In the given circuit ' a ' is an arbitrary constant. The value of m for which the equivalent circuit resistance is minimum, will be √( x /2). The value of x is .

Question

In the given circuit ' a ' is an arbitrary constant. The value of m for which the equivalent circuit resistance is minimum, will be √( x /2). The value of x is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

R =(( ma /3))+(( a /2 m )) ( dR / dm )=( a /3)-( a /2 m 2)=0 ( a /3)=( a /2 m 2) m 2=(3/2) m =√(3/2) x =3

Q. In the given circuit ' $a$ ' is an arbitrary constant. The value of $m$ for which the equivalent circuit resistance is minimum, will be $\frac{x}{2}$ . The value of $x$ is _____.

$R = (\frac{ma}{3}) + (\frac{a}{2 m})$
$\frac{d R}{d m} = \frac{a}{3} - \frac{a}{2 m ^{2}} = 0$
$\frac{a}{3} = \frac{a}{2 m ^{2}}$
$m^{2} = \frac{3}{2}$
$m = \frac{3}{2}$
$x = 3$

Q. In the given circuit ' a ' is an arbitrary constant. The value of m for which the equivalent circuit resistance is minimum, will be 2x​​. The value of x is _____.

R=(3ma​)+(2ma​) dmdR​=3a​−2m2a​=0 3a​=2m2a​ m2=23​ m=23​​ x=3

Q. In the given circuit ' $a$ ' is an arbitrary constant. The value of $m$ for which the equivalent circuit resistance is minimum, will be $\frac{x}{2}$ . The value of $x$ is _____.

$R = (\frac{ma}{3}) + (\frac{a}{2 m})$
$\frac{d R}{d m} = \frac{a}{3} - \frac{a}{2 m ^{2}} = 0$
$\frac{a}{3} = \frac{a}{2 m ^{2}}$
$m^{2} = \frac{3}{2}$
$m = \frac{3}{2}$
$x = 3$