SSE JE Electrical Engineering 29 Oct 2020 Shift 2 Papers

Question : 120 of 200

Marks: +1, -0

A 15 V AC source is applied to a load impedance of (3 – j4) Ω. Find the load current.

Solution:

Given, Z = (3 – j4) Ω and V = 15 V

∴|Z|=√{3^{2}+4^{2}}=5

| \I |={15}/{5}=3 \A

\ϕ=\tan ^{-}({ \X }/{ \R })=\tan ^{-}({4}/{3})=53.13°

From the above concept,

\I ∠\ϕ=|3|(\cos (53.13)+ j \sin (53.13))

=(1.8+ j 2.4) \A

Go to Question:

Prev Question

Next Question