# Kirchhoff’s Current & Voltage Law with Solved Examples

Table of Contents

## What are Kirchhoff’s Laws?

**Kirchhoff’s Current & Voltage Law with Solved Examples-** A German physician Robert Kirchhoff introduced two laws in 1847 through which the determining of an equivalent resistance of any complex network and the current flowing through various conductors became possible. Both DC as well as AC types of circuits can be solved through the application of these laws. There are two types of the laws. Kirchhoff’s first law is known as Kirchhoff’s current law (KCL) and the second law is Kirchhoff’s voltage law (KVL).

**Kirchhoff’s ****First Law or Kirchhoff’s Current Law (KCL)**

(I). Under this law, the algebraic sum of entire currents flowing through a point (or junction) of any network, tends to be zero. This law is also known as point law or Kirchhoff’s current law(KCL).

(II). At any moment, all currents flowing towards any point equal to the entire outgoing currents.

(III). According to this law, the algebraic sum of all incoming and outgoing currents on any point of an electrical network, always tend to be zero.

(In any network the algebraic sum of the currents meeting at a point (or junction) is zero. In other words, the sum of currents flowing towards a point is equal to the sum of those flowing away from it. Or, the algebraic sum of the currents entering a node equals the algebraic sum of the currents leaving it)

Let us suppose that some conductors meet on point A, as has been illustrated in the figure 2.1 (a). Current from some of the conductors meeting on point A flows towards point A, whereas current of some of the conductors flows away from point A. Suppose, that the currents entering point A are positive, and the currents leaving currents are negative. Then,

I_{1 }+ (-1_{2}) + (-I_{3}) + (+I_{4}) + (-I_{5}) = 0

Or I_{1} + I_{4} – I_{2} -I_{3} – I5 = 0

Or I_{1} + I_{4} = I_{2} + I_{3} + I5

Incoming currents = outgoing currents

This law can also be mentioned according to the following algebraic method;

Σ1 _{Entering} = Σ1_{Leaving }

Σ1 = 0

For example, 6 ampere current is coming towards a point, and 2 amperes and 4 ampere currents are leaving outward, as has been illustrated vide figure 2.1 (b). Thus,

Σ1 _{Entering} = Σ1_{Leaving }

6A = 2A + 4A

6A = 6A

Figure 2.1 – Demonstrating Kirchhoff’s current law

**Kirchhoff’s ****Second Law or Kirchhoff’s Voltage Law(KVL)**

(I). The algebraic sum of IR products in any closed path (or circuit) in a network, is equal to the e.m.f in that path. Remember that closed path is also known as a “mesh”, i.e.,.

ΣIR = ΣE

(II). In any closed loop the algebraic sum of the emfs applied is equal to the algebraic sum of the potential voltage drops in the elements.

(III). The sum of voltage drops in any closed circuit equals to the emf provided on this circuit. This law is also known as mesh law or Kirchhoff’s voltage law(KVL).

If we multiply each conductor’s resistance with the current flowing through it by turning around each mesh or circuit, then sum of such a product will be equal to the sum of all emfs installed on this circuit. If we move along the direction of current, we take IR product and emf are taken positive. And when we move along the reciprocal direction of the current, product will always be taken as negative.

Moreover, if we move from negative to a positive terminal of the battery, then positive direction, whereas if we move from positive terminal to a negative terminal, then direction will always be taken as negative.

Remember that under Kirchhoff’s laws, the determination of direction of a current is very essential for solving questions. The current direction can either be taken as clock – wise or anti – clockwise. Once the direction has been set, that direction is retained throughout the solution of the entire question. If the answer is positive, it will mean that our chosen direction is correct, and if the answer is negative, then current’s direction is considered opposite to the selected direction.

Figure 2.2 – Kirchhoff’s voltage (second) law

In figure 2.2, a close network has been illustrated, which consists of two batteries E_{1} and E_{2}. The e.m.f sum of batteries has been shown as E_{1} – E_{2}. A hypothetical direction of current’s flow has also been illustrated in the figure. E1 operates current in the presumed direction, thus it has been considered as positive. Whereas E2 causes hindrances in the flow of current (that’s opposes the flow of current in the imaginary direction) that’s why it has been taken negative. The voltage drop caused in this closed circuit depends on the product of currents and resistors. The voltage drops occurring in the presumed direction of current is taken positive, and voltage drop occurring in the opposite direction, is taken as negative. In this figure, I_{1} R_{1} and I_{2} R_{2} are positive, whereas I_{3 }R_{3} and I_{4} R_{4} are negative.

