Question

In the circuit shown below, the current I is equal to

• A. $1.4\angle 0{}^{\circ }A$
• B. $2.0\angle 0{}^{\circ }A$
• C. $2.8\angle 0{}^{\circ }A$
• D. $3.2\angle 0{}^{\circ }A$

Answer 1

The correct option is B $2.0\angle 0{}^{\circ }A$

Converting delta into star, the circuit can be redrawn as below.


Equivalent impedance of the circuit,

$Z=(2+j4)\parallel (2-j4)+2$

$Z=\frac{(2+j4)(2-j4)}{2+j4+2-j4}+2$

$\Rightarrow Z=\frac{4+16}{4}+2=7\Omega$
Therefore,

$Current,I=\frac{14\angle 0{}^{\circ }}{7}=2\angle 0{}^{\circ }A$