In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point B with respect to the point A is :

Question

In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point B with respect to the point A is : (A) +1V (B) - 1V (C) - 2V (D) +2V

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/6379ac97fdd3c78b471dcbf3a20c1e7e-.png" /> Let us asssume the potential at A = V A =0 Now at junction C, According to KCL i 1+ i 3= i 2 1 A + i 3=2 A i3=2 A Now Analyse potential along ACDB v A +1+ i 3(2)-2= v B 0+1+2(1)-2=vB v B =3-2 v B =1 Volt

Q. In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point $B$ with respect to the point $A$ is :

+1V

- 1V

- 2V

+2V

Let us asssume the potential at $A = V_{A} = 0$
Now at junction $C$ , According to $K C L$
$i_{1} + i_{3} = i_{2}$
$1 A + i_{3} = 2 A$
$i_{3} = 2 A$
Now Analyse potential along ACDB
$v_{A} + 1 + i_{3} (2) - 2 = v_{B}$
$0 + 1 + 2 (1) - 2 = v_{B}$
$v_{B} = 3 - 2$
$v_{B} = 1$ Volt

Q. In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point B with respect to the point A is :

+1V

- 1V

- 2V

+2V

Let us asssume the potential at A=VA​=0 Now at junction C, According to KCL i1​+i3​=i2​ 1A+i3​=2A i3​=2A Now Analyse potential along ACDB vA​+1+i3​(2)−2=vB​ 0+1+2(1)−2=vB​ vB​=3−2 vB​=1 Volt

Q. In the circuit, given in the figure currents in different branches and value of one resistor are shown. Then potential at point $B$ with respect to the point $A$ is :

Let us asssume the potential at $A = V_{A} = 0$
Now at junction $C$ , According to $K C L$
$i_{1} + i_{3} = i_{2}$
$1 A + i_{3} = 2 A$
$i_{3} = 2 A$
Now Analyse potential along ACDB
$v_{A} + 1 + i_{3} (2) - 2 = v_{B}$
$0 + 1 + 2 (1) - 2 = v_{B}$
$v_{B} = 3 - 2$
$v_{B} = 1$ Volt