The potential difference VA-VB for the circuit shown in figure is -(22/x) V. The value of x is.

Question

The potential difference VA-VB for the circuit shown in figure is -(22/x) V. The value of x is. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

At junction C: i1+i2=i, where i1=(0-x/1), i2=(1-x/1) At D: 4 i+i1=0, solve to get x=(4/9) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/1af6af7d7afd901188b78c40d5e3357e-.png" /> Now VA-i1+1+i=VB ⇒ VA-VB =-(2+(4/9))=-(22/9) V

Q. The potential difference $V_{A} - V_{B}$ for the circuit shown in figure is $- \frac{22}{x} V$ . The value of $x$ is____.

At junction $C : i_{1} + i_{2} = i$ ,
where $i_{1} = \frac{0 - x}{1}, i_{2} = \frac{1 - x}{1}$
At $D : 4 i + i_{1} = 0$ , solve to get $x = \frac{4}{9}$

Now
$V_{A} - i_{1} + 1 + i = V_{B} \Rightarrow V_{A} - V_{B}$
$= - (2 + \frac{4}{9}) = - \frac{22}{9} V$

Q. The potential difference VA​−VB​ for the circuit shown in figure is −x22​V. The value of x is____.

At junction C:i1​+i2​=i, where i1​=10−x​,i2​=11−x​ At D:4i+i1​=0, solve to get x=94​ Now VA​−i1​+1+i=VB​⇒VA​−VB​ =−(2+94​)=−922​V

Q. The potential difference $V_{A} - V_{B}$ for the circuit shown in figure is $- \frac{22}{x} V$ . The value of $x$ is____.

At junction $C : i_{1} + i_{2} = i$ ,
where $i_{1} = \frac{0 - x}{1}, i_{2} = \frac{1 - x}{1}$
At $D : 4 i + i_{1} = 0$ , solve to get $x = \frac{4}{9}$

Now
$V_{A} - i_{1} + 1 + i = V_{B} \Rightarrow V_{A} - V_{B}$
$= - (2 + \frac{4}{9}) = - \frac{22}{9} V$