current ELECTRICITY Flashcards
OHMS law
current flowing thru a conductor is directly proportional to the potential difference V across the ends of th conductor, provided temp is kept constant
V= IR
RESISTIVITY DEPENDS ON:
open EYES WIDELY
𝝆
material
TEMP
RELAXATION TIME 𝜏
number density n
RESISTANCE DEPEAND ON
- material
- lenght (directly)
- area (inversely)
resistance
SI
dimension
SI: ohm
R= 𝝆 l/A
[ M^1 L^2 T^-3 A^-2]
resistivity 𝝆
SI :ohm m
𝝆 = RA/l
microscopic form of OHMS LAW
OR
VECTOR FORM OF ohms law
or
relation between J, σ , E
- from ohms law =V= IR
- V = I 𝝆 l/A
- V/l =J 𝝆
- V/I 1/𝝆 = J
- E σ = J
- J = σ E
current density** J ** and its direction ?
used in microscopic fom of OHMS LAW
LETITFLOW
it is the current flowing per unit AREA
SI: A m^-2
direction along electric current
DRIFT VELOCITY
THE average velocity with which the free e- get drifted under the INFLUENCE OF ELECTRIC FIELD
RELAXATION TIME 𝜏
THE time taken between two successive collision when e- are accelerated under the influence of E. F
mobility μ
magnitude of drift velocity of charge per unit electric field
SI : m^2 s^-1 V^-1
temperature coefficient α
SI UNIT
α = 𝝆t -𝝆o/ 𝝆o (Tf - To)
K^-1 or c^-1
1Ω
R= V/ I
the resistance of a conductor is said to be 1 Ω if 1A of current flow when potential difference of 1 volt is applied throug across the conductor
DEFINE resistivity
the resistance offered by a conductor of per unit lenght per unit area
HOW DOES RESISTIVITY VARIES WITH TEMP IN A CONDUCTOR
no time to relax
- WITH rise in temp: K.E increses
- collisions of no. of free e-‘s increases
- relaxation time decreases
- resistivity increase
HOW DOES RESISTIVITY VARIES WITH TEMP IN A SEMICONDUCTOR
Si, Ge
- when temp rises: n increases
- no. of free e- increases
- collison of free e- also increases
- relaxation time decreases
- BUT impact of increase in n is> than decrease in relaxation time
- so RESISTIVITY decreases
HOW DOES RESISTIVITY VARIES WITH TEMP IN NICHROME AND MAGNANIN
resistance bery high but weak dependence on temp
- when tempurature increases n increasa and realaxation time decreases
- but they compensate each other such a way that there is only slight INCREASE in RESISTIVITY OR RESISTANCE
LIMITATION OF OHMS LAW
draw graphs as well
- the relation between V AND I is not linear i.e. potential difference vary non linearly with current
e.g semiconductor - the variation of I with V MAY DEPEND UPON THE sign of V applied
on reversing the direction but keeping the magnitude fixed produce a current of same magnitude
e.g : junction diode
- the relation between I and V is not unique i.e. there is more than one value of V for the same current
cells
a source of energy that maintains continous flow of charge
EMF
SI UNIT
not a force
THE MAX POTENTIAL DIFFERENCE BTWN TWO ELECTRODES of a cell when no current is drawn form the cell or in an open circuit
VOLT
terminal POTENTIAL difference
use terminallll
potential difference between two terminals of the cell in a closed circuit
Internal resistance (r)
the resistance offered by the material of electrolyte to the flow of current inside the cell
Ω
factors depending on internal resistance
- nature of electrolyte
- conc of electrolyte (directly prop)
- temperature (inversely)
- distance btwn electrodes (directly)
- area of electrode (inversly)
expression of internal resistance
r = (ε/V - 1)R
relation between pd and r for discharging (V< ε)
V = ε- Ir
relation between pd and r for charging (V> ε)
ε= V - Ir
relation between pd and r for open circuit
I= 0
V = ε
