SIMPLE A.C CIRCUIT

 

TOPIC: -                                                                SIMPLE A.C CIRCUIT      

INSTRUCTIONAL MATERIALS: -                 

REFERENCES BOOK: -

1.       FARINDE O. E e tal, ESSENTIAL PHYSICS FOR SSS, Tonad Publishing Limited.

2.       M. W. ANYAKOHA (2011), NEW SCHOOL PHYSICS FOR SENIOR SECONDARY SCHOOLS,  Africana first publishers. 

3.       INTERNET           

PREVIOUS KNOWLEDGE: - Students have been familiar with gadgets uses electromagnetic principles.

 OBJECTIVES: -  At the end of the lesson students should be able to: -

i.                     Explain a.c circuit

ii.                   Define reactance and impedance

iii.                  Derived equation that connect L-R, R-C, LCR

iv.                 Solved mathematical problems involving a.c

CONTENT: -

A.C CIRCUIT

A.C circuit are circuit through which an alternating current flows such current are used to extensively in power transmission, radio and television, computer technology, telecommunication and in medicine.

 Alternating current and Alternating EMF

An alternating current is one whose magnitude changes sinusoidal with time. A.C are produced by alternating voltages. Alternating currents and alternating voltages are represented as follow.

I = I_o Sin2πft or I = I_oSinωt ;  V = V_o Sinπft or V = V_oSinωt ;

Where I = Instantaneous current (A), Io = Peak or maximum current (A), Frequency (Hz), t=time (s),

ωt=θ=Phase angle of current or voltage, V= Instantaneous voltage (V), Vo= Peak or maximum voltage.

Figure below shows the variation of A.C with time

Root Mean square value of AC

The peak value of an a.c is the maximum value of the current recorded in an a.c cycle. The r.m.s current is defined as the steady or direct current (d.c) which produces the same heating effect per second in a given resistor.

Root mean square current, I_r.m.s=  = 0.71 I_o;  I_o =  X I_r.m.s = 1.4141 I_r.m.s

Similarly, Root mean square voltage, V_r.m.s=  = 0.71V_o;  V_o =  X V_r.m.s =1.4141 V_r.m.s

Example

1. Calculate the peak voltage of a mains supply of r.m.s value 220V.

Solution

V_r.m.s = 220v, Vo = ?.

Vo = =1.4141 V_r.m.s;      Vo = 1.4141 X 220v = 311v

2. The voltage, V in an a.c circuit is given by the equation:  V = 30Sin 100πt. Where t is the time in seconds. Deduce the following from this equation.

(i) Frequency of the voltage        (ii) Peak value of the voltage      (iii) r.m.s value of the value of the voltage.

Solution

The equation V = 30Sin 100πt can be compared with

                          V = V_o Sin2πft

(i)                  100πt = 2πft; 100 = 2f; f= 100/2 = 50Hz

(ii)                Vo = 30v

(iii)               V_r.m.s= 0.71V_o = 0.71X30 = 21.21A

REACTANCE AND IMPEDANCE

Reactance is defined as the opposition offered to the passage of an a.c by either the inductor or capacitor or both. Reactance due to an inductor is called inductive reactance. Reactance due to a capacitor is called capacitive reactance.

Impedance is the overall opposition offered to the passage of an a.c mixed circuit containing a resistor any one or both of an inductor and a capacitor. Reactance and impedance are both measured in ohm (Ω)

Also reactance =

A.C through pure resistor

From instantaneous values of alternating voltage and current,  we can conclude that in pure resistor ,the current is always in phase with applied voltage.