### Active Circuit

A circuit that contains one or more than one e.m.f sources, is known as an active circuit.

### Passive Circuit

A circuit which does not have any e.m.f source, is known as passive circuit.

### Node

A junction (or any point) existing on a circuit, where two or more than two circuit elements (e.g., resistors, capacitors, and inductors etc.) connect together, is called node.

### Branch

That part of any network, which is located between two junctions, is known as branch. A branch may contain one or more than one elements connected in a series. And it has two terminals.

### Loop

A closed path found in any circuit, which contains more than two meshes, is known as a loop. That’s there are several meshes in a loop, however a mesh does not consist of a loop.

### Mesh

It is a type of loop, having no other loop inside it, or a path which does not consist of any other paths, is known as a mesh.

## Kirchhoff’s Laws Solved Examples

**Kirchhoff’s law (KVL) Example 1;** Resistors of 4 ohms, 10 ohms and 8 ohms are connected up to two batteries (of negligible resistance) as shown. Find the current through each resistor.

**Solution;**

Assume currents to flow in directions indicated by arrows.

Fig. 2.3

If current in mesh A B C = i_{1 }and in C A = i_{2}, then current in mesh C D A = i_{1} – i_{2}

In mesh A B C, 20 volts are acting in a clockwise direction. Equating the sum of the I R products, we get;

10 i_{1} + 4 i_{2} = 20 … (1)

In mesh A C D, 12 volts are acting in clockwise direction; then,

8 (i_{1} – i_{2}) 0 4i_{2} = 1_{2}

8i_{1 }– 12i_{2} = 12 … (2)

Multiply (1) by 3

30i_{1 }+ 12i_{2} = 60

38i_{1} = 72

i_{1} = 72/ 38 = 36 / 19 = 1.895 amperes = current in 10 ohms resistor

Substituting this value in (1), we get;

10 [36/19] + 4i_{2} = 20 or 360/ 19 + 4i_{2} = 20

4i_{2 }= 380 – 360 / 19 = 20 /19

i_{2} = 5 / 19 = 0.263 amperes = current in 4 ohms resistor

i_{1} – i_{2} = 36 / 19 – 5 / 19 = 31 / 19 = 1.63 amperes = current in 8 ohms resistor

**Kirchhoff’s law (KVL) Example 2;** Two batteries A and B are connected to the circuit shown. Battery A, 50 cells, Battery B, 40 cells, E.M.F e volts/cell. Internal resistance, 0.01 ohm/ cell. Find the current flowing in each battery and in the 0.8-ohm resistor.

Fig. 2.4

Solution;

Battery A, E.M.F = 100 volts

Internal resistance = 0.5 ohm

Battery B, E.M.F = 80 volts

Internal resistance = 0.4 ohm

Method 1;

Let i_{1} = Current output for battery A

i_{2} = Current output for battery B

Then, i_{1} + i_{2} = Current flowing in 0.8-ohm resistor

Fig. 2.5

Equate the sums of I R products in the mesh A B and in the mesh B C with the e.m. fs of their respective meshes.

In the mesh A B, resultant e.m.f = (100 – 80) = 20 volts in an anti-clockwise direction.

0.5 i_{1} – 0.6i_{2} = 20 … (1)

In the mesh B C, e.m.f = 80 volts in an anti-clockwise direction.

0.6i_{2} + 0.8 (i_{1} + i_{2}) = 80

0.8 i_{1} + 1.4 i_{2 }= 80 … (2)

Multiply (1) by 7, 3.5i_{1} – 4.2i_{2} = 140

Multiply (2) by 3, 2.4i_{1}+ 4.2i_{2} = 240

5.9i1 – 380

i_{1}= 380 / 5.9 = 64.4 amperes

Substituting this value in (1), 0.5 (64.4) – 0.6i_{2} = 20 or 32.2 – 0.6 i2 = 20

0.6i_{2} = 32.2 – 20 or 0.6i_{2} = 12.2 and i_{2} = 12.2 / 0.6 = 20.33 amperes

The value of i_{1} and i_{2} are positive, indicating that both batteries are discharging

Total current I = (i_{1} – i_{2}) = 64.4 + 20.33 = 84.73 amperes

Battery A – 64.4 amperes discharge

Battery B – 20.33 amperes discharge

Current in 0.8-ohm resistor, = 84.73 amperes

**Kirchhoff’s law (KCL) and KVL Example 3;** In the Wheatstone bridge network shown, find the resistor between the points A and D and the magnitude and direction of the current flowing in each resistor. E.M.F of battery 4 volts, internal resistance negligible.

