Percentage Impedance MCQ Quiz - Objective Question with Answer for Percentage Impedance - Download Free PDF

Concept

The equivalent resistance referred to the LV side is given by:

\(R_{eq}={R_{LV}}\space +\space {R_{HV}\over n^2}\)

where, n = Turn ratio

The turn ratio of the transformer is given by:

\(n={N_H\over N_L}\)

Calculation

Given, n = 2

R_LV = 0.06 Ω

R_HV = 0.2 Ω

\(R_{eq}={0.06}\space +\space {0.2\over (2)^2}\)

R_eq = 0.11 Ω

Concept:

Line voltage = √3 V_Ph

Where,

V_Ph → Voltage in each phase.

All line voltages are equal in magnitude and are displaced by 120°, and

All line voltages are 30° ahead of their respective phase voltages.

total power = 3 x power in each phase.

Power in each phase = VPh²/(2× Z)

Where;

Z → Impedance 

Calculation:

Given;

Total power = 1.8 kW

Power factor = 0.5

Line voltage = 230 V

Phase voltage (V_Ph) = 230/√3 

Power in each phase = 1.8/3 = 0.6 kW

Then;

   V²_Ph/(2× Z) = 0.6 kW

⇒(230/√3 )²/(2 × Z) = 0.6 kW

∴ Z = 14.7 Ω

Concept:

Base impedance is given by

\({Z_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)

kV_base is base voltage rating

MVA_base is base MVA rating

Calculation:

Given that, base kV = 11 kV

Base MVA = 100 MVA

Base impedance, \({Z_{base}} = \frac{{{{11}^2}}}{{100}} = 1.21\;{\rm{\Omega }}\)

Important:

The relation between new per-unit value & old per unit value impedance

\({({Z_{pu}})_{new}}\; = {({Z_{pu}})_{old}} \times {\left( {\frac{{k{V_{base}}}}{{k{V_{new}}}}} \right)^2} \times \;\left( {\frac{{MV{A_{new}}}}{{MV{A_{old}}}}} \right)\)

Also \({Z_{pu}} = \frac{{{Z_{Actual}}}}{{{Z_{base}}}}\)

 \({Z_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)

Where,

(Zpu)new = New per unit value of impedance

(Zpu)old = Old per unit value of impedance

kVbase = Old base value of voltage

kVnew = New base value of voltage

MVAnew = New base value of power

MVAold = Old base value of power

Base impedance \({Z_B} = \frac{{K{V^2}}}{{MVA}} = \frac{{{{0.4}^2}}}{{0.004}} = 40\;{\rm{\Omega }}\)

Resistance in ohm to HV side = \({r_{pu}} \times {Z_B} = 0.02 \times 40 = 0.8\;{\rm{\Omega }}\)

The thing to be noted here is that while transforming resistance the copper loss should remain the same

\(\begin{array}{l} I_1^2{R_{{q_1}}} = I_2^2{R_q}_2\\ {\left( {\frac{{\frac{{110}}{{\sqrt 3 }}}}{{{R_1}}}} \right)^2}{R_1} = {\left( {\frac{{\frac{{220}}{{\sqrt 3 }}}}{{{R_2}}}} \right)^2}{R_2} \end{array}\)

R₂ = 4 Ω

\(\therefore R_1^1 = 1\Omega\)
 = 1 + j0 Ω

Concept:

Base impedance is given by

\({Z_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)

kV_base is base voltage rating

MVA_base is base MVA rating

Calculation:

Given that, base kV = 11 kV

Base MVA = 100 MVA

Base impedance, \({Z_{base}} = \frac{{{{11}^2}}}{{100}} = 1.21\;{\rm{\Omega }}\)

Important:

The relation between new per-unit value & old per unit value impedance

\({({Z_{pu}})_{new}}\; = {({Z_{pu}})_{old}} \times {\left( {\frac{{k{V_{base}}}}{{k{V_{new}}}}} \right)^2} \times \;\left( {\frac{{MV{A_{new}}}}{{MV{A_{old}}}}} \right)\)

Also \({Z_{pu}} = \frac{{{Z_{Actual}}}}{{{Z_{base}}}}\)

 \({Z_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)

Where,

(Zpu)new = New per unit value of impedance

(Zpu)old = Old per unit value of impedance

kVbase = Old base value of voltage

kVnew = New base value of voltage

MVAnew = New base value of power

MVAold = Old base value of power

Concept

The equivalent resistance referred to the LV side is given by:

\(R_{eq}={R_{LV}}\space +\space {R_{HV}\over n^2}\)

where, n = Turn ratio

The turn ratio of the transformer is given by:

\(n={N_H\over N_L}\)

Calculation

Given, n = 2

R_LV = 0.06 Ω

R_HV = 0.2 Ω

\(R_{eq}={0.06}\space +\space {0.2\over (2)^2}\)

R_eq = 0.11 Ω

The thing to be noted here is that while transforming resistance the copper loss should remain the same

\(\begin{array}{l} I_1^2{R_{{q_1}}} = I_2^2{R_q}_2\\ {\left( {\frac{{\frac{{110}}{{\sqrt 3 }}}}{{{R_1}}}} \right)^2}{R_1} = {\left( {\frac{{\frac{{220}}{{\sqrt 3 }}}}{{{R_2}}}} \right)^2}{R_2} \end{array}\)

R₂ = 4 Ω

\(\therefore R_1^1 = 1\Omega\)
 = 1 + j0 Ω

Rated current in hv side

\(= {{30KVA} \over {2500}} = 12\ A\)

Given,

\(\eqalign{ & p.u{Z_{eq}} = 8\% \cr & {Z_{act}} = 0.08 \times {Z_{base}} \cr & {Z_{base}} = {{{{\left( {2500} \right)}^2}} \over {30000}} = 208.33 \cr &\therefore {Z_{act}} = 16.667{\rm\ {\Omega }} \cr}\)

∴ voltage needed = Z_act × rated current

= 12 × 16.667

= 200 V

Concept:

Line voltage = √3 V_Ph

Where,

V_Ph → Voltage in each phase.

All line voltages are equal in magnitude and are displaced by 120°, and

All line voltages are 30° ahead of their respective phase voltages.

total power = 3 x power in each phase.

Power in each phase = VPh²/(2× Z)

Where;

Z → Impedance 

Calculation:

Given;

Total power = 1.8 kW

Power factor = 0.5

Line voltage = 230 V

Phase voltage (V_Ph) = 230/√3 

Power in each phase = 1.8/3 = 0.6 kW

Then;

   V²_Ph/(2× Z) = 0.6 kW

⇒(230/√3 )²/(2 × Z) = 0.6 kW

∴ Z = 14.7 Ω

Concept:

Base impedance is given by

\({Z_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)

kV_base is base voltage rating

MVA_base is base MVA rating

Calculation:

Given that, base kV = 11 kV

Base MVA = 100 MVA

Base impedance, \({Z_{base}} = \frac{{{{11}^2}}}{{100}} = 1.21\;{\rm{\Omega }}\)

Important:

The relation between new per-unit value & old per unit value impedance

\({({Z_{pu}})_{new}}\; = {({Z_{pu}})_{old}} \times {\left( {\frac{{k{V_{base}}}}{{k{V_{new}}}}} \right)^2} \times \;\left( {\frac{{MV{A_{new}}}}{{MV{A_{old}}}}} \right)\)

Also \({Z_{pu}} = \frac{{{Z_{Actual}}}}{{{Z_{base}}}}\)

 \({Z_{base}} = \frac{{kV_{base}^2}}{{MV{A_{base}}}}\)

Where,

(Zpu)new = New per unit value of impedance

(Zpu)old = Old per unit value of impedance

kVbase = Old base value of voltage

kVnew = New base value of voltage

MVAnew = New base value of power

MVAold = Old base value of power

Base impedance \({Z_B} = \frac{{K{V^2}}}{{MVA}} = \frac{{{{0.4}^2}}}{{0.004}} = 40\;{\rm{\Omega }}\)

Resistance in ohm to HV side = \({r_{pu}} \times {Z_B} = 0.02 \times 40 = 0.8\;{\rm{\Omega }}\)

Concept

The equivalent resistance referred to the LV side is given by:

\(R_{eq}={R_{LV}}\space +\space {R_{HV}\over n^2}\)

where, n = Turn ratio

The turn ratio of the transformer is given by:

\(n={N_H\over N_L}\)

Calculation

Given, n = 2

R_LV = 0.06 Ω

R_HV = 0.2 Ω

\(R_{eq}={0.06}\space +\space {0.2\over (2)^2}\)

R_eq = 0.11 Ω

The thing to be noted here is that while transforming resistance the copper loss should remain the same

\(\begin{array}{l} I_1^2{R_{{q_1}}} = I_2^2{R_q}_2\\ {\left( {\frac{{\frac{{110}}{{\sqrt 3 }}}}{{{R_1}}}} \right)^2}{R_1} = {\left( {\frac{{\frac{{220}}{{\sqrt 3 }}}}{{{R_2}}}} \right)^2}{R_2} \end{array}\)

R₂ = 4 Ω

\(\therefore R_1^1 = 1\Omega\)
 = 1 + j0 Ω

% X = % voltage needed to get rated current therefore to get 1/2 rated full load be need to apply \(\left( {\frac{{4.5}}{2}} \right)\%\) of rated voltage

\(= \frac{{4.5}}{2} \times \frac{1}{{100}} \times \frac{{11000}}{1} = 247.5\ V\)