Question

In the potentiometer circuit shown in the figure, the expression for $V_x$ is

• A. $(1-2\alpha )V$
• B. $(1-\alpha )V$
• C. $(\alpha -1)V$
• D. $\alpha V$

Answer 1

The correct option is A $(1-2\alpha )V$


$+V+IR+V=0$

$\therefore I=-\frac{2V}{R}$

$V_x=\left(\alpha R\right)I+V$

$=\alpha R\frac{(-2V)}{R}+V$

$=-2\times \alpha \times V+V$

$=(1-2\alpha )V$

Method - 2 (Eliminating options)

$At\alpha =0$

$V_x=V$

So, answer is either (a) or (b)

$At\alpha =1$
$V_x=-V$

$\Rightarrow$ Answer is (a).