In a circuit shown below, if I_{1} = 1.5 A, then I_{2} will be:

This question was previously asked in

UP Jal Nigam E&M 2016 Official Paper

Option 3 : 1.5 A

**Method 1: Solving by using Nodal Analysis:**

Apply KCL at node A having voltage (VA):

I_{1} + I_{3} = I_{2} .... (1)

Given, I_{1} = 1.5 A

Now, \(\large{I_3=\frac{3-V_A}{2}}\)

And, \(\large{I_2=\frac{V_A}{2}}\)

From equation (1),

\(1.5+\frac{3-V_A}{2}=\frac{V_A}{2}\)

Multifilied both side by 2 for make simplifiaction easier,

3 + 3 - V_{A} = V_{A}

or, 2V_{A} = 6

V_{A} = 3 Volt

Now,

\(\large{I_2=\frac{V_A}{2}}=\frac{3}{2}=1.5\ A\)

**Method 2: Solving by Mesh Analysis:**

__ Note:__ I

I_{2} = I_{1} - I_{B} .... (1)

Apply KVL in loop 1,

6 = 2I_{A} + 2(I_{A} - I_{B})

6 = 2(1.5) + 2 (1.5) - 2I_{B}

2I_{B} = 0

I_{B} = 0

From equation (1),

I_{2} = I_{1} - I_{B} = 1.5 - 0 = 1.5 A

**Method 3: Solved by using Voltage drop method:**

Since, current through 2 Ω (I_{1}) is 1.5 A

∴ V_{1} = 3 Volts

Now, apply KVL in loop 1,

V_{2} = 6 - V_{1} = 6 - 3 = 3 Volts

∴ \(\large{I_2=\frac{3}{2}=1.5\ A}\)