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

Sign Convention for Junction Rule

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.

Sign Convention for Loop Rule

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.



Kirchhoff's Law | Problem Based on Kirchhoff's Law Kirchhoff's Law | Problem Based on Kirchhoff's Law Reviewed by Being Morning Star on 18:00 Rating: 5

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