Electric Current Test - 21

In liquids the current is carried by ions. Ions are charged particles (atoms or groups of atoms). Ions can either be positive or negative, the positive ions being attracted to the negative plate and the negative ions to the positive plate. A positive ion is a particle with some negative charge taken away and a negative ion is a particle with some extra negative charge added.

Thus, intensive recombinations of electrons and positive ions into neutral atoms and molecules occur. This results in the emission of light by the discharge, i.e., the negative glow is mainly due to a glow of recombination. Hence in electric discharge the electric current is due to both electrons and positive ions.
 

Current density is equal to Electric current divided by a given surface area, or it is the current flowing per a given cross section area.
The current density decreases with increase in cross section area which means they are inversely proportional.

When there is no Potential Difference across the block, the motion of electrons will not be systematic. They will move in random motion.
By considering the motion of all the free electron in every possible direction due to randomness, the Average mean velocity of free electrons turns out to be zero.
 

Superconductivity is a phenomenon of exactly zero electrical resistance and expulsion of magnetic fields occurring in certain materials when cooled below a characteristic critical temperature.
Even near absolute zero, a real sample of a normal conductor shows some resistance. In a superconductor, the resistance drops abruptly to zero when the material is cooled below its critical temperature. An electric current flowing through a loop of superconducting wire can persist indefinitely with no power source.
Example: - Liquid Nitrogen.
 

Unit of resistance should be 20ohm

Solution :- Resistance, R= ρl/A

When length l increases three times, the cross section A will decease three times of previous. So length becomes 3l and cross section becomes A/3.

Now resistance becomes ,R  = ρ3l / πA/3

​ =9R=9×20=180Ω

The resistivity of a semiconductor decreases with temperature. This is because of increasing temperature, the electrons in the valence band gain sufficient thermal energies to jump to the conduction band. As the number of electrons in the conduction band increases, so conductivity increases and resistivity decreases.

Cross sectional area of the copper tube is:
A=(π/4)​(d₁²​−d₂²​)=(π/4)​(102−92)=14.9cm2=14.9∗10⁻⁴m²
l=5m
ρ=1.7∗10⁻⁸
Resistance is given by: R=ρl/A​=5.77 x 10⁻⁵Ohm

Let, the length of wire = l unit

Fraction required =a

Then, the lenght of wire which is stretched =al

Resistance of a wire of length l and cross-sectional area A is given by R=ρl/A​
here, x=ρ(4a)​/(2a×a)= 2ρ/a​ ;
y=ρ(a)​/ 2a×4a=ρ/8a​ and
z= ρ(2a)/4a×a​=ρ/2a​
thus, x>z>y

Length of the wire, l =15 m
Area of cross-section of the wire, a = 6.0 × 10⁻⁷ m²
Resistance of the material of the wire, R = 5.0 Ω
Resistivity of the material of the wire = ρ
Resistance is related with the resistivity as,
ρ=RA/l
=5.0×6×10⁻⁷/15
=0.2×10⁻⁶Ωm

Unit of resistance should be 5ohm.

Resistance of a wire is given by the formula 




 

Let  n₁​ and n₂​ be the densities of charge carriers in the two wires, Vd₁​​ and Vd₂​​ be the drift velocities of charge carriers in the two wires,
A be the area of cross section,
e is the charge on electron and
I is the current flowing through both the wires
Hence, I=n₁​eAV_d1​​=n_1​eAV_d2​​
∴n₁​V_d1​​=n₂​V_d2​​
∴ V_d1/V_d2​​ ​​​=​n2​​/n1
∴ ​​V_d1​​​/ V_d2=4/1​

Here we are using the formula
current = neAv ...............(1)
 where n= number of electron density
A =area of cross section
e= charge of electron
v=drift velocity
As both the wires are of the same materials then the electron density is the same for both.
For the first wire, current I is given by,  
I = neAv  
 Now according to the problem another wire of half the radius and of the same material when the drift velocity is 2v
Therefore for the second wire current is = ne(A/4)(2v)
                                                                = (neAv)/2
                                                          now putting I at this place
= I/2
 

In case of semiconductor Resistivity∝1​/temperature
In this case as temperature increases electrons jump to conduction band resistance decreases R∝ρ(Resistivity) so resistivity also decreases and current increases.
Vibration of electrons increases with rise in temperature and hence collision between ions and electrons increases with temperature.

