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Select the exercise system, instructions for solving and solving Physics exercises (chapter "Alternating Current" grade 12 advanced program) - 19

C. SOME PRACTICE EXERCISES

1. TOPIC:

Lesson 1. Voltage is applied to the two ends of a non-branched R, L, C circuit. The voltage at the two ends of R is 80V, the two ends of L are 120V, and the two plates of capacitor C are 60V. The effective voltage at the two ends of the circuit is:

A. 260V B. 140V C. 100V D. 20V

Lesson 2. An unbranched AC circuit consists of R = 100 , coil

Maybe you are interested!

pure inductor

L1 H, capacitor has capacitance C = 15.9

Select the exercise system, instructions for solving and solving Physics exercises (chapter "Alternating Current" grade 12 advanced program) - 19

F. AC voltage

placed at the two ends of the circuit, the current in the circuit is:

u  200 2 cos100 t

(V). Intensity expression

A. i  2cos �100 t   � B. i  0.5 2 cos�100 t  

� 4 �(A) � 4 �(A)

� �

C. i  02cos�100 t  

� �

2

3

D. i1 cos�100 t   �(A)

� 4 �(A)

5 � 4 �

� �

Lesson 3. AC circuit with resistor R = 100

� �

, pure inductance coil has

reactance equal to 100

, capacitor has capacitance

104

C 

F in series. Put in

At both ends of the circuit, the alternating voltage consumed by the circuit is:

u  200sin100 t

(V). Capacity

A. P = 200W B. P = 400W C. P = 100W D. P = 50W

Lesson 4. A wire frame has N = 50 turns, the diameter of each turn is d = 20cm. Place the wire frame in a uniform magnetic field with magnetic induction B = 4.104 T. The normal line of the frame makes an angle B with the magnetic induction . The maximum value of the magnetic flux is:

A. o = 0.012 (Wb). B. o = 0.012 (Wb).

C. o = 6.28.104 (Wb). D. o = 0.05 (Wb).

Lesson 5. Connect a coil with inductance coefficient L and resistance r = 100 in series with a capacitor.

The circuit has a capacitance of 31.8 F. The voltage at the two ends of the circuit has the expression

u  200sin100 t (V). Adjust L so that the current intensity reaches its maximum.

The maximum effective current Imax is:

2

A. 2A B. 2 A C. 1A D. A

3

Lesson 6. Three-phase AC generator connected in star, phase voltage 127V, frequency 50Hz. People put current into a three-phase load connected in triangle, symmetrically. Each load

is a coil with pure resistance r = 12, the current passing through the loads will be:

, inductance L = 51mH. Current intensity

A. 6.35A B. 11A C. 12.63A D.4.54A

Lesson 7. Given a series circuit with R, L, C. The coil is purely inductive and the capacitor has variable C. Adjust C so that UC reaches its maximum value, then we will have:

A. uLC is perpendicular to u. B. uRL is perpendicular to u.

C. uLC is in phase with uRC. D. uRC is in phase with u.

3

Lesson 8. Given a circuit consisting of two closed boxes 1 and 2. u2 is in phase with i. Voltage u1

fast phase

 / 3

compared to u2. They have effective values

U1  U2  80

V. Angle

The phase shift between the voltage u of the entire circuit and i is:

A.  B.  C.  D. 

LC

3 4 2 6

Lesson 9. Series R, L, C circuit has the power of the circuit

2 f

 1. If R increases by 2 times, then the coefficient

A. increase 2 times B. decrease 2 times C. increase any D. unchanged

Lesson 10. Apply an alternating voltage to the two ends of a pure inductance coil.

L1 H then intensity 2

The current through the coil has the expression

i  3 2 sin �100 t  

(A). The voltage expression at the two ends of the circuit is:

6

�

�

A. u  150sin �100 t2  (V). B. u  150 2 sin �100 t2

� 3 �

3 �(V).

� � �

C. u  150 2 sin �100 t2  � D. u  100sin �100 t2

(V).

