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Question: The current (in Ampere) in an inductor is given by \[I = 5 + 6t\] where it is in sec. The self-induc...

The current (in Ampere) in an inductor is given by I=5+6tI = 5 + 6t where it is in sec. The self-induced emf in it is 10mV. Find the energy stored in the inductor and the power supplied to it at t=1t = 1 sec.

Explanation

Solution

Hint First, we will calculate the value of L using emf=LdIdtemf\, = \,L\,\,\dfrac{{dI}}{{dt}} . We will calculate energy using this value of L in U=12LI2U\, = \dfrac{1}{{2\,}}L{I^2} . We will use the value of emf and current to calculate power using formula P=eIP = eI .

Complete step by step solution
Inductance: It is a property of an inductor which changes the direction of current flowing through wire. L is the symbol of an inductor. Ratio of voltage induced to the rate of change of current is known as inductance.
Emf is an electromotive force which acts in the opposite direction to the applied voltage. It is used to decrease the flow of current.
emf=LdIdtemf = \,L\dfrac{{dI}}{{dt}} …(1)
Emf = 10mV =10×103V = \,10\, \times \,{10^{ - 3}}V
I=5+16tI\, = \,5\, + \,16t … (2)
Where I = current and t= time
Differentiating equation (2) with respect to time, we get
dIdt=16\dfrac{{dI}}{{dt}}\, = \,16
From equation (2)
L=emfdIdtL = \dfrac{{emf}}{{\dfrac{{dI}}{{dt}}}}
Putting values in above equation, we get
\Rightarrow L=10×10316L = \dfrac{{10 \times {{10}^{ - 3}}}}{{16}}
\Rightarrow L=10216L = \dfrac{{{{10}^{ - 2}}}}{{16}}
\Rightarrow L=0.625×103HL = 0.625 \times {10^{ - 3}}H
\Rightarrow L=0.625mHL = 0.625mH

Value of current (I) at 1 second
At t=1sect = 1\sec ,
\Rightarrow I=5+16(1)I = 5 + 16(1)
\Rightarrow I=21AI = 21A

Electrostatic energy d(U)
\Rightarrow U=12LI2U = \dfrac{1}{2}L{I^2}
Putting value of I and L , we get
\Rightarrow L=12(0.625×103)(21)2L = \dfrac{1}{2}(0.625 \times {10^{ - 3}}){(21)^2}
U=0.137JU = 0.137J

Power (P)
P=IeP = Ie
where e is emf
P=10×103×21 P=0.21Js  P = 10 \times {10^{ - 3}} \times 21 \\\ P = 0.21\dfrac{J}{s} \\\

Note
SI Unit of Inductance is Henry, Energy is Joule and Power is J/sec. While solving, if we replace m with M in emf value then the answer might get varied. Here; m is milli I.e. 103{10^{ - 3}} but M is mega I.e. 106{10^6} . Since current is in Ampere, SI units of time will be in seconds and not in minutes or hour.