RESISTORS IN SERIES
Req = R1+ R2
RESISTORS IN PARALLEL
1/REQ= 1/R1 +1/R2
εeq and req in SERIES COMBINATION
εeq = ε1+ ε2
req = r1 +r2
in ε if connection is reversed ε1-ε2
general case for n cells connected in SERIES combo
what happens to I when
* nr»R
* nr«R
* when diff emf and diff internal resistance is used and how can max current be drawn from it
I = nε/R +nr
- I = ε/r (less current) current due to a single cell
- I= nε/R (more current) current due to n single cell
- I = ε1 +ε2 +nε/ R + (r1 +r2 +..rn)
max current can be drawn if R»>total internal resistance
εeq and req in PARALLEL COMBO
εeq = ε1r2+ ε2r1/ r1r2
1/req= 1/r1 +1/r2
General case for n cells connected in PARALLEL combo
what happens to I when
* r/n«R
* r/n»R
* can max current be drawn from it
I = ε/R +r/n
- I = ε/R (less current) current due to a single cell
- I= nε/r (more current) current due to n single cell
- I = ε1 +ε2 +nε/ R + (r1 +r2 +..rn)
max current can be drawn if external RESISTANCE is very low than internal resistance
a cell of emf ε and internal resistance r is connected to a external variable RESISTANCE R
a. ε vs R
b. V vs R
c. V vs I
- ε is independent of R ε= V - Ir
- V is increasing R increasing and r/R increase
- V= ε- Ir
slope formila = V= - IR +ε
v= -rI + ε
y = mx + c
difference between EMF and V
- emf is a cause and V is an effect
- emf is the max pd between two electrodes when no current is drawn from it and V is the pd btwn two terminals when it is in a closed crk
- independent of R and dependent of R
electrical energy
SI
i HERT THEN KEEP HERTING
total work done W by the source of emf V in maintaining the electric current I in the given circuit for a specified time
joule
** H= I^2 R t**
H= V^2 /R t
H= V I t
1kWh
AKA 1 UNIT OF ELECTRICAL ENERGY
the amount of energy dissipated in 1 hour when th eelectric power in the circuit is 1 kilowatt
ELECTRICAL POWER
SI UNIT
PIVIR
RATE OF electrical energy supplied per unit time
Watt
P = I^2R
p = V^2 R
p = V I
**KIRCHOFF’S **
KCL
FIRST LAW
JUNCTION RULE
CURRENT LAW
what is it based on?
0
the algebraqic sum of current meeting at a juction in a closed circuit is ALWAYS** zero**
based on conservation of electric charge
KIRCHOFF’S 2nd LAW
LOOP LAW
MESH LAW
VOLTAGE LAW
in any closed part of the electrical circuit. the algebraic sum of EMF is equal to the algebraic sum of products of current and resistance flowing thru them
balanced condition in wheatstone bridge
R1/ R2 = R3/R4
no current flow thru galvanometer
WHEATSTONE BRIDGE
- wheat stone bridge aka resistance bridge is used to calculate the unknoiwn resistance by balancing the 2 legs of the bridge crk of which the one leg includes the unknown resistance
- principle : the WSB works on the principle of null deflectyion i.e. th eratio of their resistances are equal and no current flow thru the crk. the bridge is said to be in balance condn when no curent flow thru GAL i,e, P/Q = R/S
-
construction : it consist of 4 resistances P,Q,R,S arranged in the form of a bridge
a source of emf is connected between A & C
a GAL is connected between B &D
unknown resistance can be placed im any of the arms of the bridge
out of 4 cresistances one is unknown, one is variable and 2 are fixed or standard resistances - working: 1. when key is closed the current flows thru diff arms in WSB
- APPLY kirchoffs voltage law on ADBA. g is the resistance of GAL
- APPLY kirchoffs loop law on BCDB
- By adjusting the value of R the balancing conditon is obtained
- in case of balancing cond no current flows tthru GAL
- Ig = 0
- Q/P = R/S
