EXAM key phrases-
rate at which the component dissipates energy = Power
energy dissipates = energy transfer (Power-IV ✗ time = energy) or (W- Q V charge x voltage)
area of current/time graph = Charge • conversions
potential difference = Voltage
Resistance arises/increases in a metal lattice (electrons collide with ions) mm mm:&:P, millimeter
milli-✗ 10-3
Energy Transfer = work done (P = ¥ SO sub into P=I²R → F- = I²R) mm³ Tm3
Thinner wire of same length + material effects- CSA as radius is smaller 1m = 1000mm
CSA: diameter of wire measured in multiple places: an average is calculated to check the uniformly of the wire
current: the rate of flow of charge Potential difference (P. D) Voltage in a circuit Graphs - Ohms Law
measured in: Amps or coulombs per second ↳ voltage (V) • a charge (electron) will resistor → gradien
(A) (CS-1) In a circuit, p. d is measured gain energy when passing grad- constant
- = Δ Q charge from two points through a battery/cell. It ÷ R=¥ s
t gradient: Definition = P-d is the electrical will then transfer this energy
P. D/voltage
rearranged- current energy transferred to each unit to component so electron gradient:
Q= It of charge passing between 2 points. loses it's energy. Filament lamp
¥ = resistance For fixed c
t
measured in: vor so-' ↳ this causes a P. D ↳ temperat
Conductor: a material through which charge V = ¥ ← withe/energy transfer 'across battery + components' ; definition: c
can flow (all materials conduct) rearranged. ⇒ In Parallel: use a voltmeter • as current increases, and potentia
p. d/voltage
Good conductor = metal due to free electrons in P. D in terms of current, work + time: resistance increases as across a dev
Parallel series
their structure. "E" EEE •Ya, ¼%I÷Iv ""IV. circuit: gradient decreases proportional
• Electrons are charge carriers
derivation component
↳ Lightbulb
resistor '
A • If asked about a graph name the:
→ gradient, intercept, area
remains con
↳ only for an
Relationship between
V = voltmeter
☆ movement of free electrons in a metal A- ammeter
BEFORE and AFTER a potential difference) current and drift velocity: V
is applied. ↑
voltage HE e
model of conduction and resistanc
Answer I • a • Resistance when free electrons move they collide
BEFORE: free electrons move rapidly and randomly # ¼!'ge R- ¥ in ohms a with the fixed ions in the metal latt
in a metal. without a (PD) voltage there N-MAL ] ¥=%%Éty increasing electrical resistance. when
is no net motion and no net flow of charge. nA l e collide, electrons transfer their vibrati
Tid?
AFTER: When a voltage is applied the electrons notes-
- work done energy to the ions. ions vibration size
still move rapidly and randomly but A = cross sectional area I =nAve when we have a current increases and displays thermal energy in
there is a current which causes a drift N = no. of electrons we can find the work wire.
velocity in one direction (✗ 10-4ms'') e- charge done: (power
I - n Ave
✓ = drift velocity w. electrical potential
(energy lost per unit ✗ 519% POWER EQUATIONS
charge
In terms of units:
① = charge (which is same as Nx e)
substitute V-IR into power
P = VI = Electrical work done
I = CS" c-D= [MY m² ms' 19 n = free electron density
time taken ① P-IV
A = m² CS-I = (s-t ↳ n- ¥-volume- Axl so N = MAL
② If V-IR, then P=I(IR)
V: ms-' Energy DISSIPATION in a conductor I. P=I²R
n - m-3 ← density ↳ power dissipation = rate at which energy changes 3 If I = ½, then PY1R)V
from electrical to another form.
Einarcge
rate at which the component dissipates energy = Power
energy dissipates = energy transfer (Power-IV ✗ time = energy) or (W- Q V charge x voltage)
area of current/time graph = Charge • conversions
potential difference = Voltage
Resistance arises/increases in a metal lattice (electrons collide with ions) mm mm:&:P, millimeter
milli-✗ 10-3
Energy Transfer = work done (P = ¥ SO sub into P=I²R → F- = I²R) mm³ Tm3
Thinner wire of same length + material effects- CSA as radius is smaller 1m = 1000mm
CSA: diameter of wire measured in multiple places: an average is calculated to check the uniformly of the wire
current: the rate of flow of charge Potential difference (P. D) Voltage in a circuit Graphs - Ohms Law
measured in: Amps or coulombs per second ↳ voltage (V) • a charge (electron) will resistor → gradien
(A) (CS-1) In a circuit, p. d is measured gain energy when passing grad- constant
- = Δ Q charge from two points through a battery/cell. It ÷ R=¥ s
t gradient: Definition = P-d is the electrical will then transfer this energy
P. D/voltage
rearranged- current energy transferred to each unit to component so electron gradient:
Q= It of charge passing between 2 points. loses it's energy. Filament lamp
¥ = resistance For fixed c
t
measured in: vor so-' ↳ this causes a P. D ↳ temperat
Conductor: a material through which charge V = ¥ ← withe/energy transfer 'across battery + components' ; definition: c
can flow (all materials conduct) rearranged. ⇒ In Parallel: use a voltmeter • as current increases, and potentia
p. d/voltage
Good conductor = metal due to free electrons in P. D in terms of current, work + time: resistance increases as across a dev
Parallel series
their structure. "E" EEE •Ya, ¼%I÷Iv ""IV. circuit: gradient decreases proportional
• Electrons are charge carriers
derivation component
↳ Lightbulb
resistor '
A • If asked about a graph name the:
→ gradient, intercept, area
remains con
↳ only for an
Relationship between
V = voltmeter
☆ movement of free electrons in a metal A- ammeter
BEFORE and AFTER a potential difference) current and drift velocity: V
is applied. ↑
voltage HE e
model of conduction and resistanc
Answer I • a • Resistance when free electrons move they collide
BEFORE: free electrons move rapidly and randomly # ¼!'ge R- ¥ in ohms a with the fixed ions in the metal latt
in a metal. without a (PD) voltage there N-MAL ] ¥=%%Éty increasing electrical resistance. when
is no net motion and no net flow of charge. nA l e collide, electrons transfer their vibrati
Tid?
AFTER: When a voltage is applied the electrons notes-
- work done energy to the ions. ions vibration size
still move rapidly and randomly but A = cross sectional area I =nAve when we have a current increases and displays thermal energy in
there is a current which causes a drift N = no. of electrons we can find the work wire.
velocity in one direction (✗ 10-4ms'') e- charge done: (power
I - n Ave
✓ = drift velocity w. electrical potential
(energy lost per unit ✗ 519% POWER EQUATIONS
charge
In terms of units:
① = charge (which is same as Nx e)
substitute V-IR into power
P = VI = Electrical work done
I = CS" c-D= [MY m² ms' 19 n = free electron density
time taken ① P-IV
A = m² CS-I = (s-t ↳ n- ¥-volume- Axl so N = MAL
② If V-IR, then P=I(IR)
V: ms-' Energy DISSIPATION in a conductor I. P=I²R
n - m-3 ← density ↳ power dissipation = rate at which energy changes 3 If I = ½, then PY1R)V
from electrical to another form.
Einarcge
