2. EMF
P.d between the terminals of the cell when it
is not delivering any current in the external
circuit
(i.e p.d across external and internal circuit)
E = IR + Ir
P.D is the work done when one coulomb of
charge moves from one point to another.
3. Resistivity and resistance
ρ = AR/L
R = ρL/A
σ (conductivity) = 1/ρ = L/RA (siemens/m)
ρ = resistivity
A = Area
R = resistance
4. Eg find resistance of a wire of lenght 500cm
and cross-sectional area of 2m2 , if the
resistivity is 1 x 10-5 Ωm
R = ρL/A = 1x10-5 x 0.5 = 2.5 x 10-6 Ω
2
What is the conductivity?
1/ρ = 1/1 x10-5 = 1x105 Sm-1
5. OHM’S law
The electric current flowing through a
metallic conductor at constant temperature
is directly proportional to the p.d between its
ends
V α I V = Ki
V = IR *R is the constant
EMF = V +Ir = IR +Ir = I(R +r)
I = E/(R +r)
6. Eg find current in a voltmetre of 600 ohms
resistor which reads 150V when connected
to a battery of 20 0hms internal r.
V = IR I = V/R = 150/600 = ¼ A
What is the emf?
= I(R + r) = ¼ (600 + 20) = 155V
7. Resistors in series and parallel
Resistors in series
R = R1 + R2 + R3 ....
Resistors in parallel
R = 1/R1 + 1/R2 + 1/R3 ....
2 resistances R1 and R2, the equivalent =
R1R2 /R1 + R2
Eg you are given a circuit with 3 pieces of 5Ω
resistors connected in series and 2 pieces of
2 Ω resistors connected in parallel. The
resistance in the circuit =#
R = 5+5+5+(2x2/2+2) = 16Ω
8. Remember: balance conditions
1. A wheatstone bridge
R1/R2 = R3/R4
2. The metre bridge
X/R = l σ/(100 – l)σ
X and R are connected resistance
σ = resistance per unit lenght
l = lenght
Potentiometre
E1/E2 = l1/l2
9. KIRCHOFF’S laws
For current
The algebraic sum of the currents flowing
into any junction is zero. The sum of
currents flowing into a junction = sum of
currents flowing out
ΣI(in) = ΣI(out) I1 + I2 + I3 = I4 + I5
I1 + I2 + I3 - I4 - I5 = 0
I1
I2
I3
I4
I5
10. For voltage
The algerbraic sum of the p.d in any closed
loop = zero. The sum EMFs in any closed
loop = the sum of voltage IR drops in the
loop
ΣEMF – ΣIR = 0
ΣEMF = ΣIR
