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Karnataka Board PUCPUC Science 2nd PUC Class 12

Kirchhoff’s Laws

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Estimated time: 28 minutes
Maharashtra State Board: Class 11

Definition: Voltmeter

An instrument used to measure the potential difference between two points in an electrical circuit, always connected in parallel with the component across which the voltage drop is to be measured, is called a voltmeter.

CBSE: Class 12

Introduction

Kirchhoff's laws are used to solve complicated electric circuits in which simple series and parallel combination rules are not sufficient. They help in finding unknown current and potential difference in circuits containing many branches and loops. 

These laws are based on two basic physical principles: conservation of charge and conservation of energy. 

CBSE: Class 12

History/Origin

Kirchhoff's laws were given by Gustav Robert Kirchhoff (1824-1887), a German physicist. He made important contributions to spectroscopy and mathematical physics, and his circuit laws are widely used in electrical network analysis. 

Maharashtra State Board: Class 11

Law: Kirchhoff's Current Law (KCL) - Junction Rule

At any junction, the sum of currents entering = the sum of currents leaving.

\[\sum_{i=1}^nI_i=0\]

Example: I1 + I3 = I2 + I4​. Based on conservation of charge.

Maharashtra State Board: Class 11

Law: Kirchhoff's Voltage Law (KVL) - Loop Rule

The algebraic sum of potential differences in a closed loop is zero.

∑IR + ∑E = 0  OR  ∑E = ∑IR

Based on conservation of energy.

CBSE: Class 12

Key Terms

  • Junction / Node: A point where three or more conductors meet. 
  • Branch: The part of a circuit between two junctions. 
  • Loop: A closed conducting path in a circuit. 
  • Current: Rate of flow of charge through a conductor. 
  • Potential Difference: Work done per unit charge between two points. 
  • emf: Energy supplied per unit charge by a cell or battery. 
CBSE: Class 12

Kirchhoff’s First Law

Statement

At any junction in an electric circuit, the sum of currents entering the junction is equal to the sum of currents leaving the junction. 

Derivation

When the current in a circuit is steady, charge does not accumulate at any junction. Therefore, the amount of charge entering the junction per second must be equal to the amount of charge leaving the junction per second. 

If currents I1​ and I2 enter a junction and currents I3​ and I4​ leave it, then

I1 + I2 = I3 + I4

or

I1 + I2 − I3 − I4 = 0

Hence,

∑I = 0

Conclusion

Kirchhoff's First Law is a direct consequence of the conservation of charge. 

CBSE: Class 12

Kirchhoff’s Second Law

Statement

In any closed loop of an electric circuit, the algebraic sum of all changes in potential is zero. 

Derivation

Consider a charge moving around a closed loop. After completing one full loop, the charge returns to its starting point. Since electric potential depends only on position, the net change in potential over a complete loop must be zero. 

Therefore, in a closed loop,

∑V = 0

If a loop contains cells and resistors, then the total emf supplied by the sources is equal to the total potential drop across the resistors. Thus,

∑E = ∑IR

Conclusion

Kirchhoff's Second Law is a direct consequence of the conservation of energy.

CBSE: Class 12

Sign Conventions

For Kirchhoff's First Law

  • Current entering the junction is usually taken as positive. 
  • Current leaving the junction is usually taken as negative. 
  • Any other convention may also be used, but it must remain consistent. 

For Kirchhoff's Second Law

  • While moving in the direction of current through a resistor, potential decreases; hence the term is taken as negative. 
  • While moving opposite to the direction of current through a resistor, potential increases; hence the term is taken as positive. 
  • Moving from the negative terminal to the positive terminal of a cell gives a positive emf. 
  • Moving from the positive terminal to the negative terminal of a cell gives a negative emf. 
CBSE: Class 12

Example 1

Question:
Several currents meet at point P. Find the unknown current I using KCL.

Step‑by‑step explanation

1. At point P, some arrows point towards P and some away from P.

2. By convention:

  • Currents towards P are taken as positive.
  • Currents away from P are taken as negative.

3. Write KCL at P: “Algebraic sum of currents at a junction is zero”.

4. According to the source, the equation is: 0.2 − 0.4 + 0.6 − 0.5 + 0.7 − I = 0.

  • 0.2 A, 0.6 A, and 0.7 A are taken as positive (towards P).
  • 0.4 A, 0.5 A, and II are taken as negative (away from P).

5. Now group the numbers:

  • 0.2 + 0.6 + 0.7 = 1.5 A (all positive terms).
  • 0.4 + 0.5 = 0.9 A (negative terms without I).

6. So the equation becomes: 1.5 − 0.9 − I = 0.

7. Simplify: 1.5 − 0.9 = 0.6, so 0.6 − I = 0.

8. Therefore I = 0.6 A.

Final idea:
At any junction, just add the currents coming in and subtract the currents going out; the total must be zero, which lets you find the unknown current.

