Question

- A coil having inductance 2:0 H and resistance 20 2 is
connected to a battery of emf 4.0 V. Find (a) the current
at the instant 0.20 s after the connection is made and
(b) the magnetic field energy at this instant.​

Answer 1

Answer:

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Explanation:

Here L=2.0H

R=20W,emf=4.0V,

t=0.20s

i0=(eR)=(420)

andτ=(LR)=(220)=0.1

(a)i=i0(1−r−t/τ

=(420)(1−r−0.2/0.1

=(15)(1−e−2)=0.17A

(b)12Li2=12×2(0.17)2

=0.0289=0.03J.

Answer 2

Answer:  L=2HR=20Ωemf=4Vt=0.20s

      I  

0

​  

=  

R

e

​  

=  

20

4

​  

=0.2

      t=  

R

L

​  

=  

20

2

​  

=0.1

(a)

       I=I  

0

​  

(1−e  

−0.2/0.1

)=0.17A

(b)  

         

2

1

​  

LI  

2

=  

2

1

​  

×2×0.17  

2

=0.03J

Explanation: