# Lesson Notes By Weeks and Term - Senior Secondary School 2

Alternating Current

SUBJECT: PHYSICS

CLASS:  SS 2

DATE:

TERM: 2nd TERM

WEEK 3

TOPIC: Alternating Current

Alternating Current

Content

A.C in Resistor, inductor and capacitor

Energy in inductance, Reactance and impedance

Vector Diagram

Power in A/C

Resonance and its applications

At any instant, the current through the resistor ®is I an the voltage  across it is V

From ohmÂs law,we have that

V = iR

Thus the current is given by

I = Âs law,we have that

V = iR

Thus the current is given by

I = V

R.

But V = Vo sin wt

I =   V     = Vo sin wt

R              R

I  = Io sin wt

The voltmeter and ammeter connected in the circuit will read the r.m.s values of voltage and current

I r.m.s. =  Vr.m.s

R.

The voltage and the current are said to be in phase or in step with each other . This means that both of them attain their maximum, zero and minimum values at the same instant in time.

Capacitance in an a.c circuit

In the circuit above, an a.c. voltage is connected in series with a capacitor.

Ic leads Vc by  π   radians or 90o or by ¼  cycle/

2

The voltage (v) and current ( I ) are out of place ( not in step) . the current is said to lead on the voltage and the voltage is said to lad on the current. The phase difference between the current and the voltage is 90o or    ( π/2) radian

V = Vosinwt

I = Io sin (wt + π/2)

The capacitor opposes the flow of current.  This opposition to the flow of a.c. offered by the capacitor is known as capacitive reactance, xc. This is given by the relation

Xc  =  1

2πfc

when an a.c.. voltage of frequency f is applied to a capacitance, c, then

V = iXc

In other words, an ohm’s law relation applies to a capacitor.

In it, R is replaced by:

Xc  =   1  .Hence the  unit of Xc is in ohms

2πfc

Example

A 2 UF capacitor is connected directly across a 150Vrms, 60Hz a.c source. Find

1. a) the r.m.s value of the current
2. b) the peak value of current.

Xc =  1

2πfc

= 1

2π x 60 x ( 2 x 10-6)Î

= 1324.4 Î

From V = 1xc

I r.m.s.  =  Vrms    =  150 A

Xc            1324.4

= 0.113A.

Peak current, Io = √2  Ir.m.s.  = 0.160A

Inductance in A.C Circuit

VL leads IL by π/2 radians or 90o. The induced e.m.f. in the inductor L opposes the change in the current.As a result the current is delayed behing the voltage in the circuit. The current lads behind V by π/2 radian or 90o or by ¼ cycle. I and V  have a phase difference of 90o (π/2)

V = Vo sin wt

I = Io sin (wt – π/2 ). Like R and C, an inductor L opposed the flow of current; i.e it has an impedance effect known as inductive reactance, XL.

V =  1 x L

The unit of XL is in ohms

XL  = 2πfl

The unit of L is Henry (H), f is in hertz (HZ) and XL is in ohms.

Reactance is the opposition to the flow of a.c offered by a capacitor or an inductor  or both.

Find the impedance across an inductor of o.2H inductance when an a.c voltage of 60HX is applied across it, if the voltage is given by V = 150 sin 120 πt.Calcualte the  r.m.s and peak values of the current.

XL = 2    πfl

=2π x 60 XL

= 120πl

= 120 x π x 0.2

= 75.40Î

V= 150 sin 120 πt

Vo = 150V

F = 60Hz

Vrms = 0.76  = 0.7 x 150  = 105V

Irms = Vrms   =  105

XL        75.4

= 1.39A

Io = Vo 150

Xl       75.4

= 1.99A

Series Circuit Containing Resistance ®

Inductance (L) and Capacitance (C)

If an alternating voltage V = Vo sing 2πft is put across the circuit, it is observed that a steady state current given by I = Iosin2πft will flow along the circuit . The maximum or peak value of the current is given  by

Io = Vo

(R2 + (XL – Xc)2 ) ½

= Vo

√R2 + X2

X= XL  -XC.

Let Z = ( R2 + (XL –XC )2 ) ½

:. Io = Vo

Z

Ir.m.s.  = Vr.m.s

Z

Z is known as the impedance of the circuit .

Impedance (Z) is the overall opposition of a mixed circuit containing  a resistor, an inductor and or a capacitor. It is measured in ohms.