Figure 2.6

**Solution;**

Assume currents to flow in the direction indicated by the arrows. If current in AB = i_{1}, in AC = i_{2}, and in BC = i_{3}, the current in

BD = i_{1} – i_{3},

CD = i_{2 }+ i_{3} (law 1)

By Kirrchoff’s voltage law

In mesh ABC, i_{1} +5i_{3} + 4i_{2} = 0 … (1)

In mesh BDC, 2 (i_{1} – i_{3}) – 3 (i_{2} – i_{3}) – 5i_{3} = 0

Or, 2i_{1} – 3i_{2} – 10i_{3} = 0 … (2)

In mesh FACD, 4i_{2} + 3 (i_{2} + i_{3}) + 1 (i_{1} + i_{2}) = 4

i1 + 8i_{2} + 3i_{3} = 4 … (3)

Multiply (1) by 2 2i_{1} – 8i_{2} + 10 i_{3} = 0

Adding this to (2) 4i_{1 }– 11i_{2} = 0

Multiply (1) by 3 3i_{1} – 12i_{2} + 15i_{3} = 0

Multiply (3) by 5 5i_{1} +40i_{2 }+15i_{3} = 20

2i_{1} + 52i_{2} = 20

2i_{1} + 52i_{1} = 20 multiply by 2, 4i_{1} + 104i_{2} = 40

4i_{1} – 11i_{2 }= 0 −4i_{1} 11i_{2} = 0

115i_{2} = 40

i_{2} = 40 / 115 = 8 / 23 = 0.348 = 0.348 amperes

4i_{1} – 11 x 8/23 = 0 or 4i_{1} = 88/ 23, i_{1} = 22/23 = 0.957 amperes

Substituting in (1) 22/ 23 + 5i_{3} – 32/23 = 0; 5i_{3} = 10 / 23; i_{3} = 2/23 = 0.087

Current in Branch A B = i_{1} = 22 / 23 = 0.957 amperes, A to B

Current in Branch A C = i_{2} = 8 / 23 = 0.348 amperes, A to C

Current in Branch B C =i_{3} = 2/23 = 0.087 amperes, B to C

Current in Branch B D = i_{1} – i3 = 22/ 23 – 2 / 23 = 20 / 23, = 0.87 ampere, B to D

Current in Branch C D= i_{2} + i_{3} = 8/ 23 + 2/ 23 = 10/ 23 = 0.435 ampere, C to D

Total current = i_{1} + i_{2} = 22 /23 + 8/ 23 = 30 / 23 = 1.305 amperes

Resistance _{D} = _{A }p. d. _{D} / Total current = _{A }p. d. _{D }+ _{C }p. d. _{D }/ Total current

= [8 / 23 x 4 + 10 / 23 x 3] ÷ 30/ 23 = 62/23 ÷ 30/23 = 62/ 30= 2.067 ohms

**Kirchhoff’s law (KVL) Example 4;** In the circuit shown, find the magnitude and direction of the current in each resistor. Internal resistance of cells negligible.

Figure 2.7

**Solution;**

Let i_{1} and i_{2} be the currents in both section flowing in clockwise direction, i.e., current in 4 Ω is i_{1}, in 5 Ω is i_{2}, and in 10 Ω is (i_{1} – i_{2}) flowing downwards

Applying Kirchhoff’s voltage law to the circuit ABDA

4i_{1} + 10 (i_{1} – i_{2}) = 4

Or 14i_{1} – 10 i_{2} = 4 … (i)

In circuit BCDB

−10 (i_{1} – i_{2}) + 5i_{2} + 2 = 0

Or – 10i_{1} + 15i_{2} = −2 … (ii)

Multiplying (i) by 3 42i_{1} – 30i_{2} =12

Multiplying (ii) by 2 −20i_{1} + 30i_{2} = −4

22i_{1} = 8; i_{1} = 8/22 = 4 / 11 Amp.

