The resistance of platinium wire at 0°C is 4Ω What will be the resistance of the wire at 100°C. If the temperature coefficient of resistance of platinium is 0.0038/°C
We know, Resistance depend on temperature by relation.
R_{100} = 4(1 + 0.0038 × 100)
R_{100} = 4(1 + 0.38)
The correct answer is: 5.52Ω
Two metal ball of the same radius a are located in homogeneous poorly conducting medium with resistivity ρ.The resistance of the medium between the ball provided that the separation between them is much greater than the radius of the ball.
Let us impart charge q and –q to the balls respectively. Potential difference between two balls
V = V_{+} – V_{–}
Electric field at the surface of first sphere
electric field due to second sphere at the surface of first sphere is negligible
current density J = σE
The electric current
In case of insulator as the temperature decrease resistivity
The correct answer is: Increases
The law of refraction of direct current lines at the boundary between two conducting media having σ_{1} and σ_{2} are the conductivity of media 1 and 2 and α_{1} and α_{2} are the angle between the current lines and normal of the boundary surface as shown in figure.
Applying boundary condition of
Dividing (2) by (1)
The correct answer is:
A metal ball of radius a is surrounded by a thin concentric metal shell of radius b. The space between these electrodes is filled up with a poorly conducting homogeneous medium of resistivity ρ. The resistance of the interelectrode gap. assume b → ∞
As the current flow from inner to outer shell, the crosssection area increase. Assuming an element in form of spherical shell of radius r and thickness dr
The resistance
Net resistance
A long cylinder with uniformly charged surface an crosssection radius a moves with a constant velocity v along its axis. An electric field strength at the surface of the cylinder is equal to E. So current caused by mechanical transfer of a charge.
The convection current is
Here, dq = λdx. where λ is the linear charge density. But, from the gauss, theorem, electric field at the surface of the cylinder,
Hence, substituting the value of λ and subsequently of dq in Equation (i), we get
The correct answer is:
What is the drift velocity of an electron in a copper conductor having area 10 × 10^{–6} m^{2} and 10 × 10^{28} electron/m^{3}
We know, current through conductor = I = neAv_{d}
where A = Area
v_{d} = drift velocity
e = charge on electron
n = number of electron per unit volume
The correct answer is: 1.25 × 10^{–5} m/s
Two cylindrical conductor with equal crosssections and different resistivity ρ_{1} and ρ2 are put end to end. The charge at the boundary of the conductor if a current I flows from conductor 1 to conductor 2
The current density
J = I/A
Electric field in the two conductor are given by
If we consider of Gaussian pill box of crosssection area S at the interface of two conductors.
The flux of
According to Gauss Law
where σ is the surface charge density at the interface
So, charge at the boundary of two conductors
The correct answer is:
A current of 1A flows through a copper wire. How many electron pass through any crosssection of the wire in 1.6 seconds. (Charge on electron in 1.6 × 10^{–19}C)
I = 1A
t = 1.6 second
⇒ q = It = 1 × 1.6 = 1.6
q = 1.6C
q = ne
n = number of electrons
The correct answer is: 10^{19}
A battery of 6 cells, each of emf 2V and internal resistance 0.5Ω is being charged by a 220V supplying an external resistance of 10Ω is series. The potential difference across the battery is :
Total emf of the battery = 12 V
Total internal resistance of the battery = 3Ω
Current through battery (I) =
Potential difference across battery = emf of battery + Potential across the battery = 12 + (16 × 3)
= 60 V
The correct answer is: 60 V
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