In the circuit shown in figure potential difference between points A and B is 16 V. the current passing through 2Ω resistance will be

Question

In the circuit shown in figure potential difference between points A and B is 16 V. the current passing through 2Ω resistance will be (A) 2.5 A (B) 3.5 A (C) 4.0 A (D) zero

Tardigrade Admin · Accepted Answer

∴ 4 i+2(i1+i2)-3+4 i1=16 V...(i) Using Kirchhoff's second law in the closed loop 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 W resistor =2+1.5=3.5 A

Q. In the circuit shown in figure potential difference between points $A$ and $B$ is $16 V$ . the current passing through $2Ω$ resistance will be

2.5 A

3.5 A

4.0 A

zero

$∴ 4 i + 2 (i_{1} + i_{2}) - 3 + 4 i_{1} = 16 V$ ...(i)
Using Kirchhoff's second law in the closed loop we have
$9 - i_{2} - 2 (i_{1} + i_{2}) = 0$ ...(ii)
Solving equations (i) and (ii), we get
$i_{1} = 1.5 A$ and $i_{2} = 2 A$
$∴$ current through $2 W$
resistor $= 2 + 1.5 = 3.5 A$

Q. In the circuit shown in figure potential difference between points A and B is 16V. the current passing through 2Ωresistance will be