Question

For the three-phase circuit shown in the figure the ratio of the currents $I_R:I_Y:I_B$ is given by

• A. $1:1:\sqrt{3}$
• B. 1 : 1 : 2
• C. 1 : 1 : 0
• D. $1:1:\sqrt{3/2}$

Answer 1

The correct option is A $1:1:\sqrt{3}$



Assuming phase -sequence to be RYB

Taking $V_{RY}$ as the reference,

$V_{RY}$ = V$\angle$$0^0$

$V_{YB}$ = V$\angle$$-{120}^0$

$V_{BR}$ = V$\angle$$-{240}^0$

$I_R=\frac{V_{RB}}{R_1}=-\frac{V_{BR}}{R_1}$

$=-\frac{V\angle -{240}^0}{R_1}=\frac{V}{R_1}\angle -{60}^0$

$I_Y=\frac{V_{YB}}{R_1}=\frac{V\angle -{120}^0}{R_1}$

Using KCL,

$I_R+I_y+I_B=0$

$\frac{V}{R_1}\angle -{60}^0+\frac{V}{R_1}\angle -{120}_0+I_B=0$

$\Rightarrow I_B=\sqrt{3}\frac{V}{R_1}\angle {90}^0$

So,
$I_R:I_Y:I_B=\frac{V}{R_1}:\frac{V}{R_1}:\sqrt{3}\frac{V}{R_1}=1:1:\sqrt{3}$