the magnetic flux through the coil is varying according to relation ∅=5t³+4t²+2t-5. calculate the induced current through the coil at t=2sec, if resistance of coil is 5 oh
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Answer:
Explanation:
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Verified answer
Solution:-
Given
[tex] \rm \phi = 5 {t}^{3} + 4 {t}^{2} + 2t - 5[/tex]
[tex] \rm \: t \: = 2sec \: [/tex]
[tex] \rm \: resistance = 5 \Omega[/tex]
Differentiate magnetic flux with respect to time
[tex] \rm \: induced \: emf \: = \dfrac{d \phi}{dt} [/tex]
[tex] \rm \: \dfrac{d \phi}{dt} = \dfrac{d(5 {t}^{3} + 4t {}^{2} + 2t - 5)}{dt} [/tex]
[tex] \rm \: = {(15 {t}^{2} + 8t + 2) }{} [/tex]
When at t = 2 sec
we get
[tex] \rm \: 15 \times (2) {}^{2} + 8 \times 2 + 2 = 78v[/tex]
Using formula
[tex] \boxed{ \rm \:E = IR }[/tex]
[tex] \rm \: E = 78v[/tex]
[tex] \rm \: R = 5 \Omega[/tex]
Put the value
[tex] \rm \: 78 = I \times 5[/tex]
[tex] \rm \: I = \dfrac{78}{5} = 15.6A[/tex]
So induced current is 15 .6 A