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FIGURE 4(a) shows two inductances connected in parallel across an a.c. supply.

(a) Apply Kirchhoff`s voltage law to loop abef and to loop abcdef of the circuit.

See attached

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#### Solution Preview

Hi please see solutions for Q4 & 5 in the attached so just ignore those or use them to help your understanding of other techniques

Solutions:
Thevenin method
Find Thevenin equivalent voltage, E by removing the element whose current flow it is asked to find (see next diagram)

We now find the current from Ohm's Law as the sum of voltage sources
divided by the remaining sum of circuit impedances, thus

i=(V_1-V_2)/(j4+j6)=(415√2 {cos(100πt)-sin(100πt)})/j10

Thus the Thevenin equivalent voltage, E , is determined from

E=V_1-i(j4)=415√2 cos(100πt)-j4/J10 {415√2[cos(100πt)-sin(100πt)]}
E=415√2 cos(100πt)-0.4{415√2[cos(100πt)-sin(100πt)]}
E=√2 cos(100πt)+166√2 sin(100πt)

Now we calculate the Thevenin equivalent impedance from
Z_th=j4||j6 where
Z_th=j4||j6=(j4.j6)/(j4+j6)=(-24)/j10=j2.4 (using the fact that 1/j=-j)

Next we find the complex form of the impedance z = 50Ω pf 0.7 lagging. Since it is lagging it must be an inductive impedance of impedance z=50∠-〖cos〗^(-1) (0.7)=50∠〖45〗^0

So the impedance 50Ω pf 0.7 lagging may be written as

z=50.cos(45^0 )+j50.sin(〖45〗^0 )=50/√2+j 50/√2

Then the current through the element z=50Ω pf 0.7 is derived by placing z in series with the Thevenin equivalent circuit and using Ohm's Law

i=E/(Z+Z_th )=(249√2 cos(100πt)+166√2 sin(100πt))/(50/√2+j 50/√2+j2.4)

i=(249√2 √2 cos(100πt)+166√2 √2 sin(100πt))/(50+j50+j2.4√2)

i=(498cos(100πt)+332sin(100πt))/(50+j53.4)

i=({498cos(100πt)+332sin(100πt)}{50-j53.4})/((50-j53.4)(50+j53.4))

i={24900.cos(100πt)+16600.sin(100πt)-j{26593.2cos(100πt)+17728.8 sin⁡(100πt) }/(〖50〗^2+〖53.4〗^2 )

i=4.65.cos(100πt)+3.10.sin(100πt)-j{4.97.cos(100πt)+3.31.sin(100πt)}

Using the Super-position theorem

Consider only V_1 supply so we short V_2 as below

We need to then find Z_1=Z||j6 and since

Z=50 pf 0.7 lag=50/√2+j 50/√2

Z_1=(Z.j6)/(Z+j6)=j6{50/√2+j 50/√2}/(50/√2+j 50/√2+j6)

Z_1=j6{50+j50}/(50+j50+j6√2)=(-300+j300)/(50+j58.5)

Z_1=(300(-1+j)(50-j58.5))/(〖50〗^2+〖58.5〗^2 )

Z_1=0.051(-50+58.5+j108.5)

Z_1=0.434+j5.534

The Voltage across the parallel combination denoted as Z_1 is given by the voltage divider rule thus

V_Z1=Z_1/(Z_1+j4).V_1

And the current through the impedance of interest, Z, influenced
only by the V_1 supply is given by Ohm's Law as

i_1=V_Z1/Z=Z_1/(Z.(Z_1+j4)).V_1

Now consider only the V_2 supply so we short V_1 as below

We need to then find Z_2=Z||j4 and since

Z=50 pf 0.7 lag=50/√2+j 50/√2

Z_2=(Z.j4)/(Z+j4)=j4{50/√2+j 50/√2}/(50/√2+j 50/√2+j4)

Z_2=j4{50+j50}/(50+j50+j4√2)=(-200+j200)/(50+j55.66)

Z_2=(200(-1+j)(50-j55.66))/(〖50〗^2+〖55.66〗^2 )

Z_2=0.036(5.66+j105.66)

Z_2=0.204+j3.804

Voltage across the parallel combination denoted as Z_2 is given by the voltage divider rule thus

V_Z2=Z_2/(Z_2+j6).V_2
And the current through the impedance of interest, Z, only due to
the V_2 supply is given by Ohm's Law as
i_2=V_Z2/Z=Z_2/(Z.(Z_2+j6)).V_2

Total current through Z is the vector sum of currents i_1 and i_2 or

i=i_1+i_2=Z_1/(Z.(Z_1+j4)).V_1+Z_2/(Z.(Z_2+j6)).V_2

Putting in derived or the given values

i=((0.434+j5.534)(415√2.cos(100πt)))/((50/√2+j 50/√2).(0.434+j5.534+j4))+((0.204+j3.804)(415√2.sin(100πt)))/((50/√2+j 50/√2).(0.204+j3.804+j6))

i=((0.434+j5.534)(830.cos(100πt)))/((50+j50).(0.434+j9.534))+((0.204+j3.804)(830.sin(100πt)))/((50+j50).(0.204+j9.804))

i=((0.434+j5.534)(830.cos(100πt)))/((-455+j(498.4)) )+((0.204+j3.804)(830.sin(100πt)))/((10.2-490.2.25+j(10.2+490.2)) )

i=((0.434+j5.534)(830.cos(100πt)))/((-455+j498.4) )+((0.204+j3.804)(830.sin(100πt)))/((-480+j500.4) )

i=(-(0.434+j5.534)(830.cos(100πt)))/((455-j498.4) )-((0.204+j3.804)(830.sin(100πt)))/((480-j500.4) )

i=-((0.434+j5.534)(455+j498.4)(830.cos(100πt)))/(〖455〗^2+〖498.4〗^2 )-((0.204+j3.804)(480+j500.4)(830.sin(100πt)))/(〖480〗^2+〖500.4〗^2 )

i=((2560.68-j2734.28)(830.cos(100πt)))/455427.56+((1805.61-j1928)(830.sin(100πt)))/480800.16

i=4.67.cos(100πt)-j4.98.cos(100πt)+3.12.sin(100πt)-j3.33.sin(100πt)

i=4.67.cos(100πt)+3.12.sin(100πt)-j{4.98.cos(100πt)+3.33.sin(100πt)}

Norton method

To find the pair of Norton generators remove the 50 pf 0.7 lagging impedance. First find the Norton current generator equivalent due to supply v_1 removing everything to the right of AB. Short across AB as shown below

Find the short circuit current I_1SC simply from Ohm's Law
I_1SC=v_1/j4=(√2.415.cos(100πt))/j4
I_1SC=-j146.725.cos(100πt)

Next find the total impedance looking in from AB which is just Z_AB=j4
The Norton current generator equivalent, due to supply v_1 and the j4Ω impedance is thus shown below

Next find the Norton current generator equivalent, due to supply v_2 thus remove the 50 pf 0.7 lagging impedance and everything to the left of CD. Short across CD as shown below

Find the short circuit current I_1SC simply from Ohm's Law
I_2SC=v_2/j6=(√2.415sin(100πt))/j6

I_2SC=-j97.816.sin(100πt)

Next find the total impedance looking in from CD which is just Z_CD=j6
The Norton current generator equivalent to supply v_2 and the j6Ω impedance is thus shown below

The total current is thus the sum of the Norton current generators
I_SC=I_1SC+I_2SC
I_SC=-j{146.725.cos(100πt)+97.816.sin(100πt)}

Total impedance will be
〖z=Z〗_AB ||Z_CD=(Z_AB Z_CD)/(Z_AB+Z_CD )=(j4.j6)/(j4+j6)
〖z=Z〗_AB ||Z_CD=(-24)/j10=j2.4

Now add back in the Z=50Ω pf lagging=50/√2+j 50/√2 as shown

Then the current required to find, i , is determined from

i=j2.4/(Z+j2.4) I_SC
i=j2.4/(50/√2+j 50/√2+j2.4) I_SC
i=(j2.4√2)/(50+50j+j2.4√2) I_SC

i=j3.394/(50+j53.39) I_SC

i=j3.394(50-j53.39)/(〖50〗^2+〖53.39〗^2 ) I_SC

i=((181.206+j169.7))/5350.492 I_SC

i=(0.0339-j0.0317) I_SC

Substituting for I_SC

I_SC==-j{146.725.cos(100πt)+97.816.sin(100πt)}

Therefore on substitution for I_SC
i=(0.0339-j0.0317){-j{146.725.cos(100πt)+97.816.sin(100πt)}}

i=4.65.cos(100πt)+3.32.sin(100πt)-j{4.97.cos(100πt)+3.10.sin(100πt)}

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#### Solution Summary

Mutual inductance, transformers and transformer regulation two questions are analyzed.

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