Kirchhoff ‘s Laws

To maintain a steady current in a conductor, a constant potential difference must be maintained across its ends. Cells or batteries are used to accomplish this task.

If one or more conductors are connected in such a manner that it offers one or more closed paths for current to flow, the arrangement is referred to as an electric circuit.

The analysis of such circuits is accomplished with the help of two laws known as Kirchoff’s laws :

Kirchhoff’s Current law (KCL) & Kirchhoff’s Voltage law (KVL):

KCL( Kirchhoff’s current Law ): Kirchhoff’s current law states that sum of the currents entering a junction point in a circuit must be equal to the sum of the current leaving it.

In other words algebraic sum of all the currents meeting at a junction is zero.

The word algebraic here means that we have to take into account the current direction in the summation, i.e.

if the currents reaching the junction are taken as positive, the current leaving it is to be taken as negative.

For the shown diagram, KCL gives us

I1 − (I2 ) + (−I3) = 0

or, I1 = I2 + I3

KCL is a direct consequence of the principle of conservation of electric charge.

KVL(Kirchhoff’s voltage law ) : Kirchhoff’s voltage law states that the algebraic sum of the potential differences in any loop including those associated with emf must be equal to zero.

Mathematically,  ΣΔv = 0

In order to get the correct value of potential drop for different type of elements, you may consider following facts.

(i) In a resistor current flows from high potential to low potential end.
Hence for the resistor shown below.

VA − VB = IR


(ii) If E & r be the EMF and internal resistance of the cell shown below
VB − VA = E – Ir


(iii) If you encounter a capacitor with charges on its plates as shown in the diagram

VA − VB = q/C where C is its capacitance and

I = dq/dt

Example: Determine the current in each branch of the following network : Fig.
Current Electricity

Sol. The current through the various arms of the circuit have been shown in Fig

According to Kirchhoff’s second law; In a closed circuit ABCEA ;

10i1 + 5(i1-i3) -10 + 10(i1+i2) = 0

25 i1 + 10 i2-5i3 = 10

5i1 + 2i2-i3 = 2 …(i)

In a closed circuit ABDA ;

10 i1 + 5i3-5i2 = 0

2 i1 + i3-i2 = 0

i2 = 2 i1 + i3 ….(ii)

In a closed circuit BCDB ;

5 (i1 -i3)-10 (i2 + i3 )-5i3=0

5i1 – 10i2-20i3 = 0

i1 = 2i2 + 4i3 …(iii)

On solving ;

i1 =-2 (-2⁄(17))=4⁄17 A ;

i2 =-3 (-2⁄17)= 6⁄17 A

i1+i2 =(4⁄17)+(6⁄17) = 10⁄17 A

i1-i3=4⁄17-(-2⁄17)= 6⁄17 A

i2+i3 = 6⁄(17+(-2⁄(17) = 4⁄17 A

Example : Two cells having emf of 10 V and 8V, and internal resistance of 1 Ω(each) are connected as shown with an external resistance of 8 Ω. Find the current flowing through the circuit.

Solution : Suppose that a current i flows through the external resistance (8 Ω) and it divides into two branches at the node B as i1 and i2

Using KCL , i1 + i2 = i

Using KVL,

For Loop ABYXA :

10 + (−i1)1 + (i − i1) 1 − 8 = 0

For Loop ABQPA :

10 + (−i1)1 + (−i)8 = 0

Solving, we get,

i = 18/17 A ≈ 1.06A

i1 = 10 − 8i

= (10 − 8.54)A

= 1.52 A.

i2 = i − i1

= − 0.46 A

The negative sign for i2 means that its direction is opposite to the direction that we had assumed.

Example : Calculate the value of current I in the section of networks shown in figure

Solution :

This type of problem checks your understanding of KCL

At junction A : I2 + 8 = 15 ; I2 = 7 A

At junction D : I3 + 5 = 8 ; I3  = 3 A

At junction B : I2 + 3 = I4 ; I4  = 10 A

At junction C : I1 = I3 + I4 ; I1 =13 A

Example : A part of a circuit in steady state along with current flowing in the branches with value of each resistance is shown in figure. Calculate the energy stored in the capacitor C0.

Solution : Find out potential difference across the capacitor first.

Applying Kirchoff’s junction law at A and B

At A : 2 + 1 − I1 = 0 ; I1 = 3A

At B: I2 + 1 − 2 = 0 ; I2 = 1A

Now applying KVL in the mesh ADCBA (mesh means smallest loop or closed path)

VAB  = 3 × 5 + 3 × 1 + 1 × 2 = 0

VAB = 20 V

So energy stored in the capacitor

$ \displaystyle U = \frac{1}{2}CV^2$

$ \displaystyle U = \frac{1}{2}(4\times 10^{-6})(20)^2$

= 8 × 10-4 J

Example: An electrical circuit is shown in figure. Calculate the potential difference across the resistor of 400 Ω, as will be measured by the voltmeter V of resistance 400Ω by applying Kirchhoff’s rule .

Electricity

Sol:

Applying Kirchhoff’s law in loop JMGDJ, we get

Electricity

-(I3/2) × 400 + (I1 + I2-I3 )200 + (I2-I3)100 = 0

or -500I1 + 200I2 + 300I3 = 0

or 2I1 + 3I2 -5I3 = 0 …(i)

Applying Kirchhoff’s law in CDEFBC , we get

-100I2 -100(I2 -I3 )+ 100I1 = 0

or I1 -2I2 + I3 = 0 …(ii)

Multiplying Eq. (ii) by 2 and subtracting from Eq. (i), we have

2I1 + 3I2-5I3-2I1 + 4I2 -2I3 = 0

or I2 =I3 …(iii)

Applying Kirchhoff’s law in ABFEGHA, we get

-3I1 -2I2 + 2I3 + 0.1 = 0 …(iv)

Multiplying Eqs.(ii) by 3 and adding it to Eqs. (iv), we get

3I1 -6I2 + 3I3 -3I1 -2I2 + 2I3 + 0.1 = 0

or -8I2 + 5I3 + 0.1=0

or -8I3 + 5I3 + 0.1 = 0

or 3I3 = 0.1

or I3 = 0.1/3

Therefore, potential difference across JM is

(0.1/3)/2 × 400= 0.1/(2 ×3)×400

= 20/3 = 6.67V

Also Read :

→ Electric Current , Drift velocity & Deduction of Ohm’s Law
→ Combination of Resistors
→ Grouping of Cells
→Concept of EMF & Terminal P.D
→Wheatstone Bridge Principle & Metre Bridge
→Ammeter and Voltmeter
→Potentiometer & its Applications
→Sensitivity of Potentiometer
→Charging of capacitors
→Discharging of Capacitors
→ Heating effect of currents

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