Xc =  1

Wc

=  1

2πfc

XL = WL  = 2πfL

= Z = √R2 + ( wL – )2 )

wc

:. Z = √R2 + ( 2πfl -  1   )2

2πfc

in summary

V= IR

VL = I X L

Vc = I x C

V = 1Z

= 1 (R2 + (XL – Xc)2 ) ½

Example

(1) Find the r.m.s. value of an alternating current whose peak value is 5A.

Irms = Io

√2

= 0.707Io

= 0.707  = 3.53A.

(2) in a.c circuit the peak value of the potential difference is 180v. What is the instantaneous p.d when it has reached 1/8th of a cycle/

1 cycle = 360o

1/8 of a cycle = 360/8 = 45o

E = Eo sin wt = Eo sin 45

= 180sin 45

= 180/√2

= 90√2 volts.

1. A circuit consist of a resistor 500 ohms and a capacitor 5uF connected in series . if an alternating voltage of 10v and frequency 50Hz is applied across the series circuit. Calculate:
2. the reactance of the capacitor
3. the current flowing in the circuit

iii. the voltage across  the capacitor

(b) If the capacitor is replaced with an inductor of 150mH, calculate the impedance and voltage across the inductor.

Xc =  1

2πfc

1

2π x 50 x 5 x 10-6

=  636.62 ohms

1. Z = √R2 + Xc2

= √5002 + 636.622

= 809.5ohms

I  =  =  10

Z       809.5  A

= 12.35 x 10-3 A

12.35 MA.

iii.  Vc =1 X c

= 12.35  x 10-3  x 636.62

= 7.86 volts

(b)XL = WL = 2π x 50 x 150 x 10-3

= 47.12 OHMS

Z = √ r2 +XL 2

=√5002 = (47.12)2

= 502.2OHMS

I  10

502.2

= 19.9 X 10-3 A = 19.9Ma

V = 1  x L = 10.9 x 10-3 x 47.12

= 938 x 10-3 V  = 938,MA.

VECTOR DIAGRAM

When an alternating voltage is placed across a R.V.C series circuit, the resulting alternating current

1. has the same frequency as the voltage (v0 but the two differing phase or are said to be out of phase

phase is the state of vibration of a periodically varying sytems at a particular time, wt = phase angle.

Two vibrating systems with the same frequency are said to be inphase if their maximum, minimum and zero values occur at the same time; otherwise  they re said to be out of phase.

The phase difference between the voltage and the current through an RLC series circuit is given by

Tan θ = X

R

X = reactance  =Xl – Xc and R is the resistance .

For a circuit containing only a resistance R, the a.c voltage vibrates in phase or in step with the alternating current.

Thus Ç¾ = O

For a circuit containing only a capacitance C, Vc and Ic are out phase by 90o (π/2) radian. This means  that the angle by which a particular phase Ic is in advance of a similar phase of Vc is 90o or π/2 radian or ¼ cycle

If Vc = Vo sin wt

Then Ic = Io sn (wt = π/2).

iii. If only an inductor  L is connected to the a.c voltage, the current IL, lags on the voltage vL by π2 radians

VL = Vo sin wt

IL = Io sin (wt – π/2)

In a circuit containing  RLC the current is the same for all the components of the circuit, and is in phase with the voltage across R. let Vr be the reference vector, the other voltage vectors acts as shown

The effective voltage V is given by

V2 = V2R  + (VL – VC)2

Tan Ç¾  =  VL – VC

VR

VL  - XC

R.

If XL > XC, Ç¾ is positive and I lags.

If XL  < Xc,Ç¾ is negative and I leads V

For R and L series, we have

V2  = V2R + V2L.

I  =  V

√R2 + X2L

Z = √R2 + X2L

Current I, lags on the applied voltage by Ç¾ given by

Tan Ç¾  VL

VR

= XL

R

I lags V or V leads I

For R and C in series

V2 = V2R  + V2C

I  =  V

√R2 + X c2

Z = √R2 + Xc2

Tan Ç¾  Vc

VR   =  Xc

R

V lags I or I leads V.

Power in an A.C Circuit

The average power in an a.c circuit is given by;

P = IV cos Ç¾

I, V are the effective (r.m.s) values of the current and voltage respectively and Ç¾ is the angle of lag or lead between them . The quantity cos Ç¾ is known as the power factor of the device. The power factor can have any value between zero and unity for Ç¾ varying from 90o to 0o. For Ç¾ = 90o or cos Ç¾ = 0, average power P is zero. A power factor of zero means the device is either a pure reactance, inductance or capacitance. Thus no power is dissipated in an inductance or capacitance.