We know, a piece of copper is a metal while that of germanium is a semiconducting material. For metals resistance increases with increase in temperature. The semiconductor has a negative temperature coefficient of resistance. Hence, when it is cooled its resistance increases.

I₁ = n x -(e) x A x V
I₂ = n x 2e x A x V/4
   = n x e x A x V/2
Total Current = I₁ + I₂ 
=nea(V/2) - (-neav) (Since both ions will move in opposite directions,  we subtract them)
Hence Current I= (neav/2) + neav
                          = 3/2neaV
 

We have the relation between current I, n = no. of carrier per unit volume, drift velocity v_e and CSA A as.

I = neAv

Let current flowing in the 2nd wire I₂ = I thus current flowing through the 1^st wire I₁ = 4I

Let the radius of the first wire be r thus radius of the 2^nd wire = 2r

In the first wire

A₁ = πr²

In the 2^nd wire

A₂ = π(2r)² = 4πr²

I₁/ I₂ = (A₁v₁ )/(A₂v₂ )  (ne being same for two wires as they are made of the same material)

4I/I = πr² v₁/4 πr²v₂

=> v₁/v₂ = 16/1

=> v­₁:v₂ = 16:1

Thus ratio of the drift velocities are 16:1

According to Ohm's law R=V/I​ 
Where R= resistance, V= Voltage, I= Current
Also R  is directly proportional to temperature T and inversely proportional to the slope.
∴ from the above example
R₁​∝T₁​ and R₂​∝T₂​
but from given figure R₁​2​
∴T_1​2

The current flowing through the wire is given by 
I=neAv
Where I = current
n = charge carrier current density
e = charge on the carriers
A = cross sectional area
v = drift velocity
In the above example
I = current = i
n = charge carrier current density = p
e = charge on the carriers = q
A = cross sectional area = s
v = drift velocity = v
hence i=pqsv
∴v=i/pqs

Conduction of electrons in a conductor.
The motion of electrons is random but in presence of external electric field, they slowly drift opposite to  and cause a current
 

The drift speed of electrons is inversely proportional to the radius of cross section.
Thus here the cross-sectional area at point P is smaller than the cross-sectional area at point Q. Hence, the drift velocity is large for smaller area and small for larger area. The drift speed will relate as per the relation given below
vP​>vQ​
 

Between two square faces the area is 0.01×0.01=1×10⁻⁴m² and length is 0.5m. So, the resistance is (3.5×10⁻⁵×0.5)/(1×10^-4)=0.175ohm, this can be written as, (35/2) x10^-2 ohm
Between the rectangular face the area is 0.01×0.5=0.005m² and the length is 0.01m, hence the resistance is (3.5×10⁻⁵×0.01)/0.005=7×10^-5 ohm

Given A storage battery is connected to a charger for
charging with a voltage of 12.5 Volts. The internal
resistance of the storage battery is 1 ohm. When the
charging current is 0.5 A, the emf of the storage
battery is:
When the battery is charging
We know that
V = e + i r  
V = 12.5, i = 0.5, r = 1  
12.5 = e + 0.5 x 1
12.5 = e + 0.5
e = 12.5 – 0.5
e = 12 v

Potential difference(V) across the terminals of a cell
i. Discharging of cell when cell is supplying current,
Therefore, V=V_P+V_Q=E-Ir, i,r., Vii. charging of cell when cell is taking current,
V=VP-V_Q=E+Ir, i.e., V>E
iii. When the cell is short circuited i.e., external resistance is zero,
=> I=E/r
=>  V=V_P-V_Q=E-Ir=0 i.e., V=0

ε₁+ε₂=I₁R,ε₁−ε₂=I₂R
Dividing ε₁=( I₁+I₂/I₁-I₂)ε₂

Let V₁​, I₁​ be the voltage and current for Nichrome wire of length 2L and cross-sectional area A.
V_2​,I₂​ be the voltage and current for Nichrome wire of length L and cross-sectional area 2A.
According to Kirchhoff's law,
I_1​= ​(V₁​−V)​/R₁
I_2​=​(V−V2​)​/R₂
where R is thee resistance of Nichrome wire,
R=ρl/A​
Hence,
R₁​=ρ2L/A​=2R
R₂​= ρL/2A ​=0.5R
Since current entering from one end leaves from another end,
I1​=I2​
(8−V​)/2=(V−1)/0.5​
V=6/2.5​=2.4V