Mesh ANalysis

1. MESH ANALYSIS

2. KIRCHOFF’S CURRENT LAW (KCL)
• “ In any circuit or network, the algebraic sum of currents meeting at a point is 0”.
• In other words, total current leaving a junction is equal to total current entering that junction.
• Consider a junction A, where five branches carrying currents
I1, I2, I3, I4 and I5 meet. Then, according to KCL:
I1+(-I2)+(-I3)+I4+(-I5) = 0
or, I 1- I2 - I3 + I4 - I5 = 0
or, I 1 + I4 = I2 +I3 +I5
i.e. INCOMING CURRENT = OUTGOINT CURRENT

3. KIRCHOFF’S VOLTAGE LAW (KVL)
• “ The algebraic sum of product of current and resistance in each of the conductors in any closed
path or mesh plus the emf in that path is always 0”.
i.e. Ʃ IR + Ʃ emf = 0
• Sign of emf:

4. • Sign of IR
• The sign of voltage drop across a resistor
depends on the direction of the current
through that resistor and independent of
the polarity of any other emf source in
that circuit.
• From the figure:
E21 = + 45 V
E32 = IR1 = - 10 V
E43 = IR2 = - 20 V
E14 = IR3 = - 15 V
• So, from KVL : Ʃ IR + Ʃ emf = 0
i.e. -10 + (-20) + (-15) + 45 = 0, which is true.

5. MESH ANALYSIS
• Consider a circuit with 3 meshes as shown:
• Let three mesh currents be I1, I2 and I3 flowing in clockwise
direction.
• For Mesh I:
Current through R3 = (I1-I3)
Current through R2 = (I1-I2)
• For Mesh II:
Current through R5 = (I2-I3)
Current through R2 = (I2-I1)
• For Mesh III:
Current through R3 = (I3-I1)
Current through R5 = (I3-I2)

6. • Applying KVL to Mesh I:
E1 −I1R1 −R3 (I1 −I3) −R2 (I1 −I2) = 0
or, (R1 + R2 + R3) I1 −R2I2 −R3I3 = E1 ... ... ... (i)
• Applying KVL to Mesh II:
E2 −R2 (I2 −I1) −R5 (I2 −I3) −I2R4 = 0
or, −R2I1 + (R2 + R4 + R5) I2 −R5I3 = E2 ... ... ... (ii)
• Applying KVL to Mesh III:
E3 −I3R7 −R5 (I3 −I2) −R3 (I3 −I1) −I3 R6 = 0
or −R3I1 −R5I2 + (R3 + R5 + R6 + R7) I3 = E3 ... ... (iii)

7. • The matrix equivalent of the above three equations is:
(𝑅1 + 𝑅2 + 𝑅3) −𝑅2 −𝑅3
−𝑅2 (𝑅2 + 𝑅4 + 𝑅5) −𝑅5
−𝑅3 −𝑅5 (𝑅3 + 𝑅5 + 𝑅6 + 𝑅7)
𝐼1
𝐼2
𝐼3
=
𝐸1
𝐸2
𝐸3
• It can be seen that first item of first row, i.e. R1+R2+R3, is the self resistance of Mesh I, i.e. sum of
all resistances of Mesh I.
• The second item of the first row is the –ve of sum of all the resistances common to Mesh I and
Mesh II.
• The third item of the first row is the –ve of sum of all the resistances common to Mesh I and Mesh
III.
• The above matrix can now be solved using Crammer’s rule to get the values of I1, I2 and I3 as:
I1=
Δ1
Δ
, I2 =
Δ2
Δ
and I3 =
Δ3
Δ

8. • In general:
𝑅11 𝑅12 𝑅13
𝑅21 𝑅22 𝑅23
𝑅31 𝑅32 𝑅33
𝐼1
𝐼2
𝐼3
=
𝐸1
𝐸2
𝐸3
where,
𝑅11 = self-resistance of Mesh I.
𝑅22 = self-resistance of Mesh II.
𝑅33 = self-resistance of Mesh III.
𝑅12 = 𝑅21 = -ve of sum of all resistances common to Mesh I and II.
𝑅23 = 𝑅32 = -ve of sum of all resistances common to Mesh II and III.
𝑅13 = 𝑅31 = -ve of sum of all resistances common to Mesh I and III.
• In the end, it may be pointed out that assuming current direction to be clockwise:
- All the self resistance will be positive.
- All the mutual resistance will be negative.

9. • Eg-1) Write the impedance matrix of the network shown below and find the value of current I1,
I2 and I3.
• Solution:
• We have,
𝑅11 𝑅12 𝑅13
𝑅21 𝑅22 𝑅23
𝑅31 𝑅32 𝑅33
𝐼1
𝐼2
𝐼3
=
𝐸1
𝐸2
𝐸3
• So, 𝑅11 = 1 + 3 + 2 = 6 Ω ; 𝑅22 = 2 + 1 + 4 = 7 Ω
𝑅33 = 3 + 2 + 1 = 6 Ω ; 𝑅12 = 𝑅21 = − 2 Ω
𝑅23 = 𝑅31 = − 1 Ω ; 𝑅13 = 𝑅31 = − 3 Ω
E1 = + 5 V ; E2 = 0 ; E3 = 0
• So the resistance matrix is:
6 −2 −3
−2 7 −1
−3 −1 6
𝐼1
𝐼2
𝐼3
=
5
0
0