Put value of i_{1} in equation (i)

14 [4 /11] – 10i_{2} = 4 or 56 / 11 -10i_{2} = 4

And -10i_{2} =4- 56/ 11 = 44 – 56/ 11 = – 12/ 11

Or 10i_{2} = 12/ 11 and i_{2} = 12/ 11 x 10 = 12 / 110 = 6 / 55 amp.

i_{1} – i_{2} = 4/ 11 – 6 / 55 = 20 – 6 / 55 = 14/ 55 amp.

So, currents in 4 Ω resistor = 4/ 11 = 0.364 amp. Flowing A to B

In 5 Ω resistor = 6/ 55 = 0.109 amp flowing B to C

In 5 Ω resistor = 14/ 55 = 0.255 amp. Flowing B to D Ans.

**Kirchhoff’s law (KVL) and (KCL) Example 5;** Two batteries A and B are connected to the circuit as shown. Battery A 25 cells, Battery b 20 cells, emf. Per cell 2 volt, the internal resistance is 0.01 Ω/cell. Find the current flowing in each battery, and in the 5 Ω resistor.

Figure 2.8

**Solution;**

The voltage of battery A, being 25 cells of 2 volt each = 25 x 2 = 50 volts

The internal resistance of battery, 0.01 Ω/cell = 25 x 0.01= 0.25 Ω

Similarly, battery B will be of 20 x 2 = 40 V having 20 x 0.01 = 0.2 Ω internal resistance

Here, in a closed circuit RSTUVR, we have according to Kirchhoff’s law

0.25i_{1} + 5 (i_{1} +i_{2}) = 50 or 0.25i_{1} + 5i_{1} + 5i_{2} = 50

Or 5i_{2} + 5.25i_{1} = 50 …(i)

In other closed circuit VTUV

0.2i_{2} + 5 (i_{1} + i_{2}) + 2i_{2} = 40 or 0.2i_{2} + 5i_{1} + 2i_{2} = 40

Or 7.2i_{2} + 5i_{1 }= 40 … (ii)

Now solving both equations (i) and (ii) for i1 and i2

Multiply (i) by 7.2 and multiply (ii) by 5, we get (after subtraction)

36i_{2} + 37.8i_{1} = 360

− 36i_{2} 25i_{1} = −200

12.81i = 160 and i_{1} = 160 / 12.8 = 12.8 amp

Put value of i_{1} in equation (i) we get

5i_{2} + 5.25 (12.5) = 50 or 5i_{2} + 65.625 = 50

5i_{2} = 50 – 65.625 or 5i_{2} = − 15.625 and i_{2} = − 15.625 / 5 = − 3.125 amp.

The negative sign of current i_{1} indicates that the value 3.125-amp current is having opposite direction

Current in 5 Ω resistance = i_{1 }+ i_{2} = 12.5 – 3.125 = 9.375 amp

And battery A is discharging supplied 12.5 A current

Battery B is charging supplied 3.125 A current, Ans.

**Kirchhoff’s law (KVL) Example 6;** In the circuit shown, find the current in each branch and the current in the battery. What is the difference of potential between X and Y.?

**Solution;**

In the circuit let i1 and i2 be the current in A X B and A Y B. In closed circuit, A X B A

1 x i_{1} + 3i_{1} + 2 (i_{1} + i_{2}) = 4 or 1i_{1} + 311 + 2i_{1} + 2i_{2} = 4

Or 6i_{1} + 2i_{2} = 4 … (i)

Fig. 2.9

In closed circuit, A Y B A

4i_{1} + 5i_{2} + 2 (i_{1} + i_{2}) = 4 or 4i_{2} + 5i_{2} + 2i_{1 }+ 2i_{2} = 4

Or 2i_{1} + 11i_{2} = 4 … (ii)

From equation (ii) and (i) we can find the values i1 = 0.581 amp.

Potential difference of X and Y can be defined as the

V_{AX} – V_{AY }= V_{XY}

∴ 1 x 0.581 – 4 (0.258) = 0.581 – 1.032 = -0.451 volts

So, the potential difference of X and Y is 0.451 volts, Ans.

**Kirchhoff’s law (KCL) and (KVL) Example 7;** The Wheatstone Bridge network shown is connected to a 100 V supply. Determine the current in each part of the circuit and the total current taken from the supply.

