In the part of the circuit shown in the figure, the potential difference between points VA-VB=16 V . The current passing through the 2 Ω resistance will be

Question

In the part of the circuit shown in the figure, the potential difference between points VA-VB=16 V . The current passing through the 2 Ω resistance will be <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-s57nuvssrh1d.png" /> (A) 2.5 A (B) 3.5 A (C) 4.0 A (D) zero

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-s57nuvssqwp5.png" /> VA-VB=16 V ∴ 4i1+2(i1 + i2)-3+4i1=16 V ......(i) Using Kirchhoff's second law in the closed loops we have ⇒ 9-i2-2(i1 + i2)=0 ......(ii) Solving equations (i) and (ii), we get i1=1.5 A and i2=2 A ∴ Current through 2 Ω resistor =2+1.5=3.5 A

Q. In the part of the circuit shown in the figure, the potential difference between points $V_{A} - V_{B} = 16 V$ . The current passing through the $2 Ω$ resistance will be

$2.5 A$

$3.5 A$

$4.0 A$

zero

Q. In the part of the circuit shown in the figure, the potential difference between points VA​−VB​=16V . The current passing through the 2Ω resistance will be

2.5A

3.5A

4.0A

zero

VA​−VB​=16V ∴4i1​+2(i1​+i2​)−3+4i1​=16V ......(i) Using Kirchhoff's second law in the closed loops we have ⇒9−i2​−2(i1​+i2​)=0 ......(ii) Solving equations (i) and (ii), we get i1​=1.5A and i2​=2A ∴ Current through 2Ω resistor =2+1.5=3.5A

Q. In the part of the circuit shown in the figure, the potential difference between points $V_{A} - V_{B} = 16 V$ . The current passing through the $2 Ω$ resistance will be

$2.5 A$

$3.5 A$

$4.0 A$