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Electric circuits 1 Homework#3 Due date: Thursday, 26/11/2020, 4:00 PM Submission method: uploading as a PDF file on E-learning platform.

Q1) For the circuit below, determine all four nodal voltages. (2marks)

Q2) Determine the power supplied by the 1 V source as shown in the circuit below. (2 marks)

Q3) Calculate the four mesh currents labeled in the circuit below. (2 marks)

Q4) (4marks) a) Make use of source transformations to first convert all current sources in the circuit below to voltage sources, then simplify the circuit as much as possible and calculate the voltage Vx, Be sure to draw and label your simplified circuit. b) Use a superposition theorem to find Vx

 Best Wishes 

Solution

Q1) 𝑣1 − 𝑣3 = 8 ⟶ 𝑣1 = 𝑣3 + 8 ⟶ ① 𝑣2 = 5 ⟶ ② −2 =

𝑣1 −𝑣2 6

2 =

𝑣4 −𝑣2

−2 =

𝑣1 −𝑣2

3

6

+

𝑣3 2

+

𝑣4

+

𝑣3

4

2

⟶③ ⟶④

⟶ −2 =

( 𝑣3 +8 )− 5 6

+

𝑣3 2

⟶ 𝑣3 = −

15 4

= −3.75 𝑉

𝑣1 = 𝑣3 + 8 = −3.75 + 8 = 4.25 𝑉 𝑣4 − 𝑣2 𝑣4 𝑣4 − (5) 𝑣4 2 = + ⟶2 = + ⟶ 𝑣4 = 6.286𝑉 3 4 3 4 𝑣1 = 4.25 𝑉 𝑣2 = 5 𝑉 𝑣3 = −3.75 𝑉 𝑣4 = 6.286 𝑉

Q2)

−1 + (4 + 3 + 1)𝑖1 − 3 𝑖2 − 1 𝑖3 = 0 8 𝑖1 − 3 𝑖2 − 𝑖3 = 1 𝑖2 − 𝑖3 = 5 𝑖1 𝑖3 − 𝑖1 + 3 𝑖2 − 3 𝑖1 − 8 + 2 𝑖3 = 0 −4𝑖1 + 3 𝑖2 + 3 𝑖3 = 8 𝑖1 = 19 𝐴 𝑖2 = 61.5 𝐴 𝑖3 = −33.5 𝐴 𝑃1𝑉 𝑠𝑢𝑝𝑝𝑙𝑖𝑒𝑑 = 𝐼 ∗ 𝑉 = 𝑖1 ∗ (1) = 19 ∗ 1 = 19 𝑤𝑎𝑡𝑡

Q3)

𝑖1 = 3 ⟶ ① 𝑖2 − 𝑖4 = 1 ⟶ ② 𝑖4 − 𝑖3 = 3 ⟶ ③

7 + 𝑖3 (2200) + (3500)𝑖2 + (𝑖2 − 𝑖1 )(3100) + (𝑖4 − 𝑖1 )(8100) = 0 ⟶ ④

We have 4 eqns by 4 unknowns After solving 𝑖1 = 3 𝐴 𝑖2 = 2.98 𝐴 𝑖3 = −1.012𝐴 𝑖4 = 1.988𝐴

Q4) a) Using source transformation

We can convert 3A current source with parallel 1 Ω resistor into 3V voltage source with series 1Ω resistor

We can find the sum of (3V) & (6V) voltage sources, Also find the series equivalent of (1Ω)&(2Ω)

We can convert 9V voltage source with series 3Ω resistor into 3A current source with parallel 3 Ω resistor

We can find the parallel equivalent of (3Ω)&(7Ω) , Also find the parallel equivalent of (5Ω)&(5Ω) connected with current dependent source

We can convert 3A current source with parallel 2.1 Ω resistor into 6.3V voltage source with series 2.1Ω resistor

As for the dependent source, We can convert (5Vx) dependent current source with parallel 2.5 Ω resistor into (12.5Vx) dependent voltage source with series 2.5Ω resistor

At the right of the circuit, We can convert 9A current source with parallel 9 Ω resistor into 81V voltage source with series 9Ω resistor

Simplifying the circuit

Now we can assume the current I flows in clockwise, so KVL −87.3 + (13.6) 𝐼 − 12.5 𝑉𝑥 + (4) 𝐼 = 0 −87.3 + (13.6) 𝐼 − 12.5 (4 𝐼) + (4) 𝐼 = 0 87.3 −32.4 𝐼 = 87.3 ⟶ = − 2.69 𝐴 −32.4 𝑉𝑥 = 4 ∗ 𝐼 = 4 ∗ −2.69 = − 10.76 𝑉

b) Using superposition principle Making the 2 current sources open circuits Using Mesh analysis

