AC Current

Examples for Chapter 24Fundamentals of College Physics, 2nd.Ed

Dr. Peter J. Nolan

"Fundamentals of College Physics" Third EditionDr. Peter J. Nolan, SUNY Farmingdale

Chapter 24 Alternating Current CircuitsComputer Assisted InstructionInteractive Examples

Example 24.1

+ 3.50 A. Find the effective value of this current.

Initial Conditions

3.5 A

Solution. The effective value of the current, found from equation 24.5, is

0.707 ) x ( 3.5 A)

2.4745 A

Thus, even though the current is varying with time, if an AC ammeter were placed inthe circuit it would read the single value of 2.4745 A

Example 24.2

maximum voltage in a 60.0-Hz, 120-V line?

Initial Conditions

120 V

Solution. The effective voltage is 120 V and the maximum voltage, found from equation 24.7, is

120 V) / ( 0.707 )

169.7313 V

Hence, even though an AC voltmeter would read the single value 120 V,

The effective value of an AC current. An AC current in a circuit varies from -3.50 A to

imax =

ieff = 0.707 imax

ieff = (

ieff =

Finding the maximum value when the effective value is known. What is the

Veff =

Veff = 0.707 Vmax

Vmax = Veff / 0.707

Vmax = (

Vmax =


Dr. Peter J. Nolan

the actual voltage would be varying between -169.731 V and + 169.731 V.

Example 24.3

Initial ConditionsR = 400 W f = 60 HzL = 5 H V = 110 VC = 3.00E-06 F

Solution.

2 ) x ( 3.141593 ) x 60 Hz) x ( 5 H)

1884.956 W

1 ) / [( 6.283185 ) x ( 60 Hz) x ( 3.00E-06 F)]

884.1941 W

Z = Sqrt[( 400 1884.956 884.1941Z = 1077.74 W

i = V / Zi = ( 110 V) / ( 1077.74 W)

i = 0.102065 A

An RLC series circuit. The RLC series circuit shown in figure 24.7 has a resistanceR = 400 W, an inductor L = 5.00 H, a capacitor C = 3.00 mF, and they are connected

to a 110-V, 60.0-Hz line. Find (a) the inductive reactance XL, (b) the capacitive

reactance XC, (c) the impedance Z of the circuit, (d) the current i in the circuit, (e) the

voltage drop VR across R, (f) the voltage drop VL across L, (g) the voltage drop VC across C, (h) the total voltage V across RLC, and (i) the phase angle f.

a. The inductive reactance, found from equation 24.15, is

XL = 2 p f L

XL = (

XL =

b. The capacitive reactance, found from equation 24.17, is

XC = 1 / [2 p f C]

XC = (

XC =

c. The impedance of the circuit, found from equation 24.18, is

Z = Sqrt[R2 + (XL - XC)2]

W)2 + (( W) - ( W))2]

d. The effective current i in the circuit, found from equation 24.19, is

e. The voltage drop across R, found from equation 24.13, is


Dr. Peter J. Nolan

0.102065 A) x ( 400 W)

40.82617 V

0.102065 A) x ( 1884.956 W)

192.3888 V

0.102065 A) x ( 884.1941 W)

90.24566 V

V = Sqrt[( 40.82617 192.3888 V - ( 90.2457V = 110 V

which is, of course, equal to the applied voltage. Notice that the voltages are addedvectorially and not algebraically.

192.3888 V) - ( 90.24566 V)) / ( 40.8262 V) ]68.21363 degrees

This means that the applied voltage leads the current in the circuit by 68.2136 degrees

called an inductive circuit.

