In the electric network shown, when no current flows through the 4 Ω resistor in the arm EB, the potential difference between the points A and D will be :

Question

In the electric network shown, when no current flows through the 4 Ω resistor in the arm EB, the potential difference between the points A and D will be : (A) 3 V (B) 4 V (C) 5 V (D) 6 V

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/528dffba83b20061eaed1593f002491b-.png" /> i4=0 Using Kirchhoff's law: g -2 i-2-2 i-4=0 ; 3-4 i=0; i=3 / 4 A 4-i R-3=0 ⇒ i R =1 ; V AD =2 i-2-2 i=0 ⇒ V AB =2+4 i=2+((3/4))=5 V

Q. In the electric network shown, when no current flows through the $4 Ω$ resistor in the arm $EB$ , the potential difference between the points $A$ and $D$ will be :

3 V

4 V

5 V

6 V

$i_{4} = 0$
Using Kirchhoff's law:
$g - 2 i - 2 - 2 i - 4 = 0; 3 - 4 i = 0; i = 3/4 A$
$4 - i R - 3 = 0$
$\Rightarrow i R = 1;$
$V_{A D} = 2 i - 2 - 2 i = 0$
$\Rightarrow V_{A B} = 2 + 4 i = 2 + (\frac{3}{4}) = 5 V$

Q. In the electric network shown, when no current flows through the 4Ω resistor in the arm EB, the potential difference between the points A and D will be :

3 V

4 V

5 V

6 V

i4​=0 Using Kirchhoff's law: g−2i−2−2i−4=0;3−4i=0;i=3/4A 4−iR−3=0 ⇒iR=1; VAD​=2i−2−2i=0 ⇒VAB​=2+4i=2+(43​)=5V

Q. In the electric network shown, when no current flows through the $4 Ω$ resistor in the arm $EB$ , the potential difference between the points $A$ and $D$ will be :

$i_{4} = 0$
Using Kirchhoff's law:
$g - 2 i - 2 - 2 i - 4 = 0; 3 - 4 i = 0; i = 3/4 A$
$4 - i R - 3 = 0$
$\Rightarrow i R = 1;$
$V_{A D} = 2 i - 2 - 2 i = 0$
$\Rightarrow V_{A B} = 2 + 4 i = 2 + (\frac{3}{4}) = 5 V$