−6 + 𝑖1 ′ (3) + (𝑖1 ′ − 𝑖2 ′)(7) = 0 ⟶ ① 𝑖2 ′ (9 + 4) + ( 𝑖2 ′ − 𝑖1 ′ )(7) + 𝑖3 ′ (2.5) = 0 ⟶ ② 𝑖2 ′ − 𝑖3 ′ = 5 𝑉𝑥 ′ ⟶ 𝑖2 ′ − 𝑖3 ′ = 5 (4 ∗ 𝑖2 ′ ) ⟶ 𝑖2 ′ − 𝑖3 ′ = 20 𝑖2 ′ −19 𝑖2 ′ − 𝑖3 ′ = 0 ⟶ ③ 3 eqns by 3 unknowns After solving

𝑖1 ′ = 0.509 𝐴 𝑖2 ′ = −0.129 𝐴 𝑖3 ′ = 2.46 𝐴 I need only the current that flows through (4Ω) i2’ So 𝑉𝑥 ′ = 4 ∗ 𝑖2 ′ = 4 ∗ −0.129 = − 0.516 𝑉 -----------------------------------------------

Making the (9A) current source open circuit Making the (6V) voltage source short circuit Using Mesh analysis

𝑖1 ′′ = 3 ⟶ ① ( 𝑖2 ′′ − 𝑖1 ′′ )(1) + 𝑖2 ′′ (2) + ( 𝑖2 ′′ − 𝑖3 ′′ )(7) = 0 ⟶ ② ( 𝑖3 ′′ − 𝑖2 ′′ )(7) + 𝑖3 ′′ (9 + 4) + 𝑖4 ′′ (2.5) = 0 ⟶ ③ 𝑖3 ′′ − 𝑖4 ′′ = 5 𝑉𝑥 ′′ ⟶ 𝑖3 ′′ − 𝑖4 ′′ = 5 (4 ∗ 𝑖3 ′′ ) ⟶ 𝑖3 ′′ − 𝑖4 ′′ = 20 𝑖3 ′′ −19 𝑖3 ′′ − 𝑖4 ′′ = 0 ⟶ ④ 4 eqns by 4 unknowns After solving

𝑖1 ′′ = 3 𝐴 𝑖2 ′′ = 0.255 𝐴 𝑖3 ′′ = − 0.0648 𝐴 𝑖4 ′′ = 1.231 𝐴 I need only the current that flows through (4Ω) i3’’ So 𝑉𝑥 ′′ = 4 ∗ 𝑖3 ′′ = 4 ∗ − 0.0648 = −0.259𝑉 -----------------------------------------------

Making the (3A) current source open circuit Making the (6V) voltage source short circuit Using Mesh analysis

𝑖1 ′′′ = 9 ⟶ ① ( 𝑖2 ′′′ − 𝑖1 ′′′ )(9) + 𝑖2 ′′′ (4) + ( 𝑖2 ′′′ − 𝑖4 ′′′ )(7) + 𝑖3 ′′′(2.5) = 0 ⟶ ② 𝑖4 ′′′ (3) + ( 𝑖4 ′′′ − 𝑖2 ′′′ )(7) = 0 ⟶ ③

𝑖2 ′′′ − 𝑖3 ′′′ = 5 𝑉𝑥 ′′′ ⟶ 𝑖2 ′′′ − 𝑖3 ′′′ = 5 (4 ∗ 𝑖2 ′′′ ) ⟶ 𝑖2 ′′′ − 𝑖3 ′′′ = 20 𝑖2 ′ −19 𝑖2 ′′′ − 𝑖3 ′′′ = 0 ⟶ ④ 4 eqns by 4 unknowns After solving

𝑖1 ′′′ = 9 𝐴 𝑖2 ′′′ = − 2.5 𝐴 𝑖3 ′′′ = 47.5 𝐴 𝑖4 ′′′ = − 1.75 𝐴 I need only the current that flows through (4Ω) i2’’’ So 𝑉𝑥 ′′′ = 4 ∗ 𝑖2 ′′′ = 4 ∗ −2.5 = − 10 𝑉 ----------------------------------------------𝑉𝑥 = 𝑉𝑥 ′ + 𝑉𝑥 ′′ + 𝑉𝑥 ′′′ = −0.516 − 0.259 − 10 = − 10.77 𝑉 We can see as shown below the value of Vx using 2 methods

Value of Vx

Source Transformation

Superposition

-10.76 V

-10.77 V