Example 24.4

VR = i R

VR = (

VR =

f. The voltage drop across L, found from equation 24.14, is

VL = i XL

VL = (

VL =

g. The voltage drop across C, found from equation 24.16, is

VC = i XC

VC = (

VC =

h. The total voltage across R, L, and C in series, found from equation 24.11, is

V = Sqrt[VR2 + (VL - VC)2]

V)2 + ( ( V))2]

i. The phase angle, found from equation 24.12, is

f = arctan[(VL - VC) / VR ]f = arctan[((

f =

and the phase relation is shown in figure 24.8. Since f is a positive angle the circuit is

An RC series circuit. A 110-V, 60.0-Hz, AC line is connected across a resistance of1000 W and a capacitor of 1.00 mF, as shown in figure 24.9. Find (a) the capacitive

reactance, (b) the impedance, (c) the current in the circuit, (d) the voltage drop VR

across the resistor, (e) the voltage drop VC across the capacitor, (f) the total voltageacross R and C, and (g) the phase angle f between the voltage and current.

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Dr. Peter J. Nolan

Initial ConditionsR = 1000 W f = 60 HzC = 1.00E-06 F V = 110 V

Solution.

1 ) / [( 6.283185 ) x ( 60 Hz) x ( 1.00E-06 F)]

2652.582 W

Z = Sqrt[( 1000 2652.582Z = 2834.818 W

i = V / Zi = ( 110 V) / ( 2834.818 W)

i = 0.038803 A

0.038803 A) x ( 1000 W)

38.8032 V

0.038803 A) x ( 2652.582 W)

102.9287 V

V = Sqrt[( 38.8032 102.9287V = 110 V

a. The capacitive reactance XC, found from equation 24.17, is

XC = 1 / [2 p f C]

XC = (

XC =

b. The impedance Z is found from equation 24.18. Because there is no induction in

this RC circuit, XL = 0. Therefore, the impedance becomes

Z = Sqrt[R2 + XC2]

W)2 + ( W)2]

c. The effective current i in the circuit comes from Ohm's law, equation 24.19, and is

d. The voltage drop VR across the resistor, found from equation 24.13, is

VR = i R

VR = (

VR =

e. The voltage drop VC across the capacitor, found from equation 24.16, is

VC = i XC

VC = (

VC =

f. The total voltage drop across R and C in series is found from equation 24.11. Since

there is no inductance in this circuit, VL = 0. Therefore,

V = Sqrt[VR2 + VC

2]

V)2 + (( V))2]


Dr. Peter J. Nolan

Note that the voltage across R and C in series is the same as the applied voltage, which it should be. Because of the phase difference of the voltages, they add asvectors rather than as algebraic quantities.

-102.929 V)) / ( 38.8032 V) ]-69.344 degrees

This phase angle is represented in figures 24.10(a) and 24.10(b). The voltage in the

circuit lags the current in the circuit by -69.344the circuit is called a capacitive circuit.

Example 24.5

drop V across R and L, and (g) the phase angle between the voltage V and thecurrent I.

Initial ConditionsR = 1000 W f = 60 HzL = 5 H V = 110 V

Solution.

2 ) x ( 3.141593 ) x ( 60 Hz) x ( 5 H)

1884.956 W

Z = Sqrt[( 1000 1884.956Z = 2133.789 W

g. The phase angle f between the voltage and the current in the circuit is found from

equation 24.12 with VL = 0. Therefore,

f = arctan[( - VC) / VR ]f = arctan[((

f =

0. Since f is a negative quantity,

An RL series circuit. A 110-V, 60-Hz, AC line is connected across a resistance of1000 W and an inductor of 5.00 H, as shown in figure 24.11. Find (a) the inductive

reactance, (b) the impedance, (c) the current in the circuit, (d) the voltage drop VR

across the resistor, (e) the voltage drop VL across the inductor, (f) the total voltage

a. The inductive reactance XL, found from equation 24.15, is

XL = 2 p f L

XL = (

XL =

b. The impedance Z is found from equation 24.18, with XC = 0. Therefore,

Z = Sqrt[R2 + XL2]

W)2 + ( W)2]

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Dr. Peter J. Nolan

i = V / Zi = ( 110 V) / ( 2133.789 W)

i = 0.051551 A

0.051551 A) x ( 1000 W)

51.55148 V

0.051551 A) x ( 1884.956 W)

97.17225 V

Therefore,

V = Sqrt[( 51.55148 97.17225V = 110 V

Note that the voltage drop across R and L is the same as the applied voltage V, as isexpected.