CBSE: Class 12

Example 2

Question:
There is a network with a 9 V battery and several resistors (including a 1 Ω resistor). You must find the current in the 1 Ω resistor using KCL and KVL.

Step‑by‑step explanation

1. Let the current from the 9 V battery be I1.

2. At junction E, this current splits: one branch has current I2, the other branch has current I1 − I2.

  • This comes from KCL: entering current I1 equals outgoing currents I2​ and I1 − I2​.

3. Loop EFCBE (contains the 1 Ω resistor and the 9 V battery):

  • Move around the loop following a chosen direction.
  • Add all potential drops and rises.
  • According to the source, the equation becomes:
    1I2 + 3I1 + 2I1 = 9.
  • Combine like terms: 5I1 + I2 = 9. Call this equation (1).

4. Loop EADFE (the other loop):

  • This loop includes the branch with current I1 − I2​ and another resistor.
  • Applying KVL gives: 3(I1 − I2) − I2 = 6.
  • Expand and simplify:
    3I1 − 3I2 − I2 = 6.
    So, 3I1 − 4I2 = 6. Call this equation (2).

5. Solve the two equations:

  • From (1): 5I1 + I2 = 9.
  • From (2): 3I1 − 4I2 = 6.

Solve simultaneously (the usual method of linear equations). The source gives:

  • I1 = 1.83 A
  • I2 = −0.13 A

6. Interpretation of signs:

  • I1​ is positive: assumed direction is correct.
  • I2 is negative: actual current in the 1 Ω branch flows opposite to the initially assumed direction (from F to E instead of E to F).

Key idea:
Use KCL to write current relations at junctions, use KVL to get loop equations, solve the simultaneous equations, and then use the sign of each answer to check direction.

CBSE: Class 12

Example 3

Question:
In a given network with several resistors and two batteries, find the currents I1, I2, and I3 in the three main branches using Kirchhoff’s rules.

Step‑by‑step explanation

1. Assign currents:

  • Draw arrows in each branch and name currents I1, I2, and I3. These are unknowns to be solved.

2. Use KCL at junctions:

  • Apply Kirchhoff’s first rule at each junction to express some branch currents in terms of I1, I2, and I3.
  • After this step, only three independent unknown currents remain: I1, I2, and I3.

3. Apply KVL to loop ADCA:

  • Move around the loop ADCA, adding emf and subtracting drops IR.
  • 10 − 4(I1 − I2) + 2(I2 + I3 − I1) − I1 = 0.
  • Simplify to get:
    7I1 − 6I2 − 2I3 = 10. Call this equation (a).

4. Apply KVL to loop ABCA:

  • Around loop ABCA, similarly, add the emf and the resistor drops.
  • Equation: 10 − 4I2 − 2(I2 + I3) − I1 = 0
  • Simplify to: I1 + 6I2 + 2I3 = 10. Call this equation (b).

5. Apply KVL to loop BCDEB:

  • Around loop BCDEB:
  • Equation: 5 − 2(I2 + I3) −2(I2 + I3 − I1) = 0.
  • Simplify to: 2I1 − 4I2 − 4I3 = −5. Call this equation (c).

6. Now there are three simultaneous linear equations (a), (b), (c) in three unknowns I1, I2, I3​.

7. Solve using algebra (elimination/substitution):

  • I1 = 2.5 A
  • I2 = 5/8 A
  • I3 = 17/8 A

8. Substitute these values back to obtain currents in each branch (AB, CA, DEB, AD, CD, BC) as listed in the text.

9. As a check, they verify KVL in another loop (BADEB) – the sum of voltage changes is indeed zero.

Key idea:
In general networks, KCL reduces unknowns, KVL on enough independent loops gives as many equations as unknowns, and solving them gives all branch currents.

CBSE: Class 12

Real-Life Application

  • Kirchhoff's laws are used in analysing household and laboratory electric circuits. 
  • They help in finding unknown currents and voltages in complex electrical networks. 
  • These laws are also useful in understanding energy transfer in electric circuits. 

Simple Analogy

  • KCL is like water flowing at a pipe junction: total water entering equals total water leaving. 
  • KVL is like going around a hill path and returning to the same point: the net change in height is zero. 
CBSE: Class 12

Key Points

  • Kirchhoff's laws are used for complex circuits. 
  • Kirchhoff's First Law: Total current entering a junction = total current leaving a junction. 
  • Kirchhoff's Second Law: Total potential rise in a closed loop = total potential drop in the loop. 
  • KCL is based on conservation of charge. 
  • KVL is based on conservation of energy. 
  • Mathematical forms are ∑I = 0 and ∑V = 0. 
  • The correct sign convention is essential in numericals. 

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