Voltage on capacitor

uC (t)= = = UmC× cos(W t +j0 C),	(17)
UmC= ImXC – peak voltage,
j0 C = j0 I –p / 2= j0e+DF–p / 2 – initial phase.

The peak current as well as peak voltages on circuit devices are strongly depends on generator frequency. Magnitude of a current and distribution between the voltages in series RLC-circuit at three various regions of frequency spectrum are shown below.

Low frequency	Resonance Ω R	High frequency
Reactance:
XC > XL. Considerably predominate capacitive reactance.	XC = XL; Lowest resistance: Z=R	XC < XL. Considerably predominate inductive reactance.
Reactive voltage:
UmL< UmC =XC× e m / Z. Predominate capacitive voltage.	UmC = UmL =Q× e m;	UmC < UmL =XL× e m / Z. Predominate inductive voltage.
Current:
. Low current.	Im = e m /R; Heavy current.	. Low current.
Phase difference (DF=j0 I – j0e) between the current and generator voltage:
DF > 0;	DF = 0;	DF < 0.
Current and voltage phasors at t= 0 are represented on vector voltage diagram:

EXAMPLE OF PROBLEM SOLUTION

Example 3. Simple harmonic external EMF with the frequency 10³ Hz applied to the series oscillatory circuit of RLC -filter. The elements of filter have a nominal value: resistance 100 W, inductance 40 mH and capacity 1 m F. The voltage on capacitor varies with time by the law: u_C (t) = U_mC× cos(Ω t), with the reading of an voltmeter U _{RMS C} = 20 V.

1) To rebuild the equation of changing of current in circuit, of voltage on resistor, voltage on capacitor, of voltage on inductor and EMF, applied to circuit with numerical coefficients.

2) To build the vector voltage diagram at t= 0.

3) To find the values of external EMF – ε, voltages – u_R, u_C, u_L at the moment of time of t ₁ = Т/ 8 (Т – period of oscillations). To build the voltage diagram at t ₁ = Т/ 8.

Input data: R = 100 Ω; L = 40 mH =0, 04 H; С = 1 m F = 10 – 6 F; uC (t) = Umc cos(Ω t); U RMS C = 20 V; f = 103 Hz; t 1 = T/ 8.
Find: 1) i (t), uR (t), uL (t), uC (t), ε (t) –? 2) Phasors at t= 0 –? 3) phasors at t 1, uR (t 1), uL (t 1), uC (t 1), ε (t 1) –?

Solution:

1) Let’s evaluate the peak voltage on capacitor U_mC and cyclic frequency W of oscillations in circuit:

U_mC = U _{rms C} × Ö 2; [ U_mC ]= V; U_mC = 20 × Ö 2=28, 3 V;

W=2p f; [W] =rad× Hz= rad /s; W= 6, 28 × 10³ rad /s.

Here U _{rms C} – root mean square value of voltage on capacitor voltmeter of which is indicated; f –frequency of external EMF of generator.

From the view of the given equation u_C (t) = U_mc cos(Ω t) we can conclude then j_{0 C} =0.

Finally we obtain the equation of oscillations of voltage on capacitor with numerical coefficients:

u_С (t) = 28, 3cos(6, 28 × 10³ t) V.

From Ohm’s law for the AC voltage on capacitor with the peak voltage and initial phase there is a view:

where capacitive reactance is defined by the equation:

X_C = 1 / W C; (3.2)

From this we obtain: I_m= U_mC / X_C ; j_{0 I} = j_{0 C}+ p / 2.

Let’s check dimensionality and make the calculations:

[ X_C ]=1 / (F× rad /s)=W; X_C = 1 / (6, 28 × 10³ × 10 ^{– 6})= 159W;

[ I_m ]= V / W= A; I_m = 28, 3 / 159=0, 178 A;

[j_{0 I} ] = rad; j_{0 I} = 0+p / 2=p / 2.

From Ohm’s law for the AC public current in circuit with the peak value and initial phase there is a view:

where impedance of series oscillatory circuit and phase difference between the current and generator voltage are defined by the equations:

; (3.4)

DF= arctg[(X_C – X_L) / R ]. (3.5)

Let’s substitute numerical values in (3.3) and obtain the equation of current oscillations with numerical coefficients:

i (t) = 0, 178 × cos(6, 28 × 10³ t+ p / 2) А.

Thus current in circuit has phase lead relative to the voltage on capacitor on p / 2.

From Ohm’s law for the AC voltage on resistor with the peak voltage and initial phase there is a view:

Let’s check dimensionality and make the calculations:

[ U_mR ]= A× W= V; U_mR = 0, 178× 100=17, 8 V;

[j_{0 R} ] = rad; j_{0 R} = p / 2.

Let’s substitute numerical values in (3.6) and obtain the equation of oscillations of voltage on resistor with numerical coefficients:

u_R (t) = 17, 8 × cos(6, 28 × 10³ t+ p / 2) V.

Thus voltage on resistor has phase lead relative to the voltage on capacitor on p / 2.

