Electromagnetic Induction and Alternating Current 4 Question 22

####22. A circuit containing a two position switch S is shown in figure.

(1991, 4+4M)

(a) The switch S is in position 1. Find the potential difference VAVB and the rate of production of joule heat in R1.

(b) If now the switch S is put in position 2 at t=0. Find

(i) steady current in R4 and

(ii) the time when current in R4 is half the steady value. Also, calculate the energy stored in the inductor L at that time.

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Answer:

Correct Answer: 22. (a)-5V , 24.5W $

(b) (i) 0.6A

(ii) 1.386×103s, 4.5×104J

Solution:

  1. (a) In steady state, no current will flow through capacitor. Applying Kirchhoff’s second law in loop 1

2i2+2(i1i2)+12=02i14i2=12 or i12i2=6

Applying Kirchhoff’s second law in loop 2

122(i1i2)+32i1=0

4i12i2=9

Solving Eqs. (i) and (ii), we get

i2=2.5A and i1=1A Now, VA+32i1=VB or VAVB=2i13=2(1)3=5VPR1=(i1i2)2R1=(12.5)2(2)=24.5W

(b) In position 2 : Circuit is as under

(i) Steady current in R4 :

i0=33+2=0.6A

(ii) Time when current in R4 is half the steady value

t1/2=τL(ln2)=LRln(2)=(10×103)5ln(2)=1.386×103sU=12Li2=12(10×103)(0.3)2=4.5×104J



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