**Solution;**

As shown in the network let us consider the i_{1} current flowing in branch AB, (i_{1} – i_{3}) in branch BC, i_{3} in branch BD, i_{2} in branch AD and (i_{2} + i_{3}) in branch DC.

Fig. 2.10

Let us take closed mesh A B D

10i_{1} + 5i_{3} – 4i_{2} = 0 … (i)

In mesh ABCEA,

10i_{1} + 2 (i_{1} – i_{3}) = 100 or 10i_{1} + 2i_{1} – 2i_{3} = 100

Or 12i_{1} – 2i_{3} = 100 … (ii)

In mesh ADCEA

4i_{2} + 8 (i_{2} – i_{3}) = 100 or 4i_{2} + 8i_{2} + 8i_{3} = 100

Or 12i_{2} + 8i_{3} = 100 … (iii)

From equation (ii) i_{1} = 100 + 2i_{3}/ 12 amp.

From equation (iii) i_{2} = 100 – 8i_{3}/12 amp.

Now substituting the values of i_{1} and i_{2} in equation (i), we have

10 x [100 + 2i_{3}/12] -4 x [100.8i_{3}/12] + 5i_{3} = 0

1000 + 2i_{3} – 400 + 32i_{3} + 60 i_{3} = 0

112i_{3} = -600; i_{3} = -600/ 112 or i_{3} = 05.357 amp.

It indicates that current is flowing from D to B not from B to D.

i_{1} = 100 + 2i_{3}/ 12 = 100 + 2 (-5.357)/ 12 = 100 – 2 x 5.357 / 12 = 7.44 amp.

i_{2} = 100 – 8i_{3} /12 = 100 – 8 (-5.357)/ 12 = 100 + 8 x 5.357/ 12 = 11.904 amp.

And total current = i_{1} + i_{2} = 7.44 + 11.904 = 19.344 amp. Ans

**Kirchhoff’s law (KVL) and (KCL) Example 8;** In the Wheatstone Bridge network shown, find the current in each resistor.

**Solution;**

In the closed circuit mark the direction of current as shown in fig. considering the closed mesh ABDA

Fig. 2.11

2i_{1} + 2 + 3i_{3} – 4i_{2} = 0

Or 2i_{1} – 4i_{2} + 3i_{3} = -2 … (i)

In closed mesh ABCEA

2i_{1} + 2 + 5(i_{1} – i_{3}) = 4 or 2i_{1} + 2 + 5i_{1} – 5i_{3} = 4

Or 7i_{1} – 5i_{3} = 2 … (ii)

In mesh ADCEA

4i_{2} + 1 (i_{2} + i_{3}) = 4 or 4i_{2} + 1i_{2} + 1i_{3} = 4

Or 5i_{2} + i_{3} = 4

From equation (ii) i_{1} = 2 + 5i_{3}/ 7 amp.

From equation (iii) i_{2} = 4 – i_{3}/ 5

Now substituting in eq. (i) the values of i_{1} and i_{2},

2 [2 + 5i_{3}/ 7] – 4 [4 – i_{3}/ 5] + 3i_{3} = -2

4 + 10i_{3}/ 7 – 16 + 4i_{3}/ 5 + 3i_{3} = -2

5 (4 + 10i_{3}) – 7 (16 + 4i_{3})/ 35 + 3i_{3} = -2

Or 20 + 50i_{3} – 112 + 28i_{3} + 105i_{3} = -70

183i_{3} = 070 + 112 – 20 = 22

Or i_{3} = 22/ 183 = 0.1201 amp.

i_{1} = 2 + 5i_{3}/ 7 = 2 + 5 x 0.1201/ 5 = 0.7759 amp.

Now current in different branches

AB = 0.371 amp

BD = 0.1201 amp

BC = 0.2514 amp, i.e., i1 – i_{3} = 0.3715 – 0.1201 = 0.2514 amp

AD = 0.7759 amp

DC = 0.896 amp i.e., i_{2} + i_{3} = 0.7759 + 0.1201 = 8.96 amp

∴ 0.3715-amp, 0.1201-amp, 0.2514-amp, 0.7795 amp, and 0.896-amp, Ans.

**Kirchhoff’s law (KCL) and (KVL) Example 9;** In the Wheatstone Bridge network shown, find the currents in each resistor.

**Solution;**

Mark the direction of branch currents in the network shown above.

