Physics 12th EMF and Electric Measurement

EMF AND ELECTRIC MEASUREMENT

(1) Introduction

• In previous chapter we have already studied about electric current and resistance.
• We know that a force must be applied on free charges of a conductor in order to maintain a continous current in the conductor.
• Here a question arises how can we maintain this force in order to maintain a continous flow of current. You will find answer to this question while studying this chapter.
• In this chapter we will learn about ElectroMotive Force(emf) and sources of emf ( responsible for driving charge round the closed circuit). We'll also learn about electric circuits and measurements. 

(2) ElectroMotive Force(emf)

• Consider a conductor lying in presence of electric field as shown below in the figure such that an electric field exists inside the conductor.
  
   
• We know that when electric field exists in a conductur electric current begins to flow inside the conductor. Now a question arises what happens to the
  charge carriers when they reach the ends of the conductor and would this current remains constant with the passage of time.
• We can easily conclude that for an open ended conductor as shown in the figure , charges would accumulate at the ends of the conductor resulting a change in electric field with the passage of time. Due to this electric current would not remain constant and would flow only for a very short interval of time , digrametically shown below in the figure.
  
  
  
   
• Thus, in order to maintain a steady current throughout a conducting path the path must be in the form of a closed loop forming a complete circuit. Even this condition is not sufficient to maintain a steady current in the circuit.
• This is because charge always moves in the direction of decreasing potential and electric field always does a positive work on the charge.
• Now after travelling through a complete circuit when charge returns to a point where it has started, potential at that point must be same as the potential at that point in the begning of the journey but flow of current always involves loss of potential energy.
• Hence we need some external source in the circuit in which maintains a potential difference at its terminals by increasing the potential energyof the electric charge.
• Such a source make charge travel from lower potential to higher potential energy in direction opposit to the electrostatic force trying to push charge from higher potential to lower potential.
• This force that makes charge move from lower potential to higher potential is called electro mative force (EMF).
• The source or device which provide emf in a complete circuit is known as source of EMF and examples of such devices are generator, batteries, thermocuples etc.
• The source of EMF are basically energy converters that convert mechanical, thermal, chemical or any other form of energy into electrical potential energy and transform it into the circuit to which the source of emf is connected.
• Now we know that a source of emf or battery maintains a potential difference between its two terminals as shown in below figure
  
   
• Generally a battery consist of two terminals one positive and other is negative.
• Some internal force F_n generally non electric in nature is exerted on the charges of the material of the battery. This non-electric force depends on the nature of source of EMF.
• These Force(F_n) drives the positive charges of the material towards P and nigative charges of the material towards Q. This battery force
  F_n is directed Q to P.
• Positive charge accumulate on plate Pand negative charge accumulate on plate Q and a potential developes between plates P and Q. Thus an electric field would set up inside the battery from P to Q which excert an electric force on the charge of the material.
• When a steady state is reached, the the electric force and battery force F_n would become equal and opposit
  mathematically,
  qE=F_n                                (1)
  and after a steady state is reached no further accumulation of charhe takes place.
• Workdone by battery force F_n in taking poisitive charge from terminal Q to terminal P would be
  W=F_nd
  where d is distance between plates P and Q.
• Workdone by force F_n per unit charge is
  EMF=W/q=F_nd/q                                (2)
  where the quantity E is known as E.M.F. of the battery.
• For steady state
  ENF=qEd/q=ED=V                                (3)
  where V_eq=Ed is the potential difference across the terminals of the battery when nothing is connected externally between P and (i.e. when circuit is open)

(3)Internal Resistance of Battery (or cell)

• The resistance offered by medium in between plates of battery (electrolytes and electrodes of the cell) to the flow of current within the battery is called internal resistance of the battery.
• Internal resistance of a battery usually d branch containing batteryenoted by r and in electric circuit its representation is shown below in the figure
  
   
• Internal resistance of a battery depends on factors like seperation between plates, plate area, nature of material of plate etc. For an ideal cell r=0 , but real batteries or sourcesof emf always has same finite internal resistance.
• If P and Q are two terminals of the battery shown below in the figure
  
  
  then potential difference between terminals P and Q is
  V_P=(V_P-V_x) - (V_Q - V_x) = E-Ir
  let V_P-V_Q=V
  V=E-IR
  now for I=0 and V=EMF
  and this potential difference V is called the terminal difference of the cell or battery and defined as the emf of the batterywhen no current drawn from it.
• For real battery equation(4) which gives V=E-Ir whereI is the current in the branch containing battery.
• From figure(4) potential difference across the external resistance R of the circuit would be equal to terminal potential difference of the cell. Thus
  V=IR also V=E-Ir
  or, IR=E-Ir
  which gives
  I=E/(R+r) =Net EMF/Net resistance
• From equation(4) we can calculate that when current is drawn from the battery terminal potential difference is less than the EMF of the battery.

