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As we have already seen, circuits with sinusoidal excitation can be solved using *complex impedances* for the elements and *complex peak* or *complex* *rms values* for the currents and voltages. Using the complex values version of Kirchhoff’s laws, nodal and mesh analysis techniques can be employed to solve AC circuits in a manner similar to DC circuits. In this chapter we will show this through examples of Kirchhoff’s laws.

**Example 1**

Find the amplitude and phase angle of the current i_{vs}(t) if

v_{S}(t) = V_{SM} cos 2pft; i(t) =I_{SM} cos 2pft; V_{SM} = 10 V; I_{SM} = 1 A; f = 10 kHz;

_{1}= 10 mF; C

_{2}= 5 mF

**Click/tap the circuit above to analyze on-line or click this link to Save under Windows**

Altogether we have 10 unknown voltages and currents, namely: i, i_{C1}, i_{R}, i_{L}, i_{C2}, v_{C1}, v_{R}, v_{L}, v_{C2} and v_{IS}. (If we use complex peak or rms values for the voltages and currents, we have altogether 20 real equations!)

The equations:

Loop or mesh equations: for M_{1 }– **V**_{SM} +**V**_{C1M}+**V**_{RM} = 0

M_{2} – **V**_{RM} + **V**_{LM} = 0

M_{3} – **V**_{LM} + **V**_{C2M} = 0

M_{4} – **V**_{C2M} + **V**_{IsM} = 0

Ohm’s laws** V**_{RM} = R***I**_{RM}

**V**_{LM} = **j***w*L***I**_{LM}

**I**_{C1M} = **j***w*C_{1}***V**_{C1M}

**I**_{C2M} = **j***w*C_{2}***V**_{C2M}

Nodal equation for N_{1} – **I**_{C1M} – **I**_{SM }+ **I**_{RM }+ **I**_{LM} +**I**_{C2M} = 0

**I**=

**I**

_{C1M}

Solving the system of equations you can find the unknown current:

i_{vs} (t) = 1.81 cos (wt + 79.96°) A

Solving such a large system of complex equations is very complicated, so we haven’t shown it in detail. Each complex equation leads to two real equations, so we show the solution only by the values calculated with TINA’s Interpreter.

The solution using TINA’s Interpreter:

{Solution by TINA’s Interpreter} om:=20000*pi; Vs:=10; Is:=1; Sys Ic1,Ir,IL,Ic2,Vc1,Vr,VL,Vc2,Vis,Ivs Vs=Vc1+Vr {M1} Vr=VL {M2} Vr=Vc2 {M3} Vc2=Vis {M4} Ivs=Ir+IL+Ic2-Is {N1} {Ohm’s rules} Ic1=j*om*C1*Vc1 Vr=R*Ir VL=j*om*L*IL Ic2=j*om*C2*Vc2 Ivs=Ic1 end; Ivs=[3.1531E-1+1.7812E0*j] abs(Ivs)=[1.8089] fiIvs:=180*arc(Ivs)/pi fiIvs=[79.9613] |

The solution using TINA:

**Click/tap the circuit above to analyze on-line or click this link to Save under Windows**

**Click/tap the circuit above to analyze on-line or click this link to Save under Windows**

To solve this problem by hand, work with the complex impedances. For example, R, L and C

_{2}are connected in parallel, so you can simplify the circuit by computing their parallel equivalent. || means the parallel equivalent of the impedances:

Numerically:

**Click/tap the circuit above to analyze on-line or click this link to Save under Windows**

The simplified circuit using the impedance:

The equations in ordered form : **I** + **I**_{G1} = **I**_{Z }

_{ }**V**_{S} = **V**_{C1} +**V**_{Z}

**V**_{Z} = **Z** · **I**_{Z}

**I** = **j **w C_{1}· **V**_{C1}

There are four unknowns- **I**; **I**_{Z}; **V**_{C1}; **V**_{Z} – and we have four equations, so a solution is possible.

Express **I** after substituting the other unknowns from the equations:

Numerically

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According to TINA’s Interpreter’s result.

{Solution using the impedance Z} om:=20000*pi; Vs:=10; Is:=1; Z:=replus(R,replus(j*om*L,1/j/om/C2)); Z=[2.1046E0-2.4685E0*j] sys I I=j*om*C1*(Vs-Z*(I+Is)) end; I=[3.1531E-1+1.7812E0*j] abs(I)=[1.8089] 180*arc(I)/pi=[79.9613] |

The time function of the current, then, is:

i(t) = 1.81 cos (wt + 80°) A

You can check Kirchhoff’s current rule using phasor diagrams. The picture below was developed by checking the node equation in i

_{Z}= i + i

_{G1 }form. The first diagram shows the phasors added by parallelogram rule, the second one illustrates the triangular rule of the phasor addition.

Now let’s demonstrate KVR using TINA’s phasor diagram feature. Since the source voltage is negative in the equation, we connected the voltmeter “backwards.” The phasor diagram illustrates the original form of the Kirchhoff’s voltage rule.