Fig. 2.12

In a closed-circuit ABDA

2i_{1} + 2i_{3} – i2 = 2 … (i)

In closed mesh ABCEA

2i_{1} + 3 (i_{1 }– i_{3}) + 4 (i_{1} + i_{2}) = 4 or 2i_{1} + 3i_{1} – 3i_{3} + 4i_{1} + 4i_{2} = 4

9i_{1} + 4i_{2} – 3i_{3} = 4 … (ii)

In closed mesh ADCEA

i_{2} + 2 (i_{2} + i_{3}) + 4 (i_{1} + i_{2}) = 4 or i_{2} + 2i_{2} + 2i_{3} + 4i_{1}+ 4i_{2} = 4

4i_{1} + 7i_{2} + 2i_{3 }= 4 … (iii)

Solving the equation (i) and (ii) we get

17i_{1} + 5i_{3} = 12 … (iv)

Solving equation (ii) and (iii), we have

47i_{1} – 29i_{3} = 12 … (v)

Again, solving equation (v) and (iv), we get

i1 = 408 / 728 = 0.56043 amp., i_{3} = 47i_{1} -12 / 29 = 0.4945 amp

Put the values of i_{1} and i_{3} in eq. (i) and calculate i_{3}, i.e.,

2i_{1} + 2i_{3} – i_{2} = 2

And i_{2} = 2i_{1} + 2i_{3 }-2

= 2 x 0.56043 + 2 x 0.4945 – 2 = 0.10984 ≅ 0.11 amp.

Now current in different branches

AB = 0.56 amp

BC = 0.066 amp. *I* in BC = i_{1} – i_{3}

= 0.56043 – 0.4945 = 0.06593

≅ 0.066 amp

AD = 0.11 amp and BD = 0.49 amp

DC = i_{2} + i_{3} = 0.11 + 0.4945 = 0.6045 amp

CEA = i_{1} + i_{2} = 0.56043 + 0.11 = 0.67 amp

So, current is

0.56 A, 0.066 A, 0.11 A, 0.049 A, 0.604 A, and 0.67-amp, Ans.

**Kirchhoff’s law (KCL) and (KVL) Example 10;** determine the currents in unbalanced bridge circuit of fig. below. Also determine the p.d across BD and the resistance from B to D.

Figure 2.13

**Solution;**

Assumed current directions are shown in figure 2.13

Applying Kirchhoff’s voltage law to circuit DACD, we get

-x – 4z + 2y = 0 or x – 2y + 4z = 0 … 91)

Circuit ABCA gives

-2 (x – z) + 3 9y + z) + 4 z = 0 or 2x – 3y – 9z = 0 … (2)

Circuit DABED gives

-x – 2 (x – z) -2 (x + y) + 2 = 0 or 5x + 2y – 2z = 2 … (3)

Multiplying (1) by 2 and subtracting (2) from it, we have

-y + 17z = 0 … (4)

Similarly, multiplying (1) by 5 and subtracting (3) from it, we have

-12y +

22z = -2 or -6 + 11z = -1 … (5)

Eliminating y from 94) and (5), we have

91z = 1 or z = 1/ 91 A

From (4); y = 17 / 91 A. Putting these values of y and z in (1), we get x = 30/91 A

Current in DA = x = 30/ 91 A, Current in DC = y = 17/ 91 A

Current in AB = x – z = 30 / 91 – 1/ 91 = 29 / 91 A

Current in CB = y + z = 17/ 91 + 1/ 91 = 18/ 91 A

Current in external circuit = x + y = 30 / 91 + 17 / 91 = 47 / 91 A

Current in AC = z = 1/ 91 A

Internal voltage drops in the cell = 2 (x + y) = 2 x 47 / 91 = 94 / 91 V

∴ P.D across points D and B = 2 – 94/ 91 = 88/ 91 V

Or P.D between D and B = drop across DC + drop across CD = 2 x 17/ 91 + 3 x 18 / 91 = 88 / 91 V

Equivalent resistance of the bridge between points D and B

= P.D between points B and D/ current between points B and D = 88/91 ÷ 47/ 91 = 88/ 47 = 1.87 Ω (approx.)

**Kirchhoff’s law (KCL) and (KVL) Example 11;** Determine the branch currents in the network of fig. below when the value of each branch resistance is one ohm.

