![• 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](https://image.slidesharecdn.com/1-231011231020-a41ff187/85/1-Mesh-Analysis-pdf-13-320.jpg)
1.Mesh Analysis.pdf
- 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