COE Flashcards
Defn of electric current + symbol + unit
Rate of flow of charge wrt time
Symbol: I
Unit: A , amphere
Defn of electric charge + symbol + unit
When a constant current I flows thro a cross-sectn of a conductor for a duration t, the amt of electric charge Q that passes thro it is given by Q=It
Symbol: Q
Unit: C, coulomb
Defn of coulomb + symbol
One coulomb is defined as the amt of electric charge that passes thro a pt in a circuit in 1 sec when there is a const current of 1 amphere (1c = 1As)
Symbol: C
Defn of potential diff + symbol + unit
p.d. b/w 2 pts in a circuit is defined as the amt of electrical energy that is converted TO OTHER forms when 1 coulomb of charge passes from 1 pt to the other
Symbol: V
Unit: Voltage
Defn of volt + symbol
One volt is defined as the pd b/w 2 pts in a circuit when 1 joule of electrical energy is converted to other forms when one coulomb of charge passes from one pt to the other.
Symbol: V
Defn of electromotive force + symbol + unit
E.m.f. of a source is defined as the amt of electrical energy that is converted FROM OTHER forms of energy when the source drives unit charge thro a complete circuit.
Symbol: ℰ
Unit: volt
Defn of resistance + symbol + unit
Ratio of the pd across it to the current flowing thro it
Symbol: R
Unit: Ω (ohm)
Defn of ohm + symbol
One ohm is defined as the resistance of a conductor when a pd of 1 volt across it causes a current of one ampere to flow thro it
Symbol: Ω
Defn of ohm’s law
States that the pd across a conductor is directly proportional to the electric current passing thro it, provided that its temp remains const
Defn of power + symbol + unit
Rate at which energy conversion takes place wrt time
Symbol: P
Unit: W, Watt
Eqn of current (2)
I=Q/t
I = nevA OR nqvA
Eqn for potential diff
V = W/Q
Eqn for emf
ℰ = V/Q
Eqn for resistance (2)
R = V/I R = ρl / A
Eqns for power (4)
what do the V represent
P = W/t = IV = I^2R = V^2/R
The V are V across the device so it’s actually giving u Pdissipated
Psupplied = IVsupplied
Derivation of I = nevA
n = no. of charge carriers per unit vol = N (total no. of e) / vol
I = ΔQ/Δt = N e(charge of e-) / t = n (vol) e / t = n (A x) e / t = nevA
Internal resistance eqns
ℰ = Ir + IR = Ir + VR
V = ℰ - Ir
OR
Potential divider principle
VR = (R / R+r) ℰ
When is V = ℰ?
When internal resistance = 0 OR circuit is OPEN & I = 0
How to find r from V against I graph
V = ℰ - Ir
-r = grad
Diode defn + circuit symbol + I-V graph + simple explanation
Diode is a semiconductor device which allows current to flow in 1 direction only
Reverse bias: Current thro diode is very very small, R is very high
Forward bias: Current thro diode increases very rapidly when V > 0.7V, R is very low
How does an increase in current affect a non-ohmic conductor
As current increases, more charge carriers move thro the conductor per unit time which results in more collisions with the lattice ions. This increases the temp of the conductor which affects its resistivity in 2 ways:
- E- gain enough energy to break free, which results in an increase in the no. density of charge carriers -> Increases conductivity
- Lattice ions gain thermal energy and start to vibrate faster with greater amplitude. This causes more collisions b/w the charge carriers and the lattice ions which in turn slows down the movement of charge carriers -> Increases resistivity
Filament lamp + I-V graph + effect on resistivity & conductivity + how to show resistance increases/decreases
Conductivity: E- are already free and mobile at rm temp so as temp increases, there is no appreciable increase in no. of conducting free e-
Resistivity: freq of collision b/w lattice ions and e- increases so movement of charge carriers slow down
Hence, resistivity increases more than conductivity
How to show resistance increases/decreases?
R = V1/I1 < V2/I2
RATIO V/I increases -> curve gets more gentle -> resistivity and hence resistance increases
- DON’T SAY GRADIENT X
Thermistor + I-V graph + effect on resistivity & conductivity + how to show resistance increases/decreases
Conductivity: As temp increases, more e- acquire energy to break free from their atoms -> no. of charge carriers increases significantly
Resistivity: freq of collision b/w lattice ions and e- increases so movement of charge carriers slow down
Hence, conductivity increases more than resistivity
How to show resistance increases/decreases?
R = V1/I1 > V2/I2
RATIO V/I decreases -> curve gets more steep -> resistivity and hence resistance decreases
- DON’T SAY GRADIENT X
Graph of resistance against temp for a metal wire & NTC thermistor
metal wire is a str line
What happens to the brightness of a bulb?
Brightness is dependent on POWER
- Total R of circuit (eg decreases)
- Total I (eg increases)
- Look at spectators
E.g.
If bulb a, b & c are in series
Va = Ra I (I increases, so V increases)
Pa = Va^2 / Ra (V increases, so P increases)
=> A brighter
..”.. => Vb also increases
SO Since others all increase, Vc decreases
=> C is dimmer
Max power theorem & when max power is delivered, what is the efficiency?
When R = r
Efficiency: 50%
What contributes to the current if there’s a pair of +ve ions & e- b/w 2 charged plates?
both e- and +ve ions contribute to the current (LOOK FOR EXPLANATION)
Explain why the R at either of these voltages is not the reciprocal of the grad of the I-V graph at the same voltage values.
R of a conductor is defined as the ratio of the pd across it to the current flowing thro it (V/I), whereas the reciprocal of the grad of the graph is the change in pd divided by the change in current (dV/dI)
These quantities are not equal in general, and are equal only if the I-V characteristic graph is proportional (i.e. linear and passes the origin). The current I-V curve is not proportional, hence the 2 quantities are not equal.
How to describe graphs? (2)
- How y varies with x
2. How dy/dx [grad] varies with x
When using R = ρl / A, how to determine l & A in more confusing diagrams
A & l is always PARALLEL to flow of I
When switch is open, what’s the V
when switch is closed, what are the eqn for V
Cell + internal resistor + voltmeter = switch is open
Voltmeter has high R -> I=~0 so V = ε - Ir = ε
When switch closed,
Circuit:
|——voltmeter———|
|—–cell – internal resistor–|
|—– resistor —–|
Vab = E - Ir Vdc = IR = Vab = E-Ir Vdc = E(R/R+r)