**Solution;**

Let the current directions be as shown in fig. 2.14

Apply Kirchhoff’s voltage law to the closed-circuit ABDA, we get

5 – x – z + y = 0 or x – y + z = 5 … (i)

Similarly, circuit BCDB gives

(x – z) +5 + (y + z) + z = 0

Or x -y -3z = 5 … (ii)

Lastly, from circuit ADCEA, we get

-y – (y + z) + 10 – (x + y) = 0

Or x + 3y + z = 10 … (iii)

Fig 2.14

From eq. (i) and (ii), we get, z= 0

Substituting z = 0 either in eq. (i) or (ii) and in eq. (iii), we get

X – y = 5 … (v)

X + 3y = 10 … (v)

Subtracting eq. (v) from (iv), we get

-4y = -5 or y = 5/4 = 1,25A

Eq (iv) gives x = 25 / 4 A = 6.25 A

Current in branch AB = current in branch BC = 6.25 A, Current in branch DC = 1.25 A, current in branch CEA = x + y = 6.25 + 1.25 = 7.5A

**Kirchhoff’s law (KCL) and (KVL) Example 12;** A two conductor distributor Ab, 1400 yards long, is fed at end A at 220 volts. The loads are as follows;

60 amperes at 200 yards, 40 amperes at 700 yards, and 80 amperes at 1000 yards from the feeding end. Resistance of cable 0.1 ohm/1000 yards. Find the current in each section of the cable and the p.d at each load point.

Fig. 2.15

**Solution;**

Total current fed into distributor = 60 + 40 + 80 – 180 amperes

Current, A to M = 180 amperes = current M to A

Current M to N = 180 – 60 = 120 amp = current N to M

Current N to P = 120 – 40 = 80 amps = current P to N

P.D at MM’ = P.D at AA’ – _{A} (I R drop) _{A}’ – _{M} (I R drop) _{A’}

= P.D at AA’ – 2 x A (I R drop) _{M}

In this type of problem, it is usual to draw the diagram as shown below, but it must be borne in mind that when finding the total resistance of a section we must take twice the resistance of the single conductor.

Fig. 2.16

Total resistance of section AM = 2 x 200 / 1000 x 0.1 = 0.04 ohm

Total resistance of section MN = 2 x 500 / 1000 x 0.1 = 0.1 ohm

Total resistance of section NP = 2 x 300/ 1000 x 0.1 = 0.06 ohm

_{A} Total I R drop _{M} = 180 x 0.04 = 7.2 volts

∴ P.D at 60 amperes load = 220 – 7.2 = 212.8 volts

_{M }Total I R drop _{N }= 120 x 0.1 = 12 volts

∴ P.D at 40 amperes load = 212.8 – 12 = 200.8 volts

_{N }Total I R drop _{P} = 80 X 0.06 = 4.8 volts

∴ P.D at 80 amperes load = 200.8 – 4.8 = 196 volts

**Kirchhoff’s law (KVL) and (KCL) Example 13;** As in above example (12) but the distributor is now fed at both ends at 220 volts. Find the current in each section of the cable and the p.d at each lead point.

Fig. 2.17

**Solution;**

It can no longer be assumed that the total current of 180 A is being delivered from end AA’.

Assume a current *i* to be fed from end A; then the currents in the several sections will be as indicated above.

In the mesh ABB’A’, the resultant e.m.f is 220 – 220 = 0

Equating the sum of the I R products in this mesh, we get

0.04i + 0.1 (*i* – 60) + 0.06 (*i* – 100) + 0.08 (I – 180) = 0

Multiply through by 100

4i x 10 (*i* – 60) + 6 (I – 100) + 8 (I – 180) = 0

4i + 10i – 600 + 6i – 600 + 8i – 1440 = 0

28i = 2640 ∴ *i* = 2640 / 28 = 94. 29 amperes

∴ (*i* – 60) = 94.29 – 60 = 34.29 amperes

(*i* – 100) = 94.29 – 100 = – 5.71 amperes

The – sign indicates that the current is being supplied from end B.

