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If you are a student using this Manual, you are using it without permission. We require Thus, a i 1 Referring to Fig.
It takes 2 pJ to move —1. Reading from the graph, this corresponds to approximately 3. Reading from the graph, this corresponds to roughly 0. We select the bottom node as our reference terminal and define two nodal voltages: Next, we write the two required nodal equations: We begin by selecting the bottom node as the reference terminal, and defining two nodal voltages VA and VB, as shown.
Note if we choose the upper right node, v1 becomes a nodal voltage and falls directly out of the solution. VA VB Ref. We next designate the bottom node as the reference terminal, and define VA and VB as shown: So, we obtain v1 by KVL: Simplifying Eqs. The bottom node has the largest number of branch connections, so we choose that as our reference node. This also makes vP easier to find, as it will be a nodal voltage. Working from left to right, we name our nodes 1, P, 2, and 3.
NODE 1: The logical choice for a reference node is the bottom node, as then vx will automatically become a nodal voltage. The bottom node is chosen as the reference node. Only one nodal equation is required: At the node where three resistors join, 0.
Thus, we require three additional equations: We choose the bottom node as ground to make calculation of i5 easier. We can redraw this circuit and eliminate the 2.
We need concern ourselves with the bottom part of this circuit only. We choose the bottom node as the reference terminal. We choose the center node for our common terminal, since it connects to the largest number of branches. We next form a supernode between nodes A and B. At the supernode: The power supplied by the dependent current source is therefore 0. At node x: Making use of Eq.
We first number the nodes as 1, 2, 3, 4, and 5 moving left to right. We next select node 5 as the reference terminal. To simplify the analysis, we form a supernode from nodes 1, 2, and 3. At node 4: At this point we need to seek an additional equation, possibly in terms of v2.
Solving Eqs. We begin by selecting the bottom node as the reference, naming each node as shown below, and forming two different supernodes as indicated.
Voltages in v3 v5 volts. Resistances v7 in ohms. At node 2: Proceeding with nodal analysis, At node 1: Note that we could also have made use of the supernode approach here.
Mesh 1: Define three mesh currents as shown: We begin by defining three clockwise mesh currents i1, i2 and i3 in the left-most, central, and right-most meshes, respectively.
Define three mesh currents as shown. Then, Mesh 1: The remaining mesh current is clearly 8 A. We may then write: MESH 1: We define four clockwise mesh currents. The top mesh current is labeled i4. The bottom left mesh current is labeled i1, the bottom right mesh current is labeled i3, and the remaining mesh current is labeled i2.
We begin our analysis by defining three clockwise mesh currents. We will call the top mesh current i3, the bottom left mesh current i1, and the bottom right mesh current i2.
It would be nice to be able to express the dependent source controlling variable v1 in terms of the mesh currents. We begin by defining four clockwise mesh currents i1, i2, i3 and i4, in the meshes of our circuit, starting at the left-most mesh.
At Mesh 2: Returning to the circuit diagram, we note that it is possible to express the current of the dependent source in terms of mesh currents. We might also choose to obtain an expression for vdep in terms of mesh currents using KVL around mesh 2 or 3. Thus, 1. We define a clockwise mesh current i1 in the bottom left mesh, a clockwise mesh current i2 in the top left mesh, a clockwise mesh current i3 in the top right mesh, and a clockwise mesh current i4 in the bottom right mesh.
In a perfect world, it would simplify the solution if we could express these two quantities in terms of the mesh currents.
Thus, substituting these expressions into our four mesh equations and creating a matrix equation, we arrive at: We define a clockwise mesh current i3 in the upper right mesh, a clockwise mesh current i1 in the lower left mesh, and a clockwise mesh current i2 in the lower right mesh. It would help in the solution of Eqs. As a result, Eq. Then KVL allows us to write: Define three clockwise mesh currents i1, i2 and i3. The power absorbed by each resistor may now be calculated: The sources supply a total of 4 We begin by naming each mesh and the three undefined voltage sources as shown below: Starting with the left-most mesh and moving right, we define four clockwise mesh currents i1, i2, i3 and i4.
MESH 2: The power generated by each source is: This circuit does not require the supermesh technique, as it does not contain any current sources. We form a supermesh with meshes 3 and 4 as defined above.
The power supplied by the 2. We begin by defining six mesh currents as depicted below: We define a mesh current ia in the left-hand mesh, a mesh current i1 in the top right mesh, and a mesh current i2 in the bottom right mesh all flowing clockwise.
Thus, Forming one supermesh from the remaining two meshes, we may write: Thus, we define three mesh currents, i1, i2, and i3, beginning with the left-most mesh. We next create a supermesh from meshes 1 and 2 note that mesh 3 is independent, and can be analysed separately. The final mesh current is easily found: We begin by redrawing the circuit as instructed, and define three mesh currents: Nodal analysis is probably best then- the nodes can be named so that the desired voltage is a nodal voltage, or, at worst, we have one supernode equation to solve.
On the other hand, the dependent current source depends on the desired unknown, which would lead to the need for another equation if invoking mesh analysis. Trying nodal analysis, 0. Rearranging so that we may eliminate v1 in Eq. This is a three-mesh circuit, or a four-node circuit, depending on your perspective. Either approach requires three equations…. Except that applying the supernode technique reduces the number of needed equations by one.
At the 1, 3 supernode: Nodal analysis, on the other hand, would require only two equations, and the desired voltage will be a nodal voltage. At the b, c, d supernode: Nodal analysis requires 1 supernode, 1 KVL equation, 1 other nodal equation, and one equation to express i1 in terms of nodal voltages.
Thus, mesh analysis has an edge here. This circuit consists of 3 meshes, and no dependent sources.
