Question

The potential difference $V_A-V_B$ for the circuit shown in Figure is $-\left(22/x\right)V$. Find the value of $x$.

Answer 1

From figure, $i_1=\frac{0-x}{1}$ $i_2=\frac{1-x}{1}$

$KCL$ at junction C : $I_1+I_2=i$

$\therefore$ $\frac{0-x}{1}$ $+\frac{1-x}{1}=i$ $⟹$ $i=1-2x$

Also $KCL$ at junction D : $4i+i_1=0$

$\therefore$ $4(1-2x)+$ $\frac{0-x}{1}=0$ $⟹x=\frac{4}{9}$

Now, $V_A-V_B-i_1\left(1\right)+1+1=0$

$\therefore$ $V_A-V_B\frac{4}{9}\left(1\right)+2=0$ $⟹V_A-V_B=\frac{-22}{9}$