**Click/tap the circuit above to analyze on-line or click this link to Save under Windows**

The first phasor diagram uses the parallelogram rule, while the second uses the triangular rule.

To illustrate KVR in the form V_{C1} + V_{Z} – V_{S} = 0, we again connected the voltmeter to the voltage source backwards. You can see that the phasor triangle is closed.

**Example 2**

Find the voltages and currents of all the components if:

v_{S}(t) = 10 cos wt V, i_{S}(t) = 5 cos (w t + 30°) mA;

C_{1} = 100 nF, C_{2} = 50 nF, R_{1} = R_{2} = 4 k; L = 0.2 H, f = 10 kHz.

**Click/tap the circuit above to analyze on-line or click this link to Save under Windows**

Let the unknowns be the complex peak values of the voltages and currents of ‘passive’ elements, as well as the current of the voltage source ( i

_{VS }) and the voltage of the current source ( v

_{IS }). Altogether, there are twelve complex unknowns. We have three independent nodes, four independent loops ( marked as M

_{I}), and five passive elements which can be characterized by five “Ohm’s laws” – altogether there are 3+4+5 = 12 equations:

Nodal equations for N_{1} I_{VsM} = I_{R1M} + I_{C2M}

for N_{2 } I_{R1M} = I_{LM} + I_{C1M}

for N_{3} I_{C2M} + I_{LM} + I_{C1M} +I_{sM} = I_{R2M}

Loop equations for M_{1 } V_{SM} = V_{C2M} + V_{R2M}

for M_{2} V_{SM} = V_{C1M} + V_{R1M}+ V_{R2M}

for M_{3} V_{LM} = V_{C1M}

for M_{4} V_{R2M} = V_{IsM}

Ohm’s laws V_{R1M} = R_{1}*I_{R1M}

V_{R2M} = R_{2}*I_{R2M}

I_{C1m} = j*w*C_{1}*V_{C1M}

I_{C2m} = j*w*C_{2}*V_{C2M}

V_{LM} = j*w*L*I_{LM}

Don’t forget that any complex equation might lead to two real equations, so Kirchhoff’s method requires many calculations. It’s much simpler to solve for the time functions of the voltages and currents using a system of differential equations (not discussed here). First we show the results calculated by TINA’s Interpreter:

{Solution by TINA’s Interpreter} f:=10000; Vs:=10; s:=0.005*exp(j*pi/6); om:=2*pi*f; sys ir1, ir2, ic1, ic2, iL, vr1, vr2, vc1, vc2, vL, vis, ivs ivs=ir1+ic2 {1} ir1=iL+ic1 {2} ic2+iL+ic1+Is=ir2 {3} Vs=vc2+vr2 {4} Vs=vr1+vr2+vc1 {5} vc1=vL {6} vr2=vis {7} vr1=ir1*R1 {8} vr2=ir2*R2 {9} ic1=j*om*C1*vc1 {10} ic2=j*om*C2*vc2 {11} vL=j*om*L*iL {12} end; abs(vr1)=[970.1563m] abs(vr2)=[10.8726] abs(ic1)=[245.6503u] abs(ic2)=[3.0503m] abs(vc1)=[39.0965m] abs(vc2)=[970.9437m] abs(iL)=[3.1112u] abs(vL)=[39.0965m] abs(ivs)=[3.0697m] 180+radtodeg(arc(ivs))=[58.2734] abs(vis)=[10.8726] radtodeg(arc(vis))=[-2.3393] radtodeg(arc(vr1))=[155.1092] radtodeg(arc(vr2))=[-2.3393] radtodeg(arc(ic1))=[155.1092] radtodeg(arc(ic2))=[-117.1985] radtodeg(arc(vc2))=[152.8015] radtodeg(arc(vc1))=[65.1092] radtodeg(arc(iL))=[-24.8908] radtodeg(arc(vL))=[65.1092] |

Now try to simplify the equations by hand using substitution. First substitute eq.9. into eq 5.

V_{S} = V_{C2} + R_{2} I_{R2 }a.)

then eq.8 and eq.9. into eq 5.

V_{S} = V_{C1} + R_{2} I_{R2} + R_{1} I_{R1} b.)

then eq 12., eq. 10. and I_{L }from eq. 2 into eq.6.

V_{C1} = V_{L} = jwL I_{L} = jwL(I_{R1} – I_{C1}) = jwL I_{R1} – jwL jwC_{1} V_{C1}

Express V_{C1}

Express V_{C2} from eq.4. and eq.5. and substitute eq.8., eq.11. and V_{C1}:

Substitute eq.2., 10., 11. and d.) into eq.3. and express I_{R2}

I_{R2} = I_{C2} + I_{R1} + I_{S} = jwC_{2} V_{C2} + I_{R1} + I_{S}

Now substitute d.) and e.) into eq.4 and express I_{R1}

Numerically:

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The time function of i_{R1} is the following:

i_{R1}(t) = 0.242 cos (wt+155.5°) mA

**Click/tap the circuit above to analyze on-line or click this link to Save under Windows**

**Click/tap the circuit above to analyze on-line or click this link to Save under Windows**