Therefore 3 simultaneous equations and 1 subtraction operation would be required to solve for the two desired currents. Thus, mesh analysis has a slight edge here. Define three clockwise mesh currents: Then our mesh equations will be: Mesh a: Approaching this problem using nodal analysis would require 3 separate nodal equations, plus one equation to deal with the dependent source, plus subtraction and division steps to actually find the current i Thus, mesh analysis has a clear edge.
Mesh analysis requires 1 mesh equation, 1 supermesh equation, 2 simple KCL equations and one subtraction step to determine the currents. Thus, define four clockwise mesh equations: At the a, b, c supermesh: With 7 nodes in this circuit, nodal analysis will require the solution of three simultaneous nodal equations assuming we make use of the supernode technique and one KVL equation.
Mesh analysis will require the solution of three simultaneous mesh equations one mesh current can be found by inspection , plus several subtraction and multiplication operations to finally determine the voltage at the central node.
Either will probably require a comparable amount of algebraic manoeuvres, so we go with nodal analysis, as the desired unknown is a direct result of solving the simultaneous equations.
Define the nodes as: Mesh analysis yields current values directly, so use that approach. We therefore define four clockwise mesh currents, starting with i1 in the left-most mesh, then i2, i3 and i4 moving towards the right. A map of individual branch currents can now be drawn: If we choose to perform mesh analysis, we require 2 simultaneous equations there are four meshes, but one mesh current is known, and we can employ the supermesh technique around the left two meshes. In order to find the voltage across the 2-mA source we will need to write a KVL equation, however.
Using nodal analysis is less desirable in this case, as there will be a large number of nodal equations needed. Thus, we define four clockwise mesh currents i1, i2, i3 and i4 starting with the left- most mesh and moving towards the right of the circuit. At the 1,2 supermesh: However, this as well as any equation for mesh four are unnecessary: Thus, we define four clockwise mesh currents ia, ib, ic, id starting with the left-most mesh and proceeding to the right of the circuit.
At the a, b supermesh: Define a voltage vx at the top node of the current source I2, and a clockwise mesh current ib in the right-most mesh. There is no constraint on the value of v1 other than we are told to select a nonzero value. First, define clockwise mesh currents ix, i1, i2 and i3 starting from the left-most mesh and moving to the right.
Next, combine the 2 A and 3 A sources temporarily into a 1 A source, arrow pointing upwards. Then, define four nodal voltages, V1, V2, V3 and V4 moving from left to right along the top of the circuit. Hand analysis: PSpice simulation results: The voltage at the center node is found to be 3. Applying nodal analysis then, Node 1: The simulation results agree with the hand calculations.
The simulated results agree with the hand calculations. We first name each node, resistor and voltage source: END And obtain the following output: R5 Verify: We expect the source to provide W. AC LIN 1 60 It is an artifact of the use of V1 0. The supplied power is then separately computed as 1.
Thus, the furthest bulbs actually have less than VAC across them, so they draw slightly less current and glow more dimly.
Open circuit the 4 A source. The contribution to v1 from the 4 A source is found by first open-circuiting the 1 A source, then noting that current division yields: One approach to this problem is to write a set of mesh equations, leaving the voltage source and current source as variables which can be set to zero. We first rename the voltage source as Vx. We next define three clockwise mesh currents in the bottom three meshes: Finally, we define a clockwise mesh current i3 in the top mesh, noting that it is equal to —4 A.
Our general mesh equations are then: Our mesh equations then become: We may solve this problem without writing circuit equations if we first realise that the current i1 is composed of two terms: We may not change K1 or K2, as only the source voltages may be changed.
If we increase both sources by a factor of 10, then i1 increases by the same amount. One source at a time: The contribution from the V source may be found by shorting the V source and open-circuiting the 2-A source. The contribution of the 8-A source is found by shorting out the two voltage sources and employing simple current division: The contribution of the V source is found by open-circuiting the 8-A source and shorting the V source.
The contribution of the 0. We find the contribution of the 4-A source by shorting out the V source and analysing the resulting circuit: Proceeding to the contribution of the V source, we analyse the following circuit after defining a clockwise mesh current ia flowing in the left mesh and a clockwise mesh current ib flowing in the right mesh. This means that the response v3 will be scaled by the same factor: We can view this in a somewhat abstract form: In a similar fashion, we find that the contribution of the 7-V source is: The contribution from the current source may be calculated by first noting that 1M 2.
The total current flowing from the voltage source with the current source open-circuited is —1. We first determine the contribution of the voltage source: We proceed to find the contribution of the current source: V" Vx" Via supernode: The 5-V source may then be increased by a factor of It is impossible to identify the individual contribution of each source to the power dissipated in the resistor; superposition cannot be used for such a purpose. Simplifying the circuit, we may at least determine the total power dissipated in the resistor: We will analyse this circuit by first considering the combined effect of both dc sources left , and then finding the effect of the single ac source acting alone right.
We may now proceed: We first consider the effect of the 2-A source separately, using the left circuit: The circuit on the right yields the contribution of the 6-A source to Vx: As can be seen from the two separate PSpice simulations, our hand calculations are correct; the pV-scale voltage in the second simulation is a result of numerical inaccuracy.
This modified circuit contains a series combination of 2. Since 7. Since the quantity v appearing across this resistor is of interest, we cannot involve the resistor in a transformation. The current through the 5. Therefore, P5. The 9-V source will force the voltage across these two terminals to be —9 V regardless of the value of the current source and resistor to its left.
Thus, we may draw: Least-squares fit results: Voltage V Current mA 1. We have therefore chosen to perform a linear fit for the three lower voltages only, as shown. Our model will not be as accurate at 1 mA; there is no way to know if our model will be accurate at 20 mA, since that is beyond the range of the experimental data.
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