Please probide a matlab code for question $5$ only dont answer the rest.

Objective

To verify the relations between phase and line quantities as well as the equation of power in $3-$ balanced circuits.

Theory and Background

In a three$-$phase Y$-$connected circuit with balanced loads:

Line Voltage $=\sqrt[\phantom{2}]{?}3$ Phase Voltage

Line Current $=$ Phase Current

In a three phase $-$connected circuit with balanced loads:

Line Voltage $=$ Phase Voltage

Line Current $=\sqrt[\phantom{2}]{3}$ Phase Current

Procedure and Observations

Apparatus needed:

One $3-$phase Resistive Load $($ CL$-311-50).$

One ac Voltmeter.

One ac Ammeter.

$($a$)$Y$-$Connected Load $$

Connect the circuit as shown above $($you can obtain the resistance $503$ by paralleling the three resistors provided$).$

Turn on the supply and measure the following:

$E_{ry}=122$ Volts, $E_{ro}=70$ Volts

$E_{yb}=123\mathrm{.}4$ Volts, $E_{yo}=70\mathrm{.}6$ Volts

$E_{br}=123\mathrm{.}1$ Volts, $E_{bo}=71\mathrm{.}5$ Volts

$I_{RO}=0\mathrm{.}140$Amps,$I_{YO}=0\mathrm{.}140$Amps,$I_{Bo}=0\mathrm{.}142$Amps

$I_{No}=0,$Amps,

$0\mathrm{.}142$

Amps

$3.$ Remove the neutral wire and measure the phase currents:

$I_{RO}=0\mathrm{.}140$Amps,$I_{YO}=0\mathrm{.}140$Amps,$I_{BO}=0\mathrm{.}142$Amps

Calculate the power for each phase using results in step $2.$

$P_{ro}=E_{ro}{I}_{Ro}=9\mathrm{.}8=700\mathrm{.}14$watts

$P_{yo}=E_{yo}{I}_{Yo}=9\mathrm{.}884=70\mathrm{.}60\mathrm{.}14$ watts

$P_{bo}=E_{bo}{I}_{Bo}=10\mathrm{.}15,3=71\mathrm{.}50\mathrm{.}14watt{s}^2$

Calculate the total power with the following method:

$P_t=\sqrt[\phantom{2}]{3}{E}_{ry}{I}_{ro}=\sqrt[\phantom{2}]{3}co{l}_{0\mathrm{.}14}(step2)=29\mathrm{.}6W$

$($b$)-$Connected Load

$0\mathrm{.}1$

IIy.$($D$)$

Connect the circuit as shown above.

Turn on the power supply and note:

$E_{ry}=115\mathrm{.}2$ Volts, $I_{ry}=0\mathrm{.}228$Amps

$E_{yb}=116\mathrm{.}3$ Volts, $I_{yb}=0\mathrm{.}228$Amps

$E_{br}=115\mathrm{.}2$ Volts, $I_{br}=0\mathrm{.}229$Amps

$I_{Rr}=0\mathrm{.}396$ Amps, $I_{Yy}=0\mathrm{.}394$Amps,$I_{Bb}=0\mathrm{.}398$Amps

Calculate the power for each phase and add to obtain the total.

$P_{ry}=E_{ry}{I}_{ry}=115\mathrm{.}20\mathrm{.}228=26\mathrm{.}3$ watts

$P_{yb}=E_{yb}{I}_{yb}=116\mathrm{.}30\mathrm{.}228=26\mathrm{.}5$ watts

$P_{br}=E_{br}{I}_{br}=115\mathrm{.}26\mathrm{.}229=26\mathrm{.}4$ watts

$P_t=P_{ry}+P_{yb}+P_{br}=79\mathrm{.}2$ watts

Calculate the power using the formula:

Calculate the power using the formula:

$P_t=\sqrt[\phantom{2}]{3}{E}_{ry}{I}_{ry},(step2)=\sqrt[\phantom{2}]{3}113\mathrm{.}20\mathrm{.}228=45\mathrm{.}5W$

Discussion and Conclusion

$($a$)$Star

In a Y$-$connected system, is the line voltage greater or smaller than the phase voltage.

In a Y$-$connected system what is the line current when the phase current is $12$ Amps $.$

Why do you measure a neutral current in step $4.$

Is the total power calculated in step $6$ equal to $3$ times the power calculated in step $5.$ Comment on any differences.

$($b$)$Delta

On a graph sketch the phasors corresponding to the results recorded in step $2$ of the delta connected load.

Compare the line currents obtained in the delta connected load to those in the star connected load. Can you name any apparatus which utilizes this difference in behaviour.

$1.$ Its gratur since $\sqrt[\phantom{2}]{3}{V}_p=V_{L-L}2.I_{p_n}=\frac{I_L}{\sqrt[\phantom{2}]{3}}+I_L=\sqrt[\phantom{2}]{3}{I}_{p_n}=\sqrt[\phantom{2}]{3}12=3.$ Because there is no $100\%$ perfect hulanad systoms so thats why ucnod to littecuerront Pven if its neyliatio $4.$ Yes its al most thes ame the difforonces are du to the error in the Instrumonts. $5.$