112topic10

SAMPLE PROBLEMS: 112-Topic 10: ELECTRIC OSCILLATIONS & AC CIRCUIT PROBLEMS: 1) A doorbell transformer has an 800 turn primary and a 20 turn secondary. When the primary is connected to a 60 Hz, 120 volt line, what is the secondary voltage? If a current of 0.5 amp is present in the secondary when the doorbell is operated, what is the current drawn in the primary circuit?

Pin = IpVp = Pout = Is Vs

AC

Solution: We have a transformer problem. Since we assume it is an 'ideal' transformer (no power loss), then we have: R

The ratio of the voltages are equal to the ratio of turns. That is: Vs/Vp = IP/IS = NS/ Np Solving for VS = (20)/(800) = 1/40 ---> Vs = Vp/40 = 120/40 = 3 volts. From the power equation we then have: Ip = (Is Vs)/Vp = (0.5)(3)/(120) = (.5/40) = .0125 A. 2) A television station operates at a frequency of 200 MHz. What inductance is needed with a capacitance of 1.5 pF to form a circuit resonant to this frequency? Solution: The 'tank circuit', as its called, is shown in the figure. We note that this is an 'ideal' circuit since we are neglecting any resistance values, and treating the circuit as an 'LC' circuit. For an LC circuit the resonant frequency is determined from: ω = 2π f = Thus we have:

1 / LC

1/LC =

ω2 = (2 π f)2 = (2 π)2(200 x 106)2.

Hence: L = 1/(ω2 C) = {(2 π )2(2 x 108)(1.5 x 10-12)}-1 = .00422 x 10-4 = 422 nH . 3) The generating capacity of a nuclear power plant is 5 Mw (Mega-watts). For calculational purposes assume the electrical energy is produced at 100 V, 60 Hz. The power station is connected to a city by a transmission line 50 miles long of total resistance 2 ohms. Compare the power loss in the transmission line and the efficiency for the following two cases: a) Output voltage of the PS is boosted by a factor of 100. b) Output voltage of the PS is boosted by a factor of 1000. Solution: This is a transformer type problem since we are told that there is a 'boost' of the voltage. We wont' have any difficulties with powers provided that we label them carefully. A schematic diagram may help. We have the power station (PS), stepup transformer (TF), transmission line (TL), & ultimately the city.

Let V1 be the 'output voltage of the PS.' This is also the 'input voltage to the TF.' We assume we have an ideal transformer, hence there is no power loss in the transformer. Thus

Pout = P in TF

TL

o

PS

CITY

6

Pout PS = 5 x 10 = I1V1 = I2V2

o

Pout

For the first case: V2/V1 = 100 Hence V2 = 100 V1 = 104 volts. Then:

o

o

PS

= P

o

o

P out = P

in TF

TL

in city

I2 = (5 x 106)/(104) = 500 A.

The power loss in the TL = I2 R = (500)2 (2) = 50 x 104 = 0.5 Mw . The power output of the TL is then 5 Mw - 0.5 Mw = 4.5 Mw, and the efficiency (output/input x 100) is: Eff(case 1) = (4.5 M)/(5 M) x100 = 90% . For case two we have: V2 = (1000)(100) = 105; I2 = (5 x 106)/(105) = 50 A. Then P(lost) = 2(50)2 = 5000 w = .005 Mw, and the efficiency is: Eff(%) = (5M - .005M)/(5 M) x 100 = 499.5/5 =

99.9% .

4). Find the reactance of a 0.15 H inductor and of a 3 nf capacitor at 60 Hz. Solution: The reactance of an inductor and capacitor is frequency dependant. We first calculate ω . ω = 2 π f = 2 π (60) = 120 π (rad)/sec . For the inductor we have:

XL = ω L = (120) π (.15) = 56.6 ohms.

For the capacitor we have: XC = 1/ωC = 1/ π (120)(3 x 10-9) = 8.84 x 105 ohms. 5). An ac emf of frequency 50 Hz has a peak value of 320 V. Find the instantaneous value of the emf at the instants: 1,2,3,4,5,8, & 10 ms after it passes through its zero value. What is the rms value of this source? Solution: The form of an ac emf is as shown. We note that its behaves as a 'sinusoidal' function, that is, as either a '+' or '-' cosine, or as a '+' or '-' sine function depending on what initial configuration is selected. We have elected to choose the emf as 0 at time t=0 (which makes it a sine function), and we have chosen to have the emf go '+' in the 1st quadrant (first quarter of the cycle). The form of the emf is then: rad/sec. Thus:

V(t) =

Vo π /2 90

-V

π 180

V o sin(

ω t)

3 π /2

2 π

270

ωt

o

V(t) = V max sin (ωt) ; where Vmax = 320 V and

ω = 2 π f = 100 π

t(sec)

Degrees

1 x 10

ωt (rad) π x 10-1

2 x 10-3 3 x 10-3 4 x 10-3 5 x 10-3 8 x 10-3

-3

V(t)

18

sin (ωt) .309

98.9

2π x 10-1 3π x 10-1 4π x 10-1 5π x 10-1

36 54 72 90

.588 .809 .951 1.00

188.1 258.9 304.3 320.0

8π x 10-1

144

.588

188.1

10 x 10-3 180 0.00 10π x 10-1 The rms value is given by: Vrms = .707 Vmax = (.707)(320) = 226.24 V.