However, if I is the r.m.s value of the current in a circuit containing a resistance R, the power absorbed in the reactance is given by

P = I2R

For an a.c circuit, the instantaneous power is given by

P =IV (instantaneous value)

Power factor

Cos Ç¾ =  Resistance

Impedance

Example

A series circuit consist of a resistance 600 ohms and an inductance 5 henry’s .An a.c voltage of 15v(rms) and frequency 50hz is applied across the circuit, calculate

i the current flowing through the circuit

1. the voltage across the inductor

iii.the phase angle between I and the applied voltage

1. the average power supplied
2. the p.d across the resistance.

XL = 2πfl  = 2 πx 50 x 5 = 500πohms

Z = √R=Xl  =  √(600)2 + (500π)2

= 1.69 X 10 3 ohms

Ir.m.s  = Vrms    =   15

Z            1.69 x 103  = 8.88 x 10-3

= 8.88mA

1. voltage across the inductor

VL = I XL = 8.88 x 10-3  x 500 π

= 4.44πvolts

= 14.95 volts

iii. tan Ç¾  = XL    =   500π  = 2.62.

R          600

Ç¾ = tan-1 ( 2.62)  = 69.10

1. Power supplied

P = I2R

= (8.88 x 10-2)2 x600

= 4.73 x 10-2 w

v.p,d across R.

VR =IR

= 8.88 x 10-3 x 600

= 5.53ohms.

RESONANCE IN RLC

Series Circuit

The current in RLC series circuit is given by:

I  = V    = V

Z       √R2 +  (XL – Xc )2

The maximum current is obtained  in the circuit when the impedance is minimum. This happens when XL = Xc

2πfl =  1

2πfc

Resonance is said to occur in an a.c series circuit when the maximum current is obtained from such a circuit.  The frequency at which this resonance occur is called the resonance frequency (fo). this is the frequency at which Xl = Xc

2πfoL = 1

2πfoC

4π2foLC = 1

fo =  1

4π2LC

fo  = 1

2π √LC

since w = 2πf

wo  =  1

√LC

At f = fo, the current is maximum

Application of Resonance

It is used to tune radios and Tvs. Its great advantage is hat it responds strongly to one particular frequency.

Examples

An a.c  voltage of amplitude 2.0 volts is connected to an RlC series circuit.  If the resistance in the circuit is 5 ohms, and the inductance and capacitance are 3mh and 0.05 uf respectively. Calculate:

1. the resonance frequency,fo
2. the maximum a.c. current at resonance.

Fo =  1

2π√LC

=  1

2π√3  x 10-3 x 0.05 x 10-6

= 1

2π√x 10-11

= 1299. 545Hz

13khz

At resonance X = R since XL =XC

I =  Vo

R

=  2

5

= 0.4A

New School physics pag 458-463

WEEKEND ASSIGNMENT

If the frequency of the a.c current above is  500Hz, what will be the reactance in the circuit .

Π

1. 0.009  ohms                         (b) 400 ohms

Π

1. 1030 ohms           (d ) 1400 ohms      (e ) 2500 ohms
2. At what frequency will 20uf capacitor have a reactance of 500 ohms?

(a)   100        (b)   50        150    (d) 100π            (e) 30

π                        π                                                          π

1. What is the peak value of a voltage whose r.m.s value is 100v

(a) 140v        (b) 70V    (c) 141.4V   (d) 50V   (e) 80.60V.

1. In an RLC series a,c circuit power is dissipated in

(a) Resistance only

(b) Reactance only

(c )Resistance and reactance

(d) Resistance, inductance and capacitance

1. The resonance frequency (fo) in an RLC series circuit is given by

(a) fo  = 2π √LC  (b) fo =

√LC

(c ) fo =  1               (d) fo =  1

2π √LC                   √LC

Theory

1. Explain what is meant by the terms impedance, phase angle and reactance as applied to  an a.c. circuit. Calculate the impedance and phase angle for an a.c. circuit having a 100ohms resistance, 5uf capacitor in series if an a.c voltage of frequency 100Hz is applied across the circuit .
2. Draw a vector diagram of the relationship of  I and V for an a.c. circuit containing

(a) a pure inductor        (b) a pure capacitor        (c ) a pure resistor