10. • Now,
Δ =
6 −2 −3
−2 7 −1
−3 −1 6
= 6(42 - 1) + 2(-12 - 3) - 3(2 + 21) = 147
And, Δ3 =
6 −2 5
−2 7 0
−3 −1 0
= 6 + 2(5) - 3( -35) = 121
So, I3 =
Δ
Δ3
=
147
121
= 0.823 A
• For I1 and I2 … … (DIY): Find Δ1 and Δ2
I1 =
Δ1
Δ
and I2 =
Δ2
Δ

11. • Eg-2) Determine the current supplied by each battery in the circuit shown below:
Solution:
Since there are three meshes, let the three mesh currents
be I1, I2 and I3 flowing in clockwise direction.
Applying KVL in Mesh I, we get:
20 − 5I1 − 3 (I1 −I2) − 5 = 0
or, 8I1 − 3I2 = 15 … … …(i)
Applying KVL in Mesh II, we get:
− 4I2 + 5 − 2 (I2 −I3) + 5 + 5 − 3 (I2 −I1) = 0
or, 3I1 − 9I2 + 2 I3 = − 15 … … …(ii)
Applying KVL in Mesh III, we get:
− 8 I3 − 30 − 5 − 2(I3 − I2) = 0
or, 2 I2 − 10 I3 = 35 ... ... ...(iii)

12. • Eliminating I1 from (i) and (ii), we get:
63 I2 − 16 I3 = 165 ... … … (iv)
• Similarly, for I2 from (iii) and (iv), we have:
I2 = 542/299 A
• From (iv), I3 = − 1875/598 A
• Here, the –ve sign of I3 indicates that the direction of current we assumed is wrong and current
actually flows in anti-clock wise direction.
• Substituting the value of I2 in (i), we get
I1 = 765/299 A
• Discharge current of B1 = 765/299A
• Charging current of B2 = I1 − I2 = 220/299 A
• Discharge current of B3 = I2 + I3 = 2965/598 A
• Discharge current of B4 = I2 = 545/299 A
• Discharge current of B5 = 1875/598 A

13. • SOLUTION BY USING MESH RESISTANCE MATRIX.
• The different items of the mesh-resistance matrix [Rm] are as under :
• R11 = 5 + 3 = 8 Ω; R22 = 4 + 2 + 3 = 9 Ω; R33 = 8 + 2 = 10 Ω
R12 = R21 = − 3 Ω ; R13 = R31 = 0 ; R23 = R32 = − 2 Ω
E1 = algebraic sum of the voltages around Mesh I = 20 − 5 = 15 V
E2 = 5 + 5 + 5 = 15 V ; E3 = − 30 − 5 = − 35 V
• Hence, the mesh equations in the matrix form are:
𝑅11 𝑅12 𝑅13
𝑅21 𝑅22 𝑅23
𝑅31 𝑅32 𝑅33
𝐼1
𝐼2
𝐼3
=
𝐸1
𝐸2
𝐸3
Or,
8 −3 0
−3 9 −2
0 −2 10
𝐼1
𝐼2
𝐼3
=
15
15
−35
• Now, Δ =
8 −3 0
−3 9 −2
0 −2 10
= 8(90 - 4) + 3( -30) = 598

14. And, Δ1 =
15 −3 0
15 9 −2
−35 −2 10
= 15(90 - 4) – 15( -30) - 35 (6) = 1530
And, Δ2 =
8 15 0
−3 15 −2
0 −35 10
= 8(150 - 70) + 3(150 + 0) = 1090
And, Δ3 =
8 −3 15
−3 9 15
0 −2 −35
= 8(-315 + 30) + 3(105 + 30) = -1875
Therefore, I1= Δ1/ Δ= 1530/598 = 765/299 A
I2= Δ2/ Δ= 1090/598 = 545/299 A
I3= Δ3/ Δ= -1875/598 A

15. • Eg-3) Determine the current in the 4-Ω branch in the circuit shown below:
Solution,
Since there are three meshes, let the three mesh currents be I1, I2
and I3 flowing in clockwise direction.
Applying KVL in Mesh I, we get:
− 1 (I1 −I2) − 3 (I1 −I3) − 4I1 + 24 = 0
or, 8I1 −I2 − 3I3 = 24 ... ... ... ... ... ... ... (i)
Applying KVL in Mesh II, we get:
12 − 2 I2 − 12 (I2 −I3) − 1 (I2 − I1) = 0
or, I1 − 15 I2 + 12I3 = − 12 ... ... ... ... ... (ii)
Applying KVL in Mesh III, we get:
− 12 (I3 − I2) − 2I3 − 10 − 3 (I3 − I1) = 0
or, 3 I1 + 12 I2 − 17I3 = 10 ... ... ... ... ... ... (iii)

16. • Eliminating I2 from Eq. (i) and (ii) above, we get:
119I1 − 57I3 = 372 ... … … … (iv)
• Similarly, eliminating I2 from Eq. (ii) and (iii), we get:
57I1 − 111I3 = 6 ... … … … … (v)
• From (iv) and (v) we have, I1 = 40,950/9,960 = 4.1 A
• SOLUTION BY USING MESH RESISTANCE MATRIX.
The three equations as found above are
8 I1 − I2 − 3I3 = 24
I1 − 15 I2 + 12I3 = − 12
3 I1 + 12 I2 − 17I3 = 10
• So, their matrix form is :
8 −1 −3
1 −15 12
3 12 −17
𝐼1
𝐼2
𝐼3
=
24
−12
10
• So, Δ =
8 −1 −3
1 −15 12
3 12 −17
= 664 and Δ1 =
24 −1 −3
−12 −15 12
10 12 −17
= 2730
• Hence, I1 = Δ1 / Δ = 2730/664 = 4.1 A