(*i* – 180) = 94.29 – 180 = -85.71 amperes

=85.71 amperes supplied from end B

Volt drop in section AM = 94.29 x 0.04 = 3.7716 volts

∴ P.D at 60 amperes load = 220 – 3.7716 = 216.2284 = 216. 2 volts

Volt drop in section MN = 34.29 X 0.1 = 3.429 volts

P.D at 40 amperes load = 216. 2284 – 3.429 = 212.7994 = 212.8 volts

Volt drop in section BP = 85.71 X 0.08 = 6.8568 volts

P.D at 80 amperes load = 220 – 6.8568 = 213.1432 = 213.1 volts

Current in section AM = 94.29 amperes

Current in section MN = 34.29 amperes

Current in section NP = 5.71 amperes

Current in section PB = 85.71 amperes

**Kirchhoff’s law (KVL) and (KCL) Example 14;** As in above example 13 but end A is now fed at 224 volts and end B at 220 volts. Find the current in each section of the cable and the p.d at each load point.

Figure 2.18

**Solution;**

Assume as before, a current *i *to be fed from end a; then the current in the several sections will be as indicated.

Resultant voltage acting in the mesh A B B’ A’ = 224 – 220 = 4 volts

Equating the sum of the I R products in this mesh we get,

0.04i + 0.1 (i – 60) + 0.06 (i – 100) + 0.08 (i – 180) = 4

Multiply through by 100

4i + 10 (i-60) + 6 (i – 100) + 8 (i – 180) = 400

4i + 10i – 600 + 6i – 600 + 8i – 1440 = 400

28i = 3040 ∴ I = 3040 / 28 = 108.57 amperes

(i – 60) = 108.57 – 60 = 48 .57 amperes

(i – 100) = 108. 57 – 100 = 8.57 amperes

(i – 180) = 108.57 – 180 = -71.43 amperes, supplied from end B

Volt drop in section AM = 108.57 x 0.04 = 4.343 volts

P.D at 60 amperes load = 224 – 4.343 = 210.657 = 219.7 volts

Volt drop in section MN = 48.57 x 0.1 = 4.857 volts

∴ P.D at 40 amperes load = 219. 657 – 4.857 = 214.8 volts

Volt drop in section BP = 71.43 x 0.08 = 5.714 volts

∴ P.D at 80 amperes load = 220 – 5.714 = 214.286 = 214.3 volts

Current in section AM = 108.57 amperes

Current in section Mn = 48.57 amperes

Current in section NP = 8.57 amperes

Current in section BP = 71.43 amperes

**Kirchhoff’s law (KVL) Example 15;** A ring main loaded as shown is fed at two points A and B at 220 volts. Find the current in each section of the cable and the p.d at each load point. The resistance of each section – both – conductors is given in ohms. Find also the total current supplied at A and b.

Fig. 2.19

**Solution;**

Assume the current distribution in the several sections to be as shown. In the mesh BAA’B’, resultant

e.m.f = 220 – 220 = 0

∴ 0.02i + 0.03 (i- 40) = 0

∴ 0.05i_{1} – 1.2 = 0

∴ i_{1} = 1.2/ 0.05 = 24 amperes

Fig. 2.20

i_{1}– 40) = 24 – 40 = -16 = 16 amperes supplied from A

In the mesh BXAA’X’B’, resultant e.m.f = 220 – 220 = 0

∴ 0.02i_{2} + 0.04 (i2 – 50) + 0.05 (i_{2} – 130) = 0

0.02i_{2} + 0.04i_{2} – 2 + 0.05i_{2} – 6.5 = 0

0.11i_{2} = 8.5

∴ i_{2} = 8.5/ 0.11 = 77.27 amperes

(i_{2}– 50) = 77.27 – 50 = 27.27 amperes

(i_{2}– 130) = 77.27 – 130 = -52.73 = 52.73 amperes supplied from A

Voltage drops from B to 40 amperes load = 24 x 0.02 = 0.48 volt

∴ P.D at 40 amperes load = 220 – 0.48 = 219.52

Voltage drops from B to 50 amperes load = 77.27 x 0.02 = 1.5454 volts

∴ P.D at 50 amperes load = 220 – 1.5454 = 218.5

∴ Volt drop from A to 80 amperes load = 52.73 x 0.05 = 2.6365 volts

∴ P.D at 80 amperes load = 220 – 2.6365 = 217.4 volts

Total current supplied at A = 16 + 52.73 = 68.73 amperes

Total current supplied at B = 24 + 77.27 = 101. 27 amperes

This problem may be solved using a single-wire equivalent circuit.

Next Topic: Superposition Theorem with 10+ Solved Examples

Previous Topic: How to Calculate Resistors in Series and Parallel: 30+ Solved Examples, Circuit diagram