Question

The Thevenin's equivalent of the network shown below is,

• A. $57.34\angle -55{}^{\circ }Vand(4.7-j6.7)\Omega$
• B. $57.34\angle -61{}^{\circ }Vand(4.7-j2.4)\Omega$
• C. $55.34\angle -57.3{}^{\circ }Vand(6.7-j4.7)\Omega$
• D. $55\angle -24{}^{\circ }Vand(4.2-j6.7)\Omega$

Answer 1

The correct option is A $57.34\angle -55{}^{\circ }Vand(4.7-j6.7)\Omega$

$V_{ab}=V_{Th}$

By applying KCL,

$\frac{V_{Th}-100}{20}+\frac{V_{Th}}{-j10}=0.02V_x$

where, $V_x=100-V_{Th}$

$\frac{V_{Th}-100}{20}+\frac{V_{Th}}{-j10}=0.02(100-V_{Th})$

$V_{Th}=\frac{7}{0.07+j0.1}=57.34\angle -55{}^{\circ }V.$

$Z_{Th}=\frac{V}{I}$
By applying KCL in the circuit given,

$\frac{V}{20}+\frac{V}{-j10}-0.02V_x=I$

and, $V_x=-V$

$\therefore \frac{V}{20}+\frac{V}{-j10}+0.02V=I$

$Z_{Th}=\frac{V}{I}=\left[\frac{1}{\frac{1}{20}+\frac{1}{-j10}+0.02}\right]$

$=(4.7-j6.7)\Omega$