This question was previously asked in

ESE Electronics 2016 Paper 1: Official Paper

Option 4 : √153

__Concept:__

When adding phasors, the real part and the imaginary part are respectively added, i.e.

If z_{1} = a + ib and z_{2} = c + id

The addition of the two phsors will give:

z = a + ib + c + id

z = (a + c) + i(b + d)

The magnitude of a phasor z = a + jb is given by:

|a + jb| = \(√{a^2 + b^2 }\)

__Calculation__:

Given P_{1} = 3 + j4 and P_{2} = 6 - j8

P_{1} - P_{2} = (3 - 6) +j (4 + 8)

P_{1} - P_{2} = -3 + 12j

The magnitude of the above phasor is now:

\(|P_1-P_2|=\sqrt {(-3)^2+(12^2)}\)

**|P _{1} - P_{2}| = √153**