# (a) Use mesh analysis to find the mesh currents. (b) Use nodal analysis to find the node voltages.?

### 2 Answers

- AshLv 71 month ago
a) Using Mesh Analysis

Let current through upper left loop be I₁, lower left loop be I₂ and right loop be I₃ . And let all flowing clockwise

For upper left loop, there is a 2mA current source that is pushing current in anticlockwise direction ,

so, we get I₁ - I₃ = - 0.002

I₁ = I₃ - 0.002 ..........(1)

Using KVL for lower left loop

12 - 2k(I₂-I₁) - 10k(I₂-I₃) = 0

12 - 2k(I₂-(I₃ - 0.002)) - 10k(I₂-I₃) = 0

12 - 2k(I₂-I₃ + 0.002) - 10k(I₂-I₃) = 0

12 - 2kI₂ + 2kI₃ - 4 - 10kI₂ + 10kI₃ = 0

- 12kI₂ + 12kI₃ = -8

Divide by -12k

I₂ - I₃ = 2/3k .............(2)

Using KVL for outer supermesh

12 - 1kI₁ - 2kI₃ = 0

plug I₁ from (1)

12 - 1k(I₃ - 0.002) - 2kI₃ = 0

12 - 1kI₃ + 2 - 2kI₃ = 0

3kI₃ = 14

I₃ = 14/3k A or 14/3 mA

plug in (1)

I₁ = 14/3k - 0.002

I₁ = (14 - 6)/3k

I₁ = 8/3k A or 8/3 mA

plug I₃ in (2)

I₂ - (14/3k) = 2/3k

I₂ = 16/3k A or 16/3 mA

b) Using Nodal Analysis

Lets consider the bottom node at 0V

then Va = 12V

Va - Vb = 2k(I₂-I₁)

Vb = Va - 2k(I₂-I₁)

Vb = 12 - 2k((16/3k) - (8/3K))

Vb = 12 - 2k(8/3k)

Vb = 12 - 16/3

Vb = 20/3 V

Vc = (2k)I₃

Vc = (2k)(14/3k)

Vc = 28/3 V

Answers

I₁ = 8/3 mA

I₂ = 16/3 mA

I₃ = 14/3 mA

Va = 12 V

Vb = 20/3 V

Vc = 28/3 V

- oldschoolLv 71 month ago
Mesh analysis:

Assume 3 cw currents i1,i2,i3 in top left, right and bottom left loops.

2i1+10i2-12i3 = -12 but i2 = i1+2. Substitute into this equation

2i1+10i1+20-12i3 = -12 so i3 = (12i1+32)/12 = i1+8/3

Apply KVL around the outside of the ckt starting at the lower left corner:

12 - i1-2i2 = 0 but i2 = i1+2

12 - i1 - 2i1 - 4 = 0 so 8 = 3i1 and

Edited, added mA units

i1 = 8/3mA <<<<<

i2 = i1 + 2 = 14/3mA <<<<<

i3 = i1 + 8/3 = 16/3mA <<<<<

Node Analysis

Let V1 be above the 12V so V1 = 12, V3+12 = V1

V2 above the 2A source and V3 the bottom of the ckt with ground above the 10Ω

KCL at V1

V1/2+(V1-V2)/1+V3/10+(V3-V2)/2 Multiply by 10

15V1-15V2+6V3 = 0 ----> V2 = (15V1+6V3)/15 = V1 + 6V1/15 - 72/15

V2 = 21V1/15 - 72/15

KCL at V2

(V2-V1)/1+(V2-V3)/2 = 2 Multiply by 2

-2V1+3V2-V3 = 4 but V3 = V1-12

-2V1+3V2-V1+12 = 4

Substitute V2 = 21V1/15 - 72/15

-2V1+21V1/5-72/5-V1+12 = 4

6V1/5 = 32/5 so V1 = 16/3 V2 = 21/15 *16/3 - 72/15 = 8/3

V3 = 16/3 - 36/3 = -20/3

V1 = 16/3 <<<<<

V2 = 8/3 <<<<<

V3 = -20/3 <<<<<

Power delivered = 12*16/3 + 8/3 * 2 = 208/3 mW

Power consumed = (8/3)²*1+(14/3)²*2+(2/3)²*10+(8/3)²*2 = 208/3mW Checks!

Hope this helps