(4)Electric Energy and Power

• To understand the process of energy transfer in a simple circuit consider a simple circuit as shown in the figure given below
  
   
• Positive terminal of the battery as we all know is always at higher potential.
• Let ΔQ amount of charge begin to flow in the circuit from point E through the battery and resistor and then back to point E.
• When Charge ΔQ moves from point E to point F through the battery electric potential potential energy of the system increased by the amount
  ΔU=ΔQ V -(6) and the electric energy of the battery decreased by the same amount.
• When charge ΔQ moves from point G to S through resistoe R, there comes a decrease in electric potential energy.
• This loss in potential energy appears as the increased in thermal energy of the resistor.
• Thermal energy of the resistor increases because when the charge moves through the resistor they loss there electrical potential energy by colliding with the atom in the resistor. This way electrical energy is transformed internal energy crossesponding to increase in vibrational motion of the atom of the resistor and this cause increase in temprature of the resistor.
• The connecting wires are assumed to have negligible resistance and no energy transfer occur for the path FG and HE.
• In time Δt charge ΔQ moves through the resistor i.e. from G to H. The rate at which it loss potential energy
  ΔU/Δt=(ΔQ/Δt)ΔV=IΔV where I is thecurrent in the resistor and ΔV is the potential difference across it.
• This charge ΔQ regain its energy when it passes through the battery at the cost of conservation of chemical energy of electrolyte to the electrical energy.
• This loss of potential energy as stated earlier appears as increased thermal energy of the resistor. If P represents the rate at which energy is delivered to the resistor then
  P=IΔV
• We know that ΔV =IR for a resistor hence alternative forms of equation(8) are
  P=I²R=ΔV²/R where I is expressed in amperes, ΔV iv volts and resustance R in ohm(Ω)
• SI unites of power is watt such that
  1watt=1volt*1ampare
  Bigger unit of electric power are Kilowatt(KW) and Megawatt(MW).

(5) Kirchoff's Rules

• We have already analyzed simple circuit using ohm's laws and reducing these circuit to series and parallel combination of resistors
• But we also come across circuits containing sources of EMF and grouping of resistors can be far more complex and can not be easily reduced to a single equivalent resistors
• Such complex circuits can be anaylzed using two kirchoff's rules

(A) The junction Rule (or point rule)

• This law states that "The algebric sum of all the currents entering juntion or any point in a circuit must be equal to the sum of currents leaving the junction"
• Alternatively this rule can also be stated as " Algebric sum of the currents meeting at a point in a electric circuit is always zero i.e
  ΣI=0 at any point in a circuit
• This law is based on the law of conservation of charge
• Consider a point P in an electric circuit at which current I₁,I₂,I₃ and I₄ are flowing through conductors in
  the direction shown below in the figure below
   
• If we take current flowing towards the junction as positive and current away from the junction as negative,then from kirchoff's law
  I₁+I₂+(-I₃)+(-I₄)=0
  or,
  I₁+ I₂=I₃+ I₄
• From this law ,we conclude that netcharge coming towards a point must be equal to the net charge going away from this point in the same interval of time

(B) The Loop Rule (or Kirchoff's Voltage Law)

• The rule states that " the sum of potential difference across all the circuit elements along a closed loop in a circuit is zero
  ΣV=0 in a closed loop
• Kirchoff's loop rule is based on the law of conservation of energy becuase total amount of energy gained and losed by a charge round a trip in a closed loop is zero
• when applying this kirchoff's loop rule in any DC circuit,we first choose a closed loop in a circuit that we are analyzing
• Next thing we have to decide is that whether we will traverse the loop in a clockwise direction or in anticlockwise direction and the answer is that ,the choice of direction of travel is arbitrary to reach the same point again
• When traversing the loop ,we will be following convention to note down drop or rise in the voltage across the resistors or battery
  i) If the resistor is being traversed in the direction of the current then change in PD across it is negative i.e -IR
  ii)If the resistor is being traversed in the direction opposite to the current then change in PD across it is negative i.e IR
  iii) If a source of EMF is traversed in the direction from -ve terminal to its positive terminal then change in electric potential is positive i.e E
  iv)If a source of EMF is traversed in the direction from +ve terminal to its negative terminal then change in electric potential is negative i.e -E
• We would now demonstrate the use of kirchoff's loop law in finding equations in simple circuit
• Consider the circuit as shown below
   