97.17225 V) / ( 51.55148 V) ]62.05331 degrees

Because this is a positive angle it means that the voltage leads the current in the circuit by 62.10 and the circuit is an inductive circuit. This is shown in figures 24.12(a) and 24.12(b).

Example 24.6

c. The current i in the circuit, found from Ohm's law, equation 24.19, is

d. The voltage drop VR across the resistor, found from equation 24.13, is

VR = i R

VR = (

VR =

e. The voltage drop VL across the inductor, found from equation 24.14, is

VL = i XL

VL = (

VL =

f. The total voltage drop V across R and L is found from equation 24.11 with VC = 0.

V = Sqrt[VR2 + VL

2]

V)2 + ( V)2]

g. The phase angle f between the current and voltage in the circuit is found from

equation 24.12 with VC = 0. Therefore,

f = arctan[ VL / VR ]f = arctan[(

f =

Resonant frequency. Find the resonant frequency of the RLC series circuit in

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Dr. Peter J. Nolan

example 24.3.

Initial ConditionsL = 5 H C = 3.00E-06 F

Solution. The resonant frequency, found from equation 24.21, is

1 ) / [( 6.283185 ) x sqrt( 5 H) x ( 3.00E-06 F)]

41.09363 Hz

Example 24.7

power factor, (b) the power consumed, and (c) the total power that must be supplied to the circuit.

Initial ConditionsSee example 24.3

Solution.

circuit, found from equation 24.23, is68.21363 ) = 0.371147 = 37.1147 %

0.102065 4004.166941 W

0.102065 A) x ( 110 V)

11.2272 W

As a check, note that4.166941 W) / ( 11.2272 W)

Power supplied= 0.371147 = 37.1147 %

which agrees with the original value, within round-off errors of the calculations.

fo = 1 / [2 p sqrt(L C)]

fo = (

fo =

Power factor in an AC circuit. In the RLC series circuit of example 24.3, find (a) the

a. In example 24.3 the phase angle f was found to be 68.40. The power factor of the

PF = cos f = cos(

b. The power dissipated in the circuit is

P = i2 R

P = ( A)2 x ( W)P =

c. The power applied to the circuit is

Papp = I Vapp = (

Papp =

Power consumed =


Dr. Peter J. Nolan

Example 24.8

Initial ConditionsR = 400 W f = 60 HzL = 5 H V = 110 VC = 3.00E-06 F

Solution.

2 ) x ( 3.141593 ) x ( 60 Hz) x ( 5 H)

1884.956 W

1 ) / ( 6.283185 ) x ( 60 Hz) x ( 3.00E-06 F)

884.1941 W

110 V) / ( 400 W)

0.275 A

110 V) / ( 1884.956 W)

0.058357 A

An RLC parallel circuit. A resistor of 400 W, an inductor of 5.00 H, and a capacitor of3.00 mF are connected in parallel to a 110-V, 60.0-Hz line. Find (a) the inductive

reactance, (b) the capacitive reactance, (c) the current through the resistor IR, (d) the

current through the inductor IL, (e) the current through the capacitor IC, (f) the total

current in the circuit IT, (g) the phase angle f, and (h) the total impedance of the circuit.

a. The inductive reactance is

XL = 2 p f L

XL = (

XL =

b. The capacitive reactance XC is found from equation 24.17 as

XC = 1 / [2 p f C]

XC = (

XC =

c. The current IR through the resistor, found from Ohm's law, equation 24.27, is

IR = V / R

IR = (

IR =

d. The current IL through the inductor, found from equation 24.28, is

IL = V / XL

IL = (

IL =

e. The current IC through the capacitor, found from equation 24.29, is


Dr. Peter J. Nolan

110 V) / ( 884.1941 W)

0.124407 A

0.275 0.124407 A) - ( 0.05836

0.282821 A

0.124407 A) - ( 0.058357 A)) / ( 0.275 A) ]13.50563 degrees

Z = ( 110 V) / ( 0.282821 A) Z = 388.9388 W

Example 24.9

impedance of the circuit.

