Vs1= 6V Vs2= 12V Measured Mesh Current: I1= 1.8mA 12=1.17mA 13=2.68mA 1.From you

Question

Vs1= 6V Vs2= 12V

Measured Mesh Current:

I1= 1.8mA

12=1.17mA

13=2.68mA

1.From your measured mesh currents, calculate the value of the branch currents ia and ib shown in Figure 1. Using top to bottom in the vertical branches and left to right in the horizontal branches, what are the currents in each of the other five branches?

2.By observation, what are the values of V1 and V4? With the given values of VS1 and VS2.   Using plus reference at the top of the vertical branches and at the left for horizontal branches, what are the seven branch voltages in terms of the node voltages?

Vs1= 6V Vs2= 12V Measured Mesh Current: I1= 1.8mA 12=1.17mA 13=2.68mA 1.From your measured mesh currents, calculate the value of the branch currents ia and ib shown in Figure 1. Using top to bottom in the vertical branches and left to right in the horizontal branches, what are the currents in each of the other five branches? 2.By observation, what are the values of V1 and V4? With the given values of VS1 and VS2. Using plus reference at the top of the vertical branches and at the left for horizontal branches, what are the seven branch voltages in terms of the node voltages?

Explanation / Answer

Hi! I am a graduate student in electrical ENgineering. I answered your question with the aid of conceptual basis so as to make the comprehension of this problem as easy as possible. I hope for the best rating. ThanKs

Solution of Part-1 :

1- Calculating Ia : Since Ia is a current in vertically down direction (as shown in figure) and it flows at the intersection of I1 Mesh and I2 Mesh which means value of Ia is the value of difference between I1 and I2. We can also observe that the direction of Ia is similar to the direction of I1 which depicts that the Ia should be positive in the direction of I1, so keeping in view these observations we can infer that : Ia = I1 - I2

   => Ia = 1.8 - 1.17 = 0.63 mA.

2- Calculating Ib : Using similar arguments as I expalined for Ia, it can easily be noted that Ib = I2 - I3

=> Ib = 1.17 - 2.68 = -1.51 mA. (-ve indicates that original flow direction of Ib vertically upward)

3- Current in Branch with voltage source Vs1 : Since Vs1 is located vertically, so according to direction convention defined in question we take its direction from top to bottom (vertically down). Now it is apparent, from figure in question, that the only current that passes through this branch is only I1 but that is in vertically upward direction. So according to the direction convention we have taken we can easily infer that the current in this branch is negative of I1.   

  => IVs1 = - I1 = - 1.8 mA.

4- Current in Branch with voltage source Vs2 : Now considering the concept depicted in previous discussion it is easily inferred that the current in this branch is same as I3 (Direction of IVs2 also matches with that of I3).

=> IVs2 = I3 = 2.68 mA

5- current in branch with resistor R1 : The direction of current IR1 is from left to right (as required by question). Now by observing circuit we can know that the only current that passes through R1 is only I1. And it is easy to observe that the direction of  IR1 is similar to that of I1.

=> IR1 = I1 = 1.8 mA.

6 & 7 - current in branches with resistor R2 and R3 : Using the same arguments as we did for the case of IR1, It is clearly visible form the given circuit that: IR2 = I2   and IR3 = I3 .

=> IR2 = I2 = 1.17 mA and IR3 = I3 = 2.68 mA

Solution of Part-2:

1- Voltage at Node V1 : Since the lower side of volatge source Vs1 is grounded (at zero potential) and the V1 lies at the positive side of Vs1 which gives the clear idea that V1 = Vs1 (referenced to ground).

=> V1 = 6 Volts.

2- Voltage at Node V4 : Using similar argument as we did for V1, it is clear that V4 = Vs2 (referenced to ground).

   => V4 = 12 Volts.

3- Branch volatge at R1 : Now using the reference convention (left = +ve and right = -ve), we can write volatge drop at R1 as: VR1 = V1 - V2

4- Branch volatge at R2: Now using the reference convention (top = +ve and bottom = -ve), we can write volatge drop at R2 (considering other side of R2 is grounded) as:   VR2 = V2 - 0 volt = V2

5- Branch volatge at R3: Now using the reference convention (left = +ve and right = -ve), we can write volatge drop at R3 as:   VR3 = V2 - V3

6- Branch volatge at R4: Now using the reference convention (top = +ve and bottom = -ve), we can write volatge drop at R4 (considering other side of R4 is grounded) as:   VR4 = V3 - 0 volt = V3

7- Branch volatge at R5: Now using the reference convention (left = +ve and right = -ve), we can write volatge drop at R5 as:   VR5 = V3 - V4

That's the end of solution. I hope for the 5-star rating. Thanks

Related Questions

Navigate

• Browse (All)
• Browse V
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.