• First consider loop ABDA.Lets traverse loop in anticlock wise direction.From kirchoff's loop law
  ΣV=0
• Neglecting internal resistance of the cell and using sign conventions stated previously we find
  -I₃R₃+E-I₁R₁-I₂R₂=0
  or
  I₁R₁+I₂R₂+I₃R₃=E
  And similarly if we traverse the loop ABCA in clock wise direction
  -I₂R₂+I₅R₅+I₄R₄=0
  or,
  I₅R₅+I₄R₄-I₂R₂=0

(6)Grouping of the cell's

• A limited ammount of current can be drawn from a single cell or battery
• There are situations where single cell fails to meet the current requirement in a circuits
• To overcome the problem cells can be grouped in series and in parallel combinations or mixed grouping of cells is done in order to obtain a large value
  of electric current

(A) Series combination

• Figure below shows the two cells of emf's E₁ and E₂ and internal resistance r₁ and r₂ respectively connected in series combination through external resistance
  
  
• Points A and B in the circuit acts as two terminals of the combination
• Applying kirchoff's loop rule to above closed circuit
  -Ir₂-Ir₁-IR+E₁+E₂=0
  or
  I=E₁+E₂/R+(r₁+r₂)
  Where I is the current flowing through the external resistance R
• Let total internal resistance of the combination by r=r₁+r₂ and also let E=E₁+E₂ is the total EMF of the two cells
• Thus this combination of two cells acts as a cell of emf E=E₁+E₂ having total internal resistance r=r₁+r₂ as shown above in the figure
  
   

(B) Parallel combinations of cells

• Figure below shows the two cells of emf E₁ and E₂ and internal resistance r₁ and r₂ respectively connected in parallel combination through external resistance
  
• Applying kirchoff's loop rule in loop containing E₁ ,r₁ and R,we find
  E₁-IR-I₁r₁=0 ------------------------(1)
  Similarly applying kirchoff's loop rule in loop containing E₂ ,r₂ and R,we find
  E₂-IR-(I-I₁)r₂=0 ------------------------(2)
• Now we have to solve equation 1 and 2 for the value of I,So multiplying 1 by r₂ and 2 by r₁ and then adding these equations results in following equation
  IR(r₁+r₂)+r₂r₁I-E₁r₂-E₂r₁=0
  which gives
  
  
  We can rewrite this as
  
  E is the resulting EMF due to parallel combination of cells and r is resulting internal resistance. 

(7) Wheat stone bridge

• Wheat stone bridge was designede by british physicist sir Charles F wheatstone in 1833
• It is a arrangement of four resistors used to determine resistance of one resistors in terms of other three resistors
• Consider the figure given below which is an arrangement of resistors and is knowns as wheat stone bridge
  
  
• Wheatstone bridge consists of four resistance P,Q,R and S with a battery of EMF E.Two keys K₁ and K₂ are connected across terminals A and C and B and D respectively
• ON pressing key K₁ fisrt and then pressing K₂ next if galvanometer does not show any deflection then wheatstone bridge is said to be balanced
• Galavanometer is not showing any deflection this means that no current is flowing through the galvanameter and terminal B and D are at the same
  potential .THus for a balanced bridge
  V_B=V_D
• Now we have to find the condition for the balanced wheatstone bridge .For this applying kirchoff's loop rule to the loop ABDA ,we find the relation
  -I₂R+I₁P=0
  or I₁P=I₂R --(a)
  Again applying kirchoff's rule to the loop BCDB
  I₁Q-I₂S=0
  or I₁Q=I₂S --(b)
  From equation a and b we get
  I₁/I₂=R/P=S/Q
  or
  P/Q=R/S                       (12)
• equation 12 gives the condition for the balanced wheatstone bridge
• Thus if the ratio of the resistance R is known then unknown resistance S can easily be calculated
• One important thing to note is that when bridge is balanced positions of cell and galvanometer can be exchanged without having any effect on the balance of the bridge
• Sensitivity of the bridge depends on the relative magnitudes of the resistance in the four arm of the bridge is maximum for same order of four resistance.