� 3 �(V). � 3

� � �

Lesson 11. Given an AC circuit with R, L, C connected in series, in which R changes

ok

L  1 H,

 2.104 F, the voltage at both ends of the circuit is kept

C

unchanged

formula u  100 2 sin100 t (V). The value of R and the maximum power of the circuit are

turn is:

A. R = 40 , P = 100W. B. R = 50 , P = 500W.

C. R = 50 , P = 200W. D. R = 50 , P = 100W.

Lesson 12. A single-phase transformer has 2000 turns in the primary and 100 turns in the secondary coils, respectively. The effective voltage and current in the primary circuit are 120V - 0.8A. Ignoring power loss, the effective voltage and power in the secondary circuit are:

A. 6V – 96W. B. 240V – 96W.

C. 6V – 4.8W. D. 120V – 4.8W.

Lesson 13. A wire frame has 200 turns, each turn has an area of ​​125cm2. Place the wire frame in a magnetic field with induction B = 0.4T. At t = 0, the normal vector of

frame fits

Bcorner

 . Let the wire frame rotate evenly around axis 6

B with luck

speed

  100

rad/s. Calculate the frequency and the effective emf in the frame at

t  1 s.

50

A. f = 100Hz, E = 444 2 (V). B. f = 50Hz, E = 222 (V). C. f = 50Hz, E = 444 2 (V). D. f = 100Hz, E = 444 (V).

Lesson 14. Given an AC circuit with R, L, C connected in series.

When L is short-circuited (and R is in series with C), i is seen to be in phase with u by an angle

 . 4

When R, L, C are in series, i lags behind u by an angle  .

4

The relationship between ZL and ZC is:

A. ZL = 2ZC. B. ZC = 2ZL.

C. ZL = ZC. D. cannot be determined.

Lesson 15. Given the circuit as shown. R1 = ZL1 =

100. X is a closed box containing only one of three pure electrical elements R, L, C. When an alternating voltage is applied to the two ends of the circuit, uAB is ahead of i by an angle

 / 3 . X is an electrical element with the value:

A. R = 73.2 B. ZL = 73.2 C. ZC = 73.2 D. R = 6.8

Lesson 16. The circuit consists of a pure inductor L = 0.318H connected in series with a resistor Rx and connected in series with a capacitor C = 0.159.104F. The current frequency f = 50Hz. For the voltage at both ends of RL to be uRL in phase with the voltage at both ends of RC to be uRC, then R has the value:

A. 100 B. 141 C. 200 D. 284

Lesson 17. Given an unbranched circuit. R = 40

, coil has r = 20

and L

= 0.0636H, the capacitor has a variable capacitance. Place a

AC voltage has f = 50Hz and U = 120V. Adjust C so that the effective voltage at both ends of the coil reaches its maximum value, that value is:

A. 40V B. 80V C. 46.57V D. 56.57V

Lesson 18. A three-phase asynchronous motor is connected in star to a three-phase AC power grid with a line voltage of 380V. The motor has a capacity of 10 kW. The power factor is 0.8. What is the value of the effective current passing through each coil?

A. 18.94A B. 56.72A C. 45.36A D. 26.35A

Lesson 19. The secondary coil of a transformer has 1500 turns and the current has f = 50Hz. The maximum value of the magnetic flux in the steel core is 0.6 Wb. The initial phase is chosen to be zero. The expression of the electromotive force in the secondary coil is:

A. e  200cos100 t (V). B.

e  200cos�100 t  

(V).

2

�

�

C. (V). D.

e  200 2 cos�100 t  

e  200 2 cos100 t

� 4 �(V).


Lesson 20. Given a series circuit of R, L, C consisting of R = 100

� �

, pure coil

The inductor has L = 0.318H. The current frequency f = 50Hz. The total impedance of the circuit is known to be 100 2 . The capacitance C of the capacitor has the value:

A. 200 FB 15.9 FC 2/ FD 1/ F.

Lesson 21. Given a series circuit of R, L, C. R = 100 2 and the applied voltage

The two ends of the circuit are

u  200 2 cos�100 t  

(V). When only R and C are connected, then i

2

�

�

fast phase

 compared to u. When only L and R are connected, i is 4 times slower than u.

 vs u. Table 4

The formula for current intensity when connecting R, L, C is:

A. i  2cos100 t (A). B. i 2 2 cos�100 t  

� 2 �(A).

C. i  2cos�100 t  

� �

(A). D. i  2 2 cos100 t (A).

�

�

Lesson 22. Voltage setting

2

uU o cos t (V) into the two ends of the circuit consisting of R, C and

Pure inductor L connected in series, L can be changed. Knowing the capacitive reactance of the capacitor

electricity by R3. Adjust L so that the effective voltage at both ends of the inductor reaches

maximum, then:

A. Voltage between the two ends of the circuit resistor.


30o phase shift from the voltage at both ends

B. Voltage between two ends of the capacitor circuit.

30o phase shift from the voltage at both ends

C. There is electrical resonance in the circuit.

D. The voltage between the two ends of the inductor is 30o out of phase with the voltage between the two ends of the circuit.