Initial ConditionsR = 1000 W f = 60 HzC = 1.00E-06 F V = 110 V

Solution.

1 ) / ( 6.283185 ) x ( 60 Hz) x ( 1.00E-06 F)

2652.582 W

IC = V / XC

IC = (

IC =

f. The total current IT in the circuit, found from equation 24.26, is

IT = Sqrt[IR2 + (IC - IL)2]

IT = Sqrt[( A)2 + (( A))2]

IT =

g. The phase angle f, found from equation 24.30, is

f = arctan[(IC - IL) / IR ]f = arctan[(

f =

The total current in the circuit leads the applied voltage by 13.50

h. The total impedance of the circuit, found from equation 24.32, is

Z = V / IT

An RC parallel circuit. A 110-V, 60.0-Hz, AC line is connected in parallel to a resistorof 1000 W and a capacitor of 1.00 mF, as shown in figure 24.17. Find (a) the

capacitive reactance, (b) the current IR through the resistor, (c) the current IC through the capacitor, (d) the total current in the circuit, (e) the phase angle f, and (f) the total

a. The capacitive reactance is found to be

XC = 1 / [2 p f C]

XC = (

XC =

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Dr. Peter J. Nolan

110 V) / ( 1000 W)

0.11 A

110 V) / ( 2652.582 W)

0.041469 A

0.11 0.041469

0.117557 A

0.041469 A) / ( 0.11 A) ]20.656 degrees

Z = ( 110 V) / ( 0.117557 A) Z = 935.7152 W

Example 24.10

connected to the primary of a transformer of 50 turns. If the secondary of the transformer has 100 turns, what voltage will be found across the secondary of the transformer?

Initial Conditions

50 turns 120 V

100 turns 3 A

b. The current in the resistor, found from equation 24.27, is

IR = V / R

IR = (

IR =

c. The current IC through the capacitor, found from equation 24.29, is

IC = V / XC

IC = (

IC =

d. The total current in the circuit is found from equation 24.26 with IL = 0, that is,

IT = Sqrt[IR2 + IC

2]

IT = Sqrt[( A)2 + ( A)2]

IT =

e. The phase angle f is found from equation 24.30 with IL = 0, that is,

f = arctan[ IC / IR ]f = arctan[(

f =

f. The total impedance of the circuit, found from equation 24.32, is

Z = V / IT

The voltage in a step-up transformer. A 120-V line from an AC generator is

N1 = V1 =

N2 = i1 =

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Dr. Peter J. Nolan

Solution. The voltage across the secondary, found from equation 24.34, is

100 ) / ( 50 ) ] x ( 120 V)

240 V

Example 24.11

3.00 A, what is the current in the secondary?

Initial ConditionsSee initial conditions in example 24.10

Solution. The current in the secondary, found from equation 24.38, is

50 ) / ( 100 ) ] x ( 3 A)

1.5 A

Example 24.12

from your wall outlet and drop it down to 24.0 V to operate a toy electric train. If thereare 100 turns of wire in the primary how many turns do you need for the secondary?

Initial Conditions

120 V 100 turns

24 V

Solution. The number of turns in the secondary can be found from equation 24.37 as

24 V) / ( 120 V) ] x ( 100 turns)

20 turns

V2 = [N2 / N1] V1

V2 = [(

V2 =

The current in a step-up transformer. If the current in the primary of example 24.10 is

I2 = [N1 / N2] i1

I2 = [(

I2 =

A step-down transformer. You wish to make a transformer that can take the 120 V

V1 = N1 =

V2 =

N2 / N1 = V2 / V1

N2 = [ V2 / V1 ] N1

N2 = [(

N2 =


Dr. Peter J. Nolan

"Fundamentals of College Physics" Third Edition

The effective voltage is 120 V and the maximum voltage, found from equation 24.7, is

AC Current

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