(8) Meterbridge (slide wire bridge)

• Meter bridge is based on the principle of wheatstone bridge and it is used to find the resistance of an unknown conductor or to compare two unknown
  resistance 
• Figure below shows a schematic diagram of a meter bridge
  
  
• In above figure AC is a 1m long wire made of maganin or constanan having uniform area of cross-section
• This wire is stretched along a scale one a wooden base
• Ends A and C of the wire are screwed to two L shaped copper strips as shown in figure
• A resistance box R and an unknown resistance S are connected as shown in figure
• One terminal of galvanometer is connected to point D and another terminal is joined to a jockey that can be slided on a bridge wire
• when we adjust the suitable resistance of value R in the resistance box and slide this jockey along the wire then a balance point is obtained sat at point B
• Since the circuit now is the same as that of wheatstone bridge ,so from the condition of balanced wheatstone bridge we have
  P/Q=R/S
  Here resistance P equals
  P=ρl₁/A
  And Q=ρl₂/A
  where ρ is the resistivity of the material of the wire and A is the area of cross-section of wire
  Now P/Q=(ρl₁/A)(A/ρl₂)=l₁/l₂

(9) Potentiometer

• Potentiometer is an accurate instruments used to compare emf's of a cells,Potential difference between two points of the electric wire
• Potentiometer is based on the principle that potential drop across any portion of th wire of uniform crossection is proportional to the length of that portion of thw wire when a constant current flows through the wire
• Figure below shows the construction of a potentiometer which consists of a number of segments of wire of uniform area of cross-section stretched on a wooden board between two copper strips .Meter scale is fixed parallel to the lenght of the wire
  
  
• A battery is connected across terminals A and B through a rehestat so that a constant currents flows through the wire
• Potentiometer is provided with a jockey J with the help of which contact can be made at any point on the wire
• Suppose A and ρ are the area of cross-section and resistivity of the material of the wire the resitance
  R=ρl/A ----------------------------(i)
  where l is the lenght of the wire
• If I is the current flowing through the wire then from Ohm's Law,
  V=IR ------------------------------(ii)
  Where V is the potential differene across the position of the wire of length l
  Thus ,from (i) and (ii)
  V=IR=I(ρl/A)=kl
  where K=ρI/A
  => V is proportional to l when current I is constant
• K=V/l is also known as potential gradient which is the fall of potential per unit length of wire
• Senstivity of a potentiometer depends on its potential gradient .If the potential gradient of a potentiometer is small then the potentiometer is more sensitive and hence more accurate

(A) Comparison of EMF's of two cells using potentiometer

• Consider the circuit arrangement of potentiometer given below used for comparison of emf's of two cells
  
  
• Positive terminals of two cells of emf's E₁ and E₂( whose emf are to be compared ) are connected to the terminals A and negative terminals are connected to jockey through a two way key K₂ and a galvanometer
• Now first key K₁ is closed to establish a potential difference between the terminals A and B then by closing key Ksub>2 introduce cell of EMF E₁ in the circuit and null point junction J₁ is dtermined with the help of jockey.If the null point on wire is at length
  l₁ from A then
  E₁=Kl₁
  Where K -> Potential gradient along the length of wire
• Similarly cell having emf E₂ is introduced in the circuit and again null point J₂ is determined .If length of this null point from
  A is l₂ then
  E₂=Kl₂
  Therefore
  E₁/E₂=l₁/l₂
  This simple relation allows us to find the ratio of E₁/E₂
• if the EMF of one cell is known then the EMF of other cell can be known easily

(B) Determination of internal resistance of the cell

• Potentiometer can also be used to determine the internal resistance of a cell
  
  
• For this a cell whose internal resistance is to be determined is connected to terminal A of the potentiometer across a resistance box through a key K₂
• First close the key K₁ and obtain the null point .Let l₁ be the length of this null point from terminal A then
  E=Kl₁
• When key K₂ is closed ,the cell sends current through resistance Box (R).If E₂ is the terminal
  Potential difference and null point is obtained at length l₂(AJ₂) then
  V=Kl₂
  Thus
  E/V=l₁/l₂
  But E=I(R+ r) and V=IR
  This gives
  E/V=(r+R)/R
  So (r+R)/R=l₁/l₂
  giving
  r=R(l₁/l₂-1)
• Using above equation we can find internal resistance of any given cell