Resistance, R =  =    =    

 A.C through pure inductor
Figure below shows the circuit in which voltage source  V=V₀sinωt is applied to pure inductor (zero resistance) coil of inductance L

Voltage across inductor is V_L = IX_L

X_L =  2πfL. Inductance L  of an inductor is given by: L =

From instantaneous values of current and voltage we see that in pure inductive circuit the current lags behind emf by a phase angle of π/2

This phase relationship is graphically shown below in the figure

AC through pure capacitor

Figure given below shows circuit containing alternating voltage source V=V₀sinωt
connected to a capacitor of capacitance C

Voltage across capacitor, Vc = I x Xc

Capacitive reactance Xc =  =

With V=V₀sinωt, we see that in a perfect capacitor current leads emf by a phase angle of π/2

This phase relationship is graphically shown below in the figure

 Circuit containing inductance and resistance in series

Figure below shows pure inductor of inductance L connected in series with a resistor of resistance R through sinusoidal voltage V=V₀sin(ωt+φ)

L-R circuit, the impedance, Z_LR=  =    ;          Z_LR=

Current in L-R circuit, Io =

Tan θ =  =  =

Effective or total circuit voltage, V =

The current I also leads on the applied voltage V  by an angle θ given by:

Circuit containing capacitance and resistance in series

L-C circuit, the impedance, Z_RC=  =    ;          Z_LR= Current in L-R circuit, Io =

The applied voltage V also lead on the current I by an angle θ given by:

Tan θ =  =  =

Effective or total circuit voltage, V =

LCR series circuit

Figure below shows a circuit containing a capacitor ,resistor and inductor connected in series through an alternating voltage source

Same amount of current will flow in all the three circuit components and vector sum of potential drop across each component would be equal to the applied voltage.

If i be the amount of current in the circuit at any time and V_L,V_C and V_R the potential drop across L,C and R respectively then   V_R=iR ⇒ Voltage is in phase with I;   V_L=iωL ⇒ Voltage is leading i by 90⁰;
V_C=i/ωC ⇒ Voltage is lagging behind i by 90⁰

Since V_L is ahead of i by 90 and V_C is behind by 90 so that phase difference between V_L and V_C is 180 and they are in direct opposition to each other as shown in the figure  below

RLC circuit, the impedance, Z_RLC=  =     

        

Io =

Phase angle is given by : Tan θ =  =  =  

Effective voltage, V = )²

 which is the minimum value of Z we can have this is the case where X_L=X_C, the circuit is said to be in electric resonance where the impedance is purely resistive and minimum and currents has its maximum value

This frequency is called resonant frequency of the circuit and peak current in this case is i₀=V₀/R and reactance is zero

Figure below shows the shape of resonance curve for various values of resistance R

For small value of R,the resonance is sharp which means that if applied frequency is lesser to resonant frequency f₀,the current is high otherwise. For large values of R, the curve is broad sided which means that those is limited change in current for resonance and non -resonance conditions.

Exercise

1. When a parallel plate capacitor was connected to a 60Hz AC supply, it was found to have a reactance of 390 ohms. Calculate the value of the capacitor in micro-farads.

Solution

             F= 60Hz,   Xc =390 ohm,

Xc =    ; C =   = 1/2x3.142x60x390 = 6.8uF

2. A 400 mH coil of negligible resistance is connected to an AC circuit in which an effective current of 6 mA is flowing. Find out the voltage across the coil if the frequency is 1000 Hz.

Solution

L = 400 x 10^-3 H; I_eff = 6 x 10^-3; f = 1000 Hz

X_L =  2πfL = 2 X 3.142 X 1000 x 400 x 10^-3 = 2512 ohm

Voltage across L V_L = IX_L = 6 x 10^-3 X 2512 = 15.07v

3. If the impedance of the circuit is 250 Ω, determine resistance of resistor R.

The impedance of the circuit (Z) = 250 Ω; Capacitor (C) = 8 m F = 8 x 10^-6 F

Inductor (L) = 0.8 H;  Voltage (V) = 200 Volt ; w = 500 rad/s;  Resistance of resistor (R)=?

Solution :

                                    

 PRESENTATION

Step I: The teacher revises the previous topic.

 Step II: The teacher introduces the new topic.

Step III: The teacher explains the meaning of a.c circuit.

Step IV: The teacher derives formula for a.c containing only resistor, inductor, capacitor and their combinations.

Step V: The teacher leads the students to solve problem on a.c circuit.

 EVALUATION

The teacher evaluates the lessons by asking these questions:

i.                     Explain a.c circuit

ii.                   Define reactance and impedance

iii.                  Derived equation that connect L-R, R-C, LCR