- (at a point P) is the force per unit positive charge (placed at that point).
Direction of the field is the direction of the force on a positive test charge placed at that point.
E=
F
q
(E) = NC-1
if we know electric field at point P and we place charge q at that point, then
F=qE
 Electric field lines: show direction of the force on positive charge
• always point away from + charges, towards – charges…
• denser lines - stronger field (closer to the charge) el. field
decreases with distance
Capacitor: uniform electric field (the one that has constant
magnitude and direction is generated between two oppositely
charged parallel plates. Edge effect is minimized when the
length is long compared to their separation.
If two sources are creating electric fields in the same place, we have to add the two vectors and get a resultant
vector representing the NET ELECTRIC FIELD.
 Electrical potential energy, U
(U) = 1J (Joule)
Work done by external force on charge against an electric field between two points is stored in the charge as the
change in electrical PE.
∆U = W
(Joules)
 Potential Difference Between Two Points
The difference between the potentials at two different points (A & B) measures the work done per unit positive
charge in order to move it from one point to the other.
V 
W U

q
q
electric potential 
electric potential energy
charge
1 Volt= 1 Joule/1 Coulomb
If a charge, q, is moved through a potential difference, ∆V, then the work done on it is equal to the change in
its electric potential energy which is converted into kinetic energy:
W = ∆U = q ∆V = ½ mv2
2
Positive charge accelerates from higher to lower potential.
Negative charge accelerates from lower to higher potential.
 electron-Volt (eV)
An electron volt is the amount of energy/work it takes to move an electron through a potential difference of
1volt.
1 eV = ∆U = W = (1.6x10-19 C) (1V)
1 eV = 1.6x10-19 J
CIRCUITS
 Current is the rate at which charge flows past a given cross-section.
𝐼 =
𝑄
𝑡
1A =
1C
1s
 Electrical resistance , R is the ratio of the potential difference across the resistor/conductor to the current that
flows through it.
𝑅=
𝑉
1 (ohm) = 1V/1A
𝐼
The resistance of a conducting wire depends on four main factors:
• length, L • cross-sectional area, A • material/resistivity, ρ • temperature, T
 Resistance of a wire when the temperature is kept constant is:
𝑅=𝜌
𝐿
𝐴
 OHM’S LAW: Current through resistor (conductor) is proportional to potential difference across the resistor if
the temperature of a resistor is constant (the resistance of a conductor is constant).
 math def: 𝐼 =
𝑉
𝑅
if resistance R is constant/ temperature is constant
I – current
V – potential difference across R
When a current is flowing through a load such as a resistor, it dissipates energy in it. In collision with lattice ions
electrons’ kinetic energy is transferred to the ions, and as a result the amplitude of vibrations of the ions increases
and therefore the temperature of the device increases. KE is transferred to thermal energy.
 DEF: Electric power is the rate at which energy is supplied to or used by a device.
 DEF: Power is the rate at which electric energy is converted into another form such as mechanical energy,
thermal energy, or light.
Power dissipated in a resistor:
P=IV
Remember: W = qV → P = qV/t
and I = q/t, so P = I V
 E.M.F. –Electromotive force is the work done per unit charge in moving charge completely around the circuit.
electric potential energy
1J
electric potential 
(coming from ∆V = ∆U/Q)
1V 
charge
1C
 E.M.F. –Electromotive force is the power supplied to the circuit per unit current
(coming from P = I V ⟹ V = P/ I )
symbol: E or 𝜀 or e.m.f.
Energy supplied by the source = ε Q
(coming from U = Q ∆V)
3
Power of the source = sum of the powers across the resistors
Pout = Σ Pi
⟹ ε I = V1 I + V 2 I
Total energy supplied by the source = energy used in the resistors
ε Q = V1 Q + V2 Q
divide it by time and you get above relationship
ex: the cell supplies 8.0 kJ of energy when 4 kC of charge moves completely across the circuit with
constant current. Find e.m.f.
ε = energy/charge = 2 V
Resistors in Series
• connected in such a way that all components have the same current through hem.
• Burning out of one of the lamp filaments or simply opening the switch could cause such a break.
Req = R1+ R2 + R3
Resistors in Parallel
• Electric devices connected in parallel are connected to the same two points of an electric circuit, so all
components have the same potential difference across them.
• The current flowing into the point of splitting is equal to the sum of the currents flowing out at that point:
I = I1 + I2 + I3.
1
1
1
1
=
+
+
Req R1 R2 R3
• A break in any one path does not interrupt the flow of charge in the other paths. Each device operates
independently of the other devices. The greater resistance, the smaller current.
 TERMINAL VOLTAGE (the actual voltage delivered) is: V = ε - Ir
4
 Ammeters and voltmeters
To measure the current, we use an AMMETER.
We have to place the ammeter in SERIES with resistor R in order that whatever current passes through the resistor
also passes through the ammeter. Ram << R, so it doesn’t change the current being measured.
• would ideally have no resistance with no potential difference across it so no energy would be dissipated in it.
To measure the voltage drop between two points, we use a VOLTMETER.
If we wanted to measure the potential drop across R, that is if we wanted to measure the potential difference
between two points we connect the leads from voltmeter to these points. That way we measure electric potential
at the point ahead of the resistor and at the point beyond the resistor, and compare two measurements. The only
way to do it is to insert our voltmeter in PARALLEL with the resistor we are measuring. Rvm >> R • has a very high
resistance so that it takes very little current from the device whose potential difference is being measured.
• an ideal voltmeter would have infinite resistance with no current passing through it and no energy would be
dissipated in it.
 A potential divider
In electronic systems, it is often necessary to obtain smaller voltages from larger voltages for the various electronic
circuits. A potential divider is a device that produces the required voltage for a component from a larger voltage. It
consists of a series of resistors or a rheostat (variable resistor) connected in series in a circuit.
Potential divider equation
I=
V
R1+R 2
V1 =

V1 = IR1
R1
V
R1+R 2
Potentiometer
5
Find power of the source, current in each resistor, terminal potential, potential drop across each resistor and
power dissipated in each resistor.
Req = 120 Ω
I = ε ∕ Req = 0.3 A
terminal potential: V = ε – Ir = 36 – 0.3x6.7 = 34 V
current through resistors 100 Ω and 50 Ω : I = I1 + I2
0.3 = I1 + I2
100 I1 = 50 I2
→ I1 = 0.1 A
potential drops
V = IR
power dissipated
P = IV
80 Ω
0.3x80 = 24 V
0.3x24 = 7.2 W
100 Ω
0.1x100 = 10 V
0.1x10 = 1 W
50 Ω
0.2x50 = 10 V
0.2x10 = 2 W
6.7 Ω
0.3x6.7 = 2 V
0.3x2 = 0.6 W
ε = Σ all potential drops:
0.6 + 2 + 1 + 7.2 = 0.3x36
I2 = 0.2 A
36 V = 2 V + 24 V + 10 V
power dissipated in the circuit = power of the source
I1 R1 = I2 R2
potential drop across 100 Ω and 50 Ω
is the same and is counted only once!!!!!