Kirchhoff's Laws

KIRCHHOFF'S LAWS

Kirchhoff's Current Law (I Law)

Statement

The sum of the currents flowing towards a junction is equal to the sum of the currents flowing away from it.

In figure 1.15, A is a junction (or node) formed by six conductors. The currents in these conductors are I₁, I₂ , ... I₆. Some of these currents are flowing towards A and others flowing away from it.

According to Kirchhoff's Law,

I₁ + I₄ + I₅ + I₆ = I₂ + I₃ ……… 25)

(flowing towards A) = (flowing away from A)

Kirchhoff's Voltage Law (II Law)

Statement

In a closed circuit, the sum of the potential drops is equal to the sum of the potential rises.

In figure 1.16, ABCDA form a closed circuit. Assuming the current direction as shown, from A to B, we have a potential drop of IR₁ volts. Writing for the entire loop ABCDA, we have,

Sum of potential drops = IR₁ + IR₂ + IR₃

Potential rise from D to A = V

IR₁+ IR₂ + IR₃ = V

Sign of EMFs

Sign of Voltage Drops

 

EXAMPLE 56: The power supplied to the load R and the voltage across it in figure are 500 W and 100 V. Determine (i) the value of the V, (ii) the power dissipated in each resistor. Also confirm that the power delivered by the source equals the total power dissipated elsewhere.

Solution

The current and voltages are marked as shown in figure.

Power delivered by the source = Power dissipated in all resistors

 = 1485 W

 

EXAMPLE 57: In the circuit shown in figure, find the value of current through 100 Ω.

Solution

The current directions are marked as shown in the figure.

First, find the equivalent resistance of the circuit. vode add

40 and 100 Ω are connected in series, .. 40 + 100 = 140 Ω

150 

Ω and 140 Ω are connected in parallel.

120 Ω and 72.41 Ω are connected in series, 120+ 72.41 = 192.41 Ω

I₃ = 0.268 A

 

EXAMPLE 58: Find the voltage across the resistor 'R' in figure.

Solution

Applying KVL in the above circuit

20 - 8 + 10 - V₁- 17 + 5 = 0

10 - V₁ = 0

V1 = 10 V

Voltage across the resistor R = 10 V

 

EXAMPLE 59: Find the current through 12 2 resistor in figure.  (AU/Mech - Dec 2004) -

Solution:

Here, incoming current is 5+ 6 = 11 A

This 11 A current flows through the 12Ω resistor.

 

EXAMPLE 60: For the circuit shown in figure, determine the unknown voltage drop V₁

Solution:

Applying KVL in the above circuit

-2 – 1 - V₁ – 3 – 5 + 30 = 0

- V₁+ 19 = 0

- V₁ = -19

V_{1 =} 19 V

 

EXAMPLE 61: Calculate the currents supplied by the two batteries in the network shown in figure.

Solution:

Here, negative sign indicates, the current direction is anticlockwise. It is shown in figure.

I₃ = I₁ – I₂ = 5 - 1 = 4A

The current supplied by the battery 125 V is 5 A.

The 90 V battery is getting the charging current of 1A.

 

EXAMPLE 62: Determine the voltages V₁, V₂, V₃ and V₄ in the circuit in figure given below.

Solution

By applying KVL in loop ABEA

-V₁ - 12 + 8 = 0

-V₁ – 4 = 0

-V₁ = – 4 V

Consider another loop BCEB

V₂ – 6 + 12 = 0

V₂ + 6 = 0

V₂ = - 6V

Consider the loop CDEC

V₃ – 10 + 6 = 0

V₃ – 4 = 0

V₃ = 4 V

Consider the loop DAED

V₄ - 8 + 10 = 0

V₄ + 2 = 0

V₂ = -2 V

 

EXAMPLE 63: Calculate the current through 50 and 100 22 resistors of circuit shown in figure.

Solution

Current through 4 Ω resistor is

I₁ = 0.666 A

I₂ = I × 100 / 150 = 2 × 100 / 150

I₂ = 1.333 A

 

EXAMPLE 64: In the circuit of figure find using "Kirchhoff's Laws" the currents in the various elements. Find also the power delivered by the battery.

Solution

The currents are marked in the circuit shown. In marking the currents, Kirchhoff's current law is applied at all possible nodes. For instance, at node B, the current I₁ splits into I₂ through the 62 resistor and (11-12) through the 82 resistor. There are two unknown currents I₁ and I₂. Closed loop equations are written for two loops. Considering ABFGA,

 

EXAMPLE 65: In the circuit of figure, find the power supplied to the load. Also find the voltage at the load.

Solution :

The currents are marked using Kirchhoff's law. We write two closed loop equations using Kirchhoff's second law and then solve for the two unknowns.

For loop ABFGA

The negative sign for I₁ implies that current flows from B to A rather than from A to B. This means that the current in the load is 20.3125 A. (23.4375 -3.125)

Voltage across the load = 20.3125 × 1 = 20.3125 V

Load power = 1² × R = 20.3125² × 1 = 412.598 W

 

EXAMPLE 66: In the circuit of figure, it is given that the current in the branch OA is zero. Find the value of R and the current in it.

Solution :

We have marked the current in BA as I1 and BO as I2. Since there is no current in OA, I₁ flows through R (AC) and I₂ through OC. There are three unknowns R, I₁ and I₂. To solve these we need to write three closed loop equations.

For BOCB,

\R = 4 Ω

 

EXAMPLE 67: In the circuit of figure, A is a milliammeter of resistance 5 Q. Find the direction and magnitude of the current through it.

Solution:

The currents are marked as in figure. Note that Kirchhoffs voltage law is applied to loops. We write two loop equations and solve for the two unknowns I₁ and I₂.

For loop ACDA

 

EXAMPLE 68: In the circuit of figure, find the power dissipated in the individual resistors.

Solutios :

The currents are marked in figure. Since there is only one unknown, we write one equation for the closed loop ABCA.

 

EXAMPLE 69: In the Wheatstone bridge circuit of figure, G is a galvanometer of resistance 102 Find the current through it.

Solution :

The currents are marked as in figure. Kirchhoff's voltage law is applied to the various loops. There are three unknowns I₁, I₂ , I₃.

For loop ABDA

-3I₁ + 6I₂ -10I₃ = 0   …. (1)

For loop BDCB

By applying Cramer's rule we can can find the current 13 (Current through galvanometer).

Current through the galvanometer = 0.0809 A

EXAMPLE 70: In the circuit of figure, find the current through R_L.

Solution :

The currents are marked as shown in figure. Applying Kirchhoff's voltage law, loop equations are written for three closed circuits,

For loop ABFGA

Current through load resistor R_L = 10 A

EXAMPLE 71: What is the difference in potential between points X and Y in the circuit at is the a shown in figure.

Solution :

So The currents in the loops ABCX and DYGF are marked as shown. There can be no current in the branch CD as otherwise the currents coming out of the positive terminals of the batteries (at B and D) will not be the currents entering the negative terminals (at A and F)

For Loop ABCXA

Therefore, V_XY = 1.2 + 4 - 1.5 (since V_DY is a drop)

= 3.7 V (rise from X to Y)

 

EXAMPLE 72: Find the power dissipated in each resistor in the circuit shown in figure.

Solution:

The currents are marked in figure. Since there is only one unknown, we write one equation for the closed loop ABCA.