Lesson 23. There are three elements R, a pure inductor with ZL = R and a capacitor ZC = R. When

If we connect them in series to an AC source with constant effective voltage and current frequency, the power of the circuit is 200W. If we keep L and C the same and replace R with a resistor Ro = 2R, what is the power of the circuit?

A. P = 200W B. P = 400W C. P = 100W D. P = 50W

Lesson 24. An AC generator consists of 8 pairs of poles, the armature consists of 22 coils connected in series. The maximum magnetic flux generated by the armature passes through each coil.

wire of maximum value

0.1 Wb. Rotor rotates at 375 rpm. Power

The maximum power that the machine can generate is:

2

A. 110V B. 110 2 V C. 220V D. 220 V

Lesson 25. Given the circuit as shown. R1 = ZL1 = 20 .

X is a closed box containing only two of the three pure elements R, L,

C. Apply alternating voltage to both ends of the circuit then

uAM quadrature uMB. ​​X are electrical elements with values

A. Contains R and C, with R = 2ZC. B. Contains R and C, with R = ZC.

C. Contains L and C, has ZL = 2ZC. D. Contains L and C, has ZL = ZC.

104

Lesson 26. Given a series circuit of R, L, C with R = 30 , C   F, current

In a circuit with a frequency of 50 Hz and behind the phase, the voltage at both ends of the circuit is

 / 6 , then ZL has the value

A. 173 B. 117.3 C. 11.73 D. 17.3


Lesson 27. Given the circuit as shown. Knowing

C

 104 F ,

L  1 H, 2

uAB

 200cos100 t (V).

uAM voltage

slow phase

 compared with the current through the circuit and the current through the circuit 6

slow phase

 compared to u 3

MB. r and R have values

A. r = 25 and R = 100 . B.

r  5

0 3

3

 and

R  100 3 .

C. r  25 3 and

R  100 3 . D.

r  50 3 and

R  10  .

0 3

3

Lesson 28. Given the circuit as shown. The coil has

degree of freedom

have a cold

L3 H, resistance

pure r = 100 .

Apply a voltage to both ends of the circuit.

u AB  100 2 cos100 t (V). Calculate the value of C so that the voltmeter has the largest value and find that largest value of the voltmeter.

4 3

A. C  

.104 F and U C max  120 V.

B. C3 .104 F and U C max  180 V.

4

C. C3 .104 F and U C max  200 V.

4

D. C3 .104 F and U C max  220 V.

Lesson 29. A 200W50V motor is connected to the two ends of the secondary coil of a step-down transformer with a primary to secondary turns ratio of k = 4. The energy loss in the transformer is negligible. If the motor operates normally and the effective current in the primary coil is 1.25A, the power factor of the motor is

A. 0.75 B. 0.8 C. 0.85 D. 0.9

Lesson 30.

Given a circuit consisting of R, L, C connected in series. R changes,

L  1 H,

 103 F.

C

4

Apply an alternating voltage to both ends of the circuit.

u  75 2 cos100 t (V).

The power of the whole circuit is P = 45W. What is the value of resistor R?

A. R = 45 B. R = 60 C. R = 80 D. sentence A or C

Lesson 31. Given the circuit as shown. Knowing

R1 = 4 ,

C1 

10-2

8

F, R2 = 100 ,

L  1

H, f = 50Hz. Change the value of C2 so that the voltage uAE is in phase with uEB. The value of C2 is:

A. 101 F B. 102 F C. 103 F D. 104 F

3 3 3 3

Lesson 32. An AC circuit consists of a pure inductance coil with self-inductance

L1 H, connected in series with a resistor R = 20 5

. Electric current flows through the segment

The circuit has the circuit expression:

i  2 2 cos100 t (A) then the voltage expression at both ends of the segment is

A. u 40 2 cos100 t (V). B. u  40 2 cos�100 t  

(V).

4

�

�

C. u  80cos�100 t   � D. u  80cos�100 t  

� 4 �(V). � 4 �(V).

