Question

Two circuits (as shown in Figure) are called Circuit $A$ and Circuit $B$. The equivalent resistance of Circuit $A$ is $x$ and that of Circuit $B$ is $y$ between 1 and 2.

• A. $y>x$
• B. $y=(\sqrt{3}+1)R$
• C. $xy=2R^2$
• D. $y-x=2R$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $y>x$
B $y=(\sqrt{3}+1)R$
C $xy=2R^2$
D $y-x=2R$

The equivalent circuits for A and B are shown in figure.
For circuit A: $x=R||(2R+x)=\frac{R(2R+x)}{R+2R+x}=\frac{2R^2+Rx}{3R+x}$
or $3Rx+x^2=2R^2+Rx$
or $x^2+2Rx-2R^2=0$
or $x=\frac{-2R_{-}^{+}\sqrt{4R^2-4\left(1\right)(-2R^2)}}{2}$
as negative resistance invalid so $x=\frac{-2R+\sqrt{12R^2}}{2}=(\sqrt{3}-1)R$
For circuit B: $y=2R+R||y=2R+\frac{Ry}{R+y}$
or $Ry+y^2-2R^2-2Ry=Ry$
or $y^2-2Ry-2R^2=0$
$y=\frac{2R_{-}^{+}\sqrt{4R^2-4\left(1\right)(-2R^2)}}{2}$
as negative resistance invalid so $y=\frac{2R+\sqrt{12R^2}}{2}=(\sqrt{3}+1)R$
thus, $y>x$; $xy=\left[\right(\sqrt{3}-1\left)R\right]\left[\right(\sqrt{3}+1\left)R\right]=(3-1)R^2=2R^2$
and $y-x=(\sqrt{3}+1)R-(\sqrt{3}-1)R=2R$