From Ohm’s law for the AC voltage on inductor with the peak voltage and initial phase there is a view:

where inductive reactance is defined by the equation:

X_L = W L; (3.8)

Let’s check dimensionality and make the calculations:

[ X_L ]=rad /s× H =W; X_L = 6, 28 × 10³ × 0, 04= 251 W;

[ U_mL ]= A× W= V; U_mL = 0, 178× 251=44, 7 V;

[j_{0 I} ] = rad; j_{0 I} = p / 2+p / 2=p.

Let’s substitute numerical values in (3.7) and obtain the equation of oscillations of voltage on inductor with numerical coefficients:

u_L ( t ) = 44, 7cos(6, 28 × 10³ t+ p) V.

Thus voltage on resistor has phase lead relative to the voltage on capacitor on p.

From Ohm’s law for the AC external EMF of a generator there is a view:

e_ext(t) = e _m× cos(W × t +j_0e), (3.9)

where the peak external voltage and the initial phase we find from equations (3.3):

e_m = I_m× Z; (3.10)

j_0e =j_{0 I} – DF; (3.11)

Let’s check dimensionality and make the calculations:

; ;

[ e_m ]= A× W= V; e_m = 0, 178× 136=24, 1 V;

[DF] = rad; DF= arc tg[(159–251) / 100]= arc tg[–0, 93]= –0, 75rad= –0, 24p»–p / 4;

[j_0e] = rad; j_0e = p / 2+p / 4=3p / 4.

Let’s substitute numerical values in (9) and obtain the equation of oscillations of external voltage with numerical coefficients:

e_ext( t ) = 24, 1cos(6, 28 × 10³ t+ 3p / 4) V.

Thus external voltage has phase lead relative to the voltage on capacitor on 3p / 4.

2) For building the vector voltage diagram at t= 0 let’s write the equation of voltages in this moment of time without calculating the value of cosine:

uС (t= 0) = 28, 3cos(0) V; uR (t= 0) = 17, 8 × cos(p / 2) V; uL (t= 0) = 44, 7cos(p) V; eext(t= 0) = 24, 1cos(3p / 4) V; i (t= 0) = 0, 178 × cos(p / 2) А. DF = –p / 4; j0e= 3p / 4.

Figure 3.4, a – The vector voltage diagram at t= 0.

In this case the argument of cosine is a phase of corresponding voltage at t= 0.

For each phasor, the angle which is numerically equal to an initial phase, we lay off counterclockwise (for positive phase) relative to the axis X (dashed axis at the Fig. 3.4, a). Initial phase of external voltage j_0e=3p / 4 and DF»–p / 4 is shown at Fig. 3.4, a.

The magnitude of each phasor is equal to peak value of corresponding voltage. If one chooses the scale of voltage equal 10 V/сm, then the phasor length, representing the oscillation of voltage on capacitor, will equal 2, 8 сm (amplitude will be U_mc = 28, 3 V), this vector will be directed along the bearing axis, for the reason that the argument of cosine is equal to zero at t= 0.

In much the same way we build phasors, representing the oscillation of voltage on resistor, on inductor and external voltage of generator.

At the result we obtain, that according to the second Kirhchoff’s rule, the phasor external EMF must be equal to the vector sum of the all voltage phasors (see Fig. 3.4, a):

.

3) For finding values of external EMF and voltages at the moment of time of t ₁ = Т/8 we calculate the period of oscillations:

T= 2p / W; [ T ] =rad / rad /s = s; T= 2× 3, 14 / 6, 28 × 10³=10^–3 s.

Let’s substitute numerical value t ₁ = Т/8 =10^–3 / 8 s in the equation of oscillations. For building the vector voltage diagram at t ₁ = Т/ 8 we underline the equation of voltages in this moment of time without calculating the value of cosine and build phasors in the same way as Fig. 3.4, a:

uС (t 1) = 28, 3cos(6, 28 / 8) = 28, 3cos(p / 4); uR (t 1) = 17, 8 × cos(p / 4 + p / 2) = 17, 8 × cos(3p / 4); uL (t 1) = 44, 7cos(p / 4 + p) = 44, 7cos(5p / 4); eext(t 1) = 24, 1cos(p / 4 + 3p / 4) = 24, 1cos(p); i (t 1) = 0, 178 × cos(p / 4 + p / 2) = 0, 178 × cos(3p / 4); DF = –p / 4; j0e= 3p / 4.

Figure 3.4, b – The vector voltage diagram at t 1 = Т/ 8.

In this case the argument of cosine is a phase of corresponding voltage at t ₁ = Т/ 8. We note, that the phase of all voltages was incremented on p / 4 and phasors have turned counterclockwise through an angle of p / 4 (see Fig. 3.4, b).

Finally we calculate numerical values of the voltages at t ₁ = Т/8:

u_С (t ₁) = 28, 3× 0, 707=20 V;

i (t ₁) = 0, 178 × 0, 707=0, 126 А;

u_R (t ₁) = 17, 8 × 0, 707=12, 6 V;

u_L (t ₁) = 44, 7× (– 0, 707)= – 31, 6 V;

e_ext(t ₁) = 24, 1× (– 1)= – 24, 1 V.

Results: