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RESISTANCE
What is resistance?
• Resistance is where electrons give up potential
energy they carry from the battery.
• EXAMPLE
The resistance of a lamp causes the electrical
energy to change into there forms such as heat
and light. If the lamp filament had no resistance,
there would be no energy change and this means
that the bulb in the lamp would no light.
Resistance in Conductors
• The resistance of conductors depends on its
length and diameter.
• A thin wire has a higher resistance than a thick
wire.
• A long wire has a greater resistance than a
shorter wire.
How is resistance found?
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Resistance is found by dividing voltage by current.
Resistance= voltage ÷ current
R =V ÷ I
Resistance of a conductor is the ratio of voltage
to which across it current flows. The unit for
resistance is known as an ohm (Ω). An ohm is the
resistance to which current is 1 ampere (A) and
the potential difference between its ends in 1 volt
(V) .
Calculating Resistance
• If a current (I) of 5A flow through a lamp when
it was connected to a battery of 20V, providing
a voltage of 20V across a lamp, what is its
resistance?
• Information given is V=20V and I=5A
• Using the formula R = V ÷ I
R =20 ÷ 5
R =4Ω
Ammeter
• An ammeter is a device used for measuring
current. It is placed in series with the
components, the current through which is
being measured. Ammeters have a low
resistance so the potential difference across
them is as small as possible.
Voltmeters
• Voltmeters re used for measuring the
potential difference. It is placed in parallel
with the component, the potential difference
is being measured. Voltmeters have a high
resistance and they do not affect the current
in a circuit.
Series Circuit
• A series circuit provides one pathway for
circuit current to flow.
• The same current flows through all
components.
• If there is a break in any point of the circuit no
current flows.
• The total resistance in a series combination of
resistance is determined from potential
difference and current.
Series Circuit
• The total power dissipated in a series current
is equal to the sum of power dissipated in the
individual resistors.
• The total resistance is calculated with the
equation R1+R2+R3 =RT
• Voltmeters are placed separately across
R1,R2,R3
Series circuits
• Resistors in a series circuit
Series Circuits
• Current in a series circuit is the same throughout the
circuit.
• EXAMPLE
• A battery of 40V is connected to a series arrangement of
4Ω, 6Ω and 10Ω resistors.
a) What is the resistance in a circuit?
using the formula R1+R2+R3 =RT
4Ω+ 6Ω + 10Ω =20 Ω
Series Circuits
b) What is the current flowing?
using the formula I= VAD ÷ R1+R2+R3
= 40 ÷ 20
= 2A
Series Circuits
c) What is the potential difference across each
resistor?
using the formula V=IR
V1 : 4Ω*2A =8V
V2 : 6Ω*2A=12V
V3 : 10Ω*2A=20V
Series Circuits
d) The total power dissipated in the circuits?
Power= VI
V TOTAL = V1+V2+V3
=8+12+20
=40V
P=VI
=40*2
=80W
Parallel Circuits
• The potential difference across each parallel
branch is the same and equal to the potential
difference of the battery.
• The total current is the sum of individual
branches.
• The total current divides at each junction
leading to parallel branches of a circuit in such
a way that potential difference across each
resistor branch is the same.
Parallel Circuits
• The total resistance for parallel combinations
of resistors is always less than the value of the
smallest resistance.
• The total power dissipated equals the sum of
powers dissipated in individual resistors.
• Adding an extra branch of a parallel circuit
does not affect other branches however other
current increases.
Parallel Circuits
• A break in the branch of the parallel circuits
does not affect other branches.
• The reciprocal of the total resistance equal
1÷RT=1÷R1+1÷R2+1÷R3
• The sum of current in the branch of parallel
circuit is equal to the current entering or
leaving the parallel section.
Parallel Circuits
• Parallel Circuit below calculate:
Parallel Circuits
a)The total resistance in the circuit.
• 1÷Rp=1÷R1+1÷R2
=1/6 + 1/12
=3/12
Rp =12/3 = 4Ω
RT =4Ω+8Ω
= 12Ω
Parallel circuit
b)The current flowing in the circuit.
using the formula V=IR
I=V/R
I=12/12
I=1A
Parallel Circuits
• What will be the value of the ammeter placed
at A be?
using the formula V=IR
I=12/4
I=3A
Parallel Circuits
d) What energy will be dissipated in 1 minute?
1minute= 60 seconds
using the formula E=Pt
P=IV
therefore E=I V t
E=1*12*60
E=720J