Three phase circuits

Q. 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

1. $33.54\angle {103.43}^{\circ }$
2. $58.1\angle {103.43}^{\circ }$
3. $19.36\angle -{138.4}^{\circ }$
4. $41.3\angle {133.43}^{\circ }$

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

1. $1:1:\sqrt{3}$
2. 1 : 1 : 2
3. 1 : 1 : 0
4. $1:1:\sqrt{3/2}$

Q. Using thevenin equivalent circuit, determine the rms value of the voltage across the 100 ohm resistor after the switch is closed in the 3-phase circuit shown in figure

1. 115.5

Q. The line current of a three-phase four wire system are square wave with amplitude of 100 A. The three currents are phase shifted by ${120}^o$ with respect to each other. The rms value of neutral current is

1. 300 A
2. 0 A
3. 100 A
4. $\frac{100}{\sqrt{3}}A$

Q. A two-phase load draws the following phase current:

$i_1\left(f\right)=I_{m}sin(\omega t-{\varphi }_1)$

$i_2\left(f\right)=I_{m}cos(\omega t-{\varphi }_2)$

These currents are balanced if ${\varphi }_1$ is equal to

1. -${\varphi }_2$
2. ${\varphi }_2$
3. $(\frac{\pi }{2}-{\varphi }_2)$
4. $(\frac{\pi }{2}+{\varphi }_2)$

Q. Three identical resistances are connected in a star fashion against a balanced three phase voltage supply. If one of the resistances be removed then the percentage reduction in power is

1. 17.32%
2. 33.33%
3. 14.14%
4. 50%

Q. In the given network $V_1=100\angle 0^{0}V,V_2=100\angle -{120}^{0}V,V_3=100\angle +{120}^{0}V.$ The phasor current i (in Ampere) is

1. $173.2\angle -{60}^0$
2. $173.2\angle {120}^0$
3. $100.0\angle -{60}^0$
4. $100.0\angle {120}^0$

Q. The value of ${}^{\prime }{Z}^{\prime }$ such that no current flows in branch $Y$ will be

1. $(3.46+j2)\Omega$
2. $(13.85+j8)\Omega$
3. $(6.93+j4)\Omega$
4. $(27.7+j16)\Omega$