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 :

Answer 1

Using KVL in loop CDEFC
$E-I_{1}R+4-6=0$...............(i)
and loop AFCBA
$4+2(I_1+I_2)+4I_2-6=0$.....................(ii)
After solving equation (i) and (ii), we have

$I_1=\frac{3E-14}{4+3R};$

$I_2=\frac{R+6-E}{4+3R}$

$\left(A\right)I_2=0,E=R+6$
$E>6(R\ne 0)$
For current from F to C

$\left(B\right)I_2>0R+6>E$
$E<R+6$
Possible for any finite value of E, because R is finite.
Ans. (p,q,r,s)

(C) For current in 2$\Omega$ from B to C
$I_1+I_2=\frac{R-8+2E}{4+3R}>0$
$R-8+2E>0$
$E>4-\frac{R}{2}$
E can take any value from zero to infinity

Ans. (p,q,r,s)

(D) Current through R is zero
$\frac{3E-14}{4+3R}=0$
$E=\frac{14}{3}$
Ans. (s)