� � � �

Lesson 33. The transformer has N1 = 250 turns and N2 = 500 turns. The primary coil is a coil with r = 1 and ZL = 3. If a voltage of 110V is applied to the primary coil, what is the voltage value at the two ends of the open secondary coil?

A. 110V B. 208.8V C. 220V D. 104.4V

Lesson 34. Given the circuit as shown. Knowing UAM =

5V, UMB = 25V,

U AB  20 2 V. System

power number

of the circuit has the value:

B. 3

A. 2

2

3

2 2


CD

Lesson 35. Given an unbranched circuit. R = 100 , the coil has inductance L

= 0.318 H, f = 50Hz, capacitor has variable capacitance. Place at both ends of the segment

The circuit has an AC voltage with an effective value of U  100 2 V. Adjust C so that the circuit has electrical resonance. The value of C and the current intensity are then:

A. C = 31.8 F and I

2 AB C = 31.8 F and

I  2 2 A.

C. C = 3.18 F and I  3 2 AD C = 63.6 F and I = 2A.

Lesson 36. Two coils connected in series have resistance and inductance R1, L1 and R2, L2 respectively. Apply effective voltage U to the two ends of the circuit. Let U1 and U2 be the effective voltages of the coils. The condition for U = U1 + U2 is:

A. L1.R1 = L2.R2 B. R1.R2 = L1.L2

C. L1.R2 = L2.R1 D. no condition required.

Lesson 37. An AC circuit consists of a coil connected in series with a capacitor.

The coil has resistance r = 30

, inductance

L0.4 H, capacitor has capacitance C.

The expression for the instantaneous voltage between the two ends of the circuit is u  120cos100 t (V).

At what value of C does the power consumption of the circuit have a maximum value and what is the maximum power value?

A.C​


C. C

 104 F and 2

 103 F and 4

P max  120 WB


Pmax  240 WD

 104 F and

C

C

 103 F and

Pmax  120 W.


2

2

Pmax  240 W.

Lesson 38. Given the circuit as shown. Voltage at both ends

The circuit is u AB  100 2 coos 100 t (V). X is a closed box.

contains inductor or capacitor. RC is variable resistor.

Adjust RC = 40 to see the current intensity i is behind phase

 compared to electricity 4

voltage across the circuit. The electrical element in X and its value are:

A. coil, has L = 0.127HB capacitor, has C = 0.796.104F.

C. coil, has L = 40mH. D. capacitor, has C = 0.459.104.

104

Lesson 39. Given an AC circuit without branches. R = 100 , C   F,

Pure inductance coil with variable inductance L. Placed at both ends of the segment

circuit with voltage u AB  200cos100 t (V). What is the inductance L?

The power consumption in the circuit is 100W.

A. L1 H B.

L  1 H C.

2

L  2 H D.

L  4 H

Lesson 40. AC circuit consists of resistor R = 100 and pure inductance coil.

L1 H, variable capacitor C, current frequency f = 50Hz. Adjust C to UCmax.

Determine the value of C then.

C

A.  104 F B.

C.  104 F D.


 104 F 2

C

 2.104 F


C 4 C 

Lesson 41. Given an electric circuit with X and Y being two closed boxes. Box X consists of two electrical elements connected in series, box Y has one electrical element. The electrical elements are pure R,

Pure L, C. Knowing uX is fast in phase

 compared to i, current i is 2nd phase faster

 vs uY. 2

Identify the elements of the circuit.

A. X contains inductor L and resistor R, Y contains capacitor C.

B. Y contains capacitor C, X contains inductor L and capacitor C.

C. Y contains inductor L, X contains resistor R and inductor L.

D. Y contains resistor R, X contains capacitor C and inductor L.

Lesson 42. Given the circuit as

drawing, L

pure

have a cold,

u  200cos �100 t   � and

AB �

2 �(V)

� �

i  I

cos�100 t  

voltmeter indication V

and V.

o � 4 �(A). Find s 1 2

� �

A. 200V B. 100V C. 200V and 100V D. 100V and 200V

Lesson 43. Given an AC circuit consisting of electrical elements R, L, C connected

continued. The voltage between the two ends of the circuit is

u AB  100 2 cos100 t (V), resistor R replaces

change; coil with Ro

= 30 ,

L  1.4 H ;

C  31.8 F . Adjust R to work

When the power consumption of resistor R reaches its maximum value, the values ​​of R and PR are: A. R = 30; PR = 125W. B. R = 50; PR = 250W.

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