Heating Effect Of Current




Cause of Heating Effect of Current 
Consider a purely resistive circuit i.e. a ciruit which consist of only some resistors and a source of emf. The energy of the source gets dissipated entirely in the form of heat produced in the resistors. The phenomenon of the production of heat in a resistor by the flow of an electric current throught it is called heating effect of electric current or Joule heating.
When a potential difference is applied across the ends of a conductor, its free electron gets accelerated in the opposite direction of the applied field. But the speed of the electron does not increase beyond a constant drift speed. This is beacause  during the course of their motion, the electron collide frequently with the positive metal ions. The kinetic energy gained by the electron during the intervals of free acceleration between collision is transferred to the metal ions at time of collision. The metal ions begin to vibrate about their mean position more and more violently. The average kinetic energy of the ions increases. This increases the temperature of the conductor. Thus the conductor gets heated due to the flow of current. Obviously, the electrical energy supplied by the source of emf converted into heat
Heat Produced by Electric Current :Joule's Law
Heat produced in a resistor. Consider a conductor AB of resistance R as shown in fig. A source of emf maintains a potential difference V between its ends. A and B and sends a steady current I from A to B. Clearly, Va> Vb and the potential difference across AB is V = Va-Vb> 0

Cause of Heating Effect of Current

The amount of charge that flows from A to B in time t is 
          q = It
As the charge q moves through a decrease of potential of magnitude V, its potential decreases by tha amount ,
          U = Final P.E. at B - Initial P.E at A
qVb - qVa = -q(Va -Vb) = qV<0
If the charges move through the conductor without suffering collisions, their kinetic energy would change so that the total energy is unchanged. By conservation of energy, the change in kinetic energy must be
K = -U = qV = It × V = VIt> 0 
Thus, in case, charges were moving freely throughb the conductor under the action of the electric field, the kinetic energy would increase as they move. However, we know that on the average, the charge carriers or electrons do not move with any acceleration but with a steady drift velocity. This is because of the collision of electrons with ions and atoms during the course of their motion. The kinetic energy gained by the electron is shared with the metal ions. These ions vibrate more vigorously and the conductor gets heated up. The amount of energy dissipated as heat in conductor in time t is
H = VIt joule = I^2Rt joule = V^2t/R joule
Or
H = VIt/4.18 cal = I^2Rt/4.18 cal = V^2t/4.18 R cal
According to this law, heat produced in the resistor is: 
 1.  Directly proportional tobthe square of current for a given R
 2. Directlt proportional to the resistance R for a given I
 3. Inversaly proportional to the resistance R for a given V
 4. Directly proportional to the time for which the current flows through the resistor.
Reference links :

Potentiometer: Construction | Principal | Application | Sensitivity

An Ideal Voltmeter which does not change the original potential difference,needs to have infinite resistance. But a Voltmeter cannot be designed to have an infinite resistance. Potentiometer is one such device which does not draw any current from the circuiy and still measure the potential difference. So it acts as an ideal voltmeter.
Construction of Potentiometer



Potentiometer consist of a long wire,usually 10 m long, of material which have high resistivity and low temperature cofficient such us constantan. Usually, 1 m long separate pieces of the wires are are fixed on a wooden board which is parallel to each other. The wires are joined in series by thick copper strips. A meter scale is fixed parallel to the wires. The end A and B are coonected to a strong battery,a plug key K and a rheostat Rh. This circuit, called driving or auxiliary circuit , sends a constant current I through wire AB. Thus, the potential gradually falls from A to B. A jockey can slide along the length of the wire.
Principal of Potentiometer
The principle of the potentiometer is that when a constant current flows from a wire of uniform cross section area and composition, the potential drop across any length of the wire is directly proportional to that length.a
Potential Gradient - The potential drop per unit length of the potentiometer wires is is known as potential gradient.
    K = V/L
S.I unit -  v/m
Application of Potentiometer

 1. Comparison of emfs of two primary cell

Fig. shows the circuit diagram for the comparing the emfs of the two cells. A constant current is maintained in the potentiometer wire AB by means of a battery of emf E through a key K and rheostat Rh. Let E1 and E2 be the emfs of the two primary cells which are to be compared. The positive terminals of these cells are connected to the end  A of the potntiometer wire and their negative terminals are connected to a high resistance box R.B. ,  a galvanometer G and a jockey J through a two way key. A high R is inserted in the circuit from resistance  box R.B to prevent excessive current flowing through the galvanometer.
As a plug is inserted between a and c , the celk E1 gets introduced in the circuit. The jockey j is moved along the wire  AB till the galvanometer shows no deflection. Let the position of the jockey be j1 and length of the wire Aj1 =  l1. If k is the potential gradient along the wire AB , then the null point,
E1 = Kl1
By inserting the plug between b and c , the null point is again obtained for cell  E2. Let the balancing length be Aj2 = l2. Then
E2 = kl2
:: E2/E1 = l2/l1
If one of the two cells is a standard cell of known emf , then emf of the other cell can be determined. E2 = l2/l1 * E1
In order to get the null point on the potentiometer wire, it is necessary that the emf, E of the auxiliary battery must be greater than both E1 and E2.
 2. Internal resistance of a primary cell by a potentiometer