0.0

6). The current in a solenoid is 2 A when it is connected to a 24 V battery, and 1.2 A when it is connected to a 24 V, 60 Hz power source. Find the resistance and inductance of the coil. Solution: The solenoid is a physical coil and hence has both resistance and inductance properties. The situation in the first case is shown. Here we have a steady dc-current (time independent) and hence the inductive property of the coil disappears. We can calculate the resistance of the coil in the usual way:

R

Vo

I

Σ I R's = Σ E's  R = V0/I = 12 Ω .

R AC

The circuit in the 2nd case is now drawn. This is an ac-problem, and hence the inductance of the coil plays an important roll. For 60 Hz, we have: ω = 2 π f = (120) π rad/sec.

L i(t)

We calculate the impedance of the circuit by: Z = V0 /I = (24)/(1.2) = 20 ohms. Note that we may assume that the '24 V' and '1.2 A' are both rms values, or we may assume they are both peak values. The result is the same. Now for an RL circuit in series:

Z = R 2 + (X L − X C ) 2

Since there is no capacitor, we have no

capacitive reactance. Thus: XL2 = Z2 - R2 = 202 - 122 = 256  XL = 16 ohms = ω L . Hence L = 16/(120π) = 0.042 H.

7) An ac voltage = 100V cos(2500 t) is applied to a series combination of a resistor (300 ohms), capacitor (8 µf), and an inductor (0.02H). a) What is the resonant frequency of this circuit?

b) What is the rms current? The 8µf capacitor is now replaced by a 4µf and the 0.02H inductor is replaced by a 0.2 H inductor. c) What is the rms current in this case? d) Draw a phasor diagram of the voltages across each element. e) Is this circuit mainly capacitive, or inductive? Solution: We first calculate the resonant frequency of the circuit. This depends on the values of L & C only. We have: ω02 = 1/LC = 1/(.02)(8 x 10-6) = 1/(16 x 10-8) ω0 = (1/4) x 104 = .25 x 104 = 2500 rad/sec. The (ordinary) frequency can then be determined: f = ω0/2 π ≈ 400 Hz. From the statement of the problem we see that this circuit is being driven by an ac source of angular frequency = 2500 rad/sec. Hence, resonance takes place, and there is no net reactance in the circuit. Thus the total impedance of the circuit equals the resistance (300 ohms), and we have: Irms = (.707) I0 = (.707) V0/R = (.7)(100)/(300) = (.7)(.33) = .231 A. In the 2nd half of the problem we change the values of the capacitance and inductance. This leads us to believe that the resonance phenomena will disappear, and we will have reactance as well as resistance in the circuit.

R AC

L

XC = 1/ωC = 1/(2.5 x 103)(4 x 10-6) = 1/(10 x10-3) = 100 Ω XL =

C

ω L = (2.5 x 103)(0.2) = 5 x 102 = 500 Ω

Hence the net reactance in the circuit is: (XL - Xc) = (500 - 100) = 400 ohms. (This provides the answer to part (e). The circuit is mainly inductive since the inductive reactance is greater than the capacitive reactance.) The total impedance is then calculated from: Z = R 2 + ( X L − X C ) 2 = 300 2 + 400 2 = 500 Ω The rms current is then calculated as before. Irms = (.7)V0/Z = (.7)(100)/(500) = .14 A. The phasor diagram is illustrated. We draw VR first (anywhere) with the current in phase with VR. Then we draw VC 900 behind I (for a capacitor the current ”leads" the voltage, hence, the voltage ”lags" the current). Similarly we draw V L 900 ahead of I. Since the inductive reactance is greater than the capacitive reactance, the net I(XL-XC) is along VL. We then complete the rectangle, draw in V (which leads the current indicating a mainly inductive circuit). The phase angle (angle between V & I) is given by:

VL - VC

Vo

ϕ

VL

I

{

ω t

VC

VR o

tan ϕ = (XL - XC)/R = 4/3. Hence,

ϕ is - 530.

Since XL > XC the circuit is mainly inductive.