# Mesh analysis techniques to determine currents

A. Using Maxwell's circulating currents, calculate the three complex currents. Assume an angular frequency of 2.2 Mhz.

b. Sketch a phasor diagram showing all the currents and the two voltages V1 and V2.

Please refer to the attachment for circuit diagram.

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

a. First of all let us determine the impedances in complex form:

For the 9nF capacitor Z(cap)

Z(cap) = 1/{j x 2.2*w; x 10^6 x 9 x 10^-9} = -j16.1 = 16.1&Angle;arctan(-16.1/0) = 16.1Angle(-90)

For the resistor 12 Ohm; resistor Z(R)

Z(R) = 12 Ohm

For the 1.1 uH inductor Z(In)

Z(In) = j x 2.2pi x 10^6 x 1.1 x 10^-6 = j8.0 = 8Angle arctan(8/0) = 8Angle(90)

Note we have two loops; call them Loop one to the LHS of circuit diagram and Loop 2 to the RHS of the circuit diagram. We ignore current i2 for the moment and just consider the Maxwell circulating currents or Mesh currents.

For the 1st Maxwell current loop: (i.e. the current that flows in loop 1, i1 in circuit diagram)

i1*{Z(cap) + Z(R)} - i3*Z(R) = V1 -V2 (1)

For the 2nd Maxwell current loop: (i.e. the current that flows in loop 2, i3 in circuit diagram)

i3*{Z(R) + Z(In)} - i1*Z(R) = V2 (2)

Thus we have simultaneous equations (1) & (2) to solve.

Multiplying (1) by {Z(R) + Z(In)}we get

i1*{Z(cap) + Z(R)}*{Z(R) + Z(In)} - i3*Z(R)*{Z(R) + Z(In)} = {Z(R) + Z(In)}*{V1 -V2} (3)

Multiplying (2) by Z(R) we get

i3*{Z(R)}*{Z(R) + Z(In)} - i1Z(R)*Z(R) = V2*Z(R) (4)

Adding (3) and (4) we get rid of the terms in i3 thus

i1*{Z(cap) + Z(R)}*{Z(R) + Z(In)} - i1Z(R)*Z(R) = V2*Z(R) + {Z(R) + Z(In)}*{V1 -V2} (5)

Expanding brackets for (5)

i1*{Z(cap)*Z(R) + Z(cap)*Z(In) + Z(R)^2 + Z(R)*Z(In) - Z(R)2 } = V2*Z(R) + V1*Z(R) + V1*Z(In) - V2*Z(R) - V2*Z(In)

i1*{Z(cap)*Z(R) + Z(cap)*Z(In) + Z(R)*Z(In) } = V1*Z(R) + V1*Z(In) - V2*Z(In)

Thus loop (mesh) current i1 is given by

i1= {V1*Z(R) + V1*Z(In) - V2*Z(In)}/{Z(cap)*Z(R) + Z(cap)*Z(In) + Z(R)*Z(In) } (6)

Now let us solve for i1 as given by (7) using Z(R) = 12, Z(In) = j8, Z(cap) = -j16.1

i1= {V1*Z(R) + V1*Z(In) - V2*Z(In)}/{Z(cap)*Z(R) + Z(cap)*Z(In) + Z(R)*Z(In) }

i1= {12*(50Angle(-6)) + (50Angle(-6)*(8Angle(90) - (30Angle(30)*(8Angle(90)}/{12*(16.1Angle(-90) + (16.1Angle(-90)*(8Angle(90) + 12*(8Angle(90)}

i1= {(600Angle(-6) + (400Angle(84) - (240Angle(120)}/{(193.2Angle(-90) + 128.8 + (96Angle(90)}

Changing from polar to Cartesian complex form

i1= {597 -j63 + 42 +j398 + 120 -j208}/{128.8 -j193 +j96}

i1= {759 +j127}/{128.8 -j97}

Converting back to polar complex form so that we can easily divide, we get

i1= Sqrt{(759)^2 + (127)^2)Angle arctan(127/759)/ Sqrt{(128.8)^2 + (97)^2} Angle arctan(-97/128.8)

i1= 769.6Angle(9.5)/206.4Angle(-37)

i1= 3.7&Angle(46.5)

Now substituting this value into (2) to solve for i3, we get

i3*{Z(R) + Z(In)} - i1Z(R) = V2

i3*{12 + j8} - i1*12 = V2

i3*{12 + j8} -12*3.7Angle(46.)5 = 30Angle(30)

i3*{12 + j8} - 44.4&Angle(46.5) = 30Angle30

Converting above relationship to Cartesian complex form i.e. converting 44.4∟46.5 & 30∟30 we get

i3*{12 + j8} - 30.6 - j32.2 = 26 + j15

i3*{12 + j8} = 26 + 30.6 + j15 + j32.2

i3*{12 + j8} = 56.6 + j47.2

i3 = {56.6 + j47.2}/{12 + j8}

And finally converting back to polar complex form so that we can easily do a division, we get

i3 = Sqrt{(56.6)^2 + (47.2)^2}Angle arctan(47.2/56.6)/Sqrt{(12)^2 + (8)^2}}Angle arctan(8/12)

i3 = 73.7Angle(39.8)/14.4Angle(36.7)

i3 = 5.1&Angle(3.1)

Finally we need to solve for the branch current i2 which flows via the 12Ω resistor.

In Loop/Mesh/Maxwellian Circulating Current Analysis, which ever you want to call it, the branch currents can be determined as the phasor sum of any loop currents common to the branch in question.

As the resistor branch has both i1 and i3 loop currents, we can say that

i2 = i1 + i3 = 3.7&Angle(46.5) + 5.1&Angle(3.1)

Converting to complex cartesian form we get

i2 = 3.7cos(46.5) + j3.7sin(46.5) + 5.1cos(3.1) + j5.1sin(3.1)

i2 = 2.55 + j2.68 + 5.09 + j0.28

i2 = 7.64 +j2.96

Or in complex polar form

i2 = Sqrt{(7.64)^2 + (2.96)^2} Angle arctan(2.96/7.64)

i2 = 8.2 Angle(21.2)

So we have three currents

i1= 3.7&Angle(46.5), i2 = 8.2 Angle(21.2) and i3 = 5.1Angle(3.1)

b. Please see the attachment for phasor diagrams for currents and voltages.

Please note that there is a lot of mathematics involved and it is easy for anyone to make a small mistake which will propagate; the main thing is the method to solve.

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