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Question

In column II, R is a nonzero variable finite resistance for each diagram. Some facts are given in column I and corresponding figures are given in column II. Match the two columns :

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Solution

Using KVL in loop CDEFC
EI1R+46=0...............(i)
and loop AFCBA
4+2(I1+I2)+4I26=0.....................(ii)
After solving equation (i) and (ii), we have

I1=3E144+3R;

I2=R+6E4+3R


(A)I2=0,E=R+6
E>6(R0)
For current from F to C


(B)I2>0R+6>E
E<R+6
Possible for any finite value of E, because R is finite.
Ans. (p,q,r,s)


(C) For current in 2Ω from B to C
I1+I2=R8+2E4+3R>0
R8+2E>0
E>4R2
E can take any value from zero to infinity
Ans. (p,q,r,s)


(D) Current through R is zero
3E144+3R=0
E=143
Ans. (s)

121329_74463_ans.jpg

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