As shown in fig., the positive terminal of the cell pf emf E whose internal resistance r is to be measured is connected to the end A of the potentiometer wire and its negative terminal to a galvanometer G and jockey j. A resistance box R.B. is connected across the cell through a key k2.
Close the key k1. A constant current flows through the potentiometer wire. With key k2 kept open, move the jockey along AB till it balance the emf E of the cell. Let l1 be the balancing length of the wire. If k is the potential gradient, then emf of the cell be: E=kl1
With the help of resistance box R.B., introduce a resistance R and close key k2. Find the balance point for the terminal potential difference V of the call. If l2 is the balancing length, then 
V= kl2
E/V = l1/l2
Let r be the internal resistance of the cell. If current I flows through cell when it is shunted with resistance R, then from ohm's law we get
E = I(R + r) 
And V= IR
E/V = R + r /R = l1/l2
r/R = l1-l2/l2
:: internal resistance =  r =R [ l1-l2/l2]
Superiority of a potentiometer to a voltmeter.
Potentiometer is a null method device. Al null point , it does not draw any current from the cell and thus there is no potential drop due to the internal resistance of the cell. It measures the p.d. in an open circuit which is equal to the actual emf of the cell. On the other hand , potentiometer draws a small current from the cell for its operation. So it measures the terminal p.d. in a closed circuit which is less than the emf of the cell . That is why potentiometer is preferred over a voltmeter for measuring the emf of a cell.
 Potentiometer  Sensitivity
 1.  If is sensitive if it is capable of measuring very small potential differences 
 2. If it shows a significant change in balancing length for a small change difference being measured.
 Sensitivity of a potentiometer can be increased in two ways:
 1. For a given potential difference , the sensitivity can be increased by increasing the length of a potentiometer wire.
 2. For a potentiometer wire of fixed length, the potential gradient can be decreased by reducing the current in the circuit with the help of a rheostat.
Reference link:  Kirchhoff's Law | Problem Based on Kirchhoff's Law

In 1942,a german physicist extended ohm's law to complicated circuit and gave two law,which is enable us to determine current in any parts of such a circuit.

Some terms related to law-

Electric Network- The term electric network is used for a complicated system of electrical conductors.

Junction-Any point in an electrical circuit where two or more than two conductors are joined together in a junction.

Loop or Mesh- Any closed conducting path in an electrical network is called a loop or mesh.

Branch-A branch is any part of the network that lies between two junctions.

Kirchhoff's First Law or Junction Rule or Kirchhoff's Law of Current.

In an electric circuit,the algebraic sum of current at any junction is zero.Or the sum of currents entering a junction us equal to the sum of current leaving that junction.

Mathematically, €I=0

1.The current flowing towards the junction are taken as positive.

2.The currents flowing away from the junction are taken as negative.

€ I=0

::I1 +I2- I3 -I4=0

::Incoming current=Outgoing current.

First law is also known as-Kirchhoff's Current law(KCL).

This law is based on the law of conservation of charge. When current in a circuit are steady,charge cannot accumulate or originate at any point of the circuit. So,whatever charge flows towards the junction in any time interval,an equal charge must flow away from that junction in the same time interval.

Kirchhoff's Second Law or Loop Rule or Kirchhoff's Law of Voltage.

Around any closed loop of a network,the algebraic sum of changes in potential must be zero.Or,the algebraic sum of the emfs in any loop of a circuit is equal to the sum of the products of current and resistance in it.

Mathematically,  €delta V=0 or € EMF= €IR.

1.we can take any direction as the direction of transversal.

2.The emf of cell is taken as positive if the transversal is from its negative to the positive terminal.

3. The emf of a cell is taken as negative if the direction of transversal is from its positive to the negative terminal.

4. The Current-resistance product is taken as positive if the resistor is transversed in the same direction of assumed current.

5. The IR product is taken as negative if the resistors is transversed in the opposite direction of assumed current.

In fig.

Algebraic sum of current resistance product=I1R1-I2R2

Algebraic sum of emfs=E1-E2

-Applying K. Law to closed path ABCFA,we get 

E1-E2=I1R1-I2R2

-Similarly applying K. Second rule to mesh CDEFC,we get

E2=I2R2+(I1+I2)R3

Second rule is also called -Kirchhoff's Voltage Law(KVL).

This law is based on the law of conservation of energy. As the electrostatic force is a conservative force,so the work done by it along any closed path must be zero.

Problem Based on Kirchhoff's Law.

1.Two cells of emfs 1.5v and 2.0v and internal resistance 1ohm and 2 ohm respectively are connected in parallel so as to send current in the same direction through an external resistance of 5 ohm.

Using Kirchhoff's Law,calculate

(a) current through each branch of the circuit.

(b) p.d. across the 5 ohm resistance.

Solution..

Let I1 and I2 be the current as shown in fig. Using kirchhoff's law for the loop AFCBA,we get

2 I2 -1 I1 = E1-E2 = 2-1.5

2 I2-I1 = 0.5.....(1)

For the loop CFEDC,we have

1 I1 +5(I1+I2) = E1-1.5

5 I2+ 6 I1 = 1.5.......(2)

Solving (1) & (2)...

I1 = 1/34 A  &   I2 = 9/34A

Current through branch BA,

        I1= 1/34 A

Current through branch CF

I2 = 9/34 A

Current through branch DE

         I1 + I2 = 10/34

(b) P.D. across 5 ohm resistance

      =( I1 +I2) *5 = 1.47 V.