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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;

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

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:

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

Numerically:

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

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

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.

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.

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

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}

c.)

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

d.)

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}

e.)

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

Numerically:

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

i_{R1}(t) = 0.242 cos (wt+155.5°) mA