Question

A balanced abc-sequence Y-connected source with $V_{an}=100\angle {10}^{\circ }$ V is connected to a $\Delta$-connected balanced load with impedence $(8+j4)\Omega$ per phase. Then the line current $I_c$ is

• A. $33.54\angle {103.43}^{\circ }$
• B. $58.1\angle {103.43}^{\circ }$
• C. $19.36\angle -{138.4}^{\circ }$
• D. $41.3\angle {133.43}^{\circ }$

Answer 1

The correct option is A $33.54\angle {103.43}^{\circ }$

$Z_{\Delta }=(8+4j)\Omega$
$Z_Y=\frac{Z_{\Delta }}{3}=(\frac{8}{3}+\frac{j4}{3})\Omega$
$V_{an}=100\angle {10}^{\circ }V$
$V_{cn}=100\angle {130}^{\circ }V$
In star; $I_{c,\text{line}}=I_{c,\text{phase}}=\frac{100\angle {130}^{\circ }}{\frac{(8+4j)}{3}}$
$=33.54\angle {103.43}^{\circ }A$