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Now, e=L(di/dt)
L=e/(di/dt)
1 henry=1 volt/(1 ampere/second)
Coefficient of mutual induction:
It is a measure of the induction between two circuits, It is the ratio of the electromotive force in a circuit to the corresponding change of current in neighboring circuits; usually measured in henries.
Coefficient of mutual induction is the ratio of induced e.m.f in secondary coil to the rate of change of current in primary coil.
If a capacitor of capacitance 9.2F has a voltage of 22.5V across it. Calculate the energy of the capacitor.
E = CV^{2}/2
= 9.2 x (22.5)^{2}/2
= 2328.75 J
A solenoid coil has 10 turns per cm along its length and a cross sectional area of 10 cm^{2}. 100 turns of another wire are wound round the first solenoid coaxially. The two coils are electrically insulated from each other. The mutual inductance between the two coils is
n_{1}=10 turns per cm, =1000 turns per metre
n_{2}l=100, A=10 c=0.1 m^{2}, M=μ_{0}n_{1}(n_{2}l) A
=4πx10^{7}×1000×100×0.1H=0.13mh
If we take the analogy between mechanical motion and electric circuits, then inductance is analogous to
Inductance in electric circuits plays the same role as moment of inertia in mechanical circuits.
Two pure inductors are connected in series, if their self inductance is L. Find the total inductance
The total inductance of inductors connected in series is given by the following formula:
L_{eq} = L_{1} + L_{2}
Substituting the values in the equation, we get the total inductance as
L_{eq }= L + L = 2L
Hence, the total inductance of two inductors connected in series is 2L.
Suppose there are two coils of length 1m with 100 and 200 turns and area of cross section of 5 x 10^{3} m^{2}. Find the mutual inductance.
The magnetic field through the secondary of N2 turns of each other of area S is given as,
N2= Φ2=N2(BS)
=μ_{0}n_{1}N_{2}i_{1}S
M= N_{2}Φ_{2}/i1
M= μ_{0}n_{1}N_{2}S
Substituting the values.
M=(4πx10^{7}) x100x200x5x10^{3}
=1287142.86x10^{10}
=12.57x10^{5}H
If the current in primary coil changes from 5 A to 2 A in 0.03 sec and the induced e.m.f. is 1000 volts, find the mutual induction
As φ =mi
now take change in flux
dφ/dt=d(mi)/dt
we can write it as
dφ/ dt= m Δi/Δt
e=m(25)/0.03
e=3m/0.03
1000=300m/3
m=10
The mutual induction does not depend on which of the following factors
Mutual inductance is the phenomena where an adjoining wire coil induces the voltage from the magnetic field produced by a wire coil. A transformer is a tool which is made up of two or more coils in close proximity to each other to create a situation of mutual inductance between the coils.The factors determining the value of mutual inductance are —
a. The number of turns of the coils
b. The geometric shape and size of the coils
c. The separation of the coils and
d. The angular orientation between the coils
So clearly it does not depend upon the current flow through both of the cells.
It is also called the Coefficient of induction. Coefficient of induction is about the ratio of current induced with respect to magnetic flux. This constant of proportionality is known as Inductance. It is a scalar quantity.
The dimension of inductance M^{1}·L^{2}·T^{−2}·A^{−2}
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