Question

The Thevenin's equivalent voltage V${}_{TH}$ appearing between the terminals A and B of the network shown in the figure is given by

• A. j16(3 - j4)
• B. j16(3 + j4)
• C. 16(3 + j4)
• D. 16(3 - j4)

Answer 1

The correct option is A j16(3 - j4)


$V=100\angle 0^0$

$\therefore V_{TH}=100\angle 0^0.\frac{4j}{3+4j}$

$=\frac{100\times 4j(3-4j)}{25}$

$V_{TH}=16j(3-4j)$ V