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Very happy! Worked great, saved money! Book Review: This is not a textbook. The total power dissipated in the circuit is W. The dependent voltage source delivers W to the circuit. This circuit has a supernode includes the nodes v1, v2 and the 25 V source. This circuit now has only one non-reference essential node where the voltage is not known — note that it is not a supernode.
Both methods give the same result but the choice of reference node in part b yielded fewer equations to solve, so is the preferred method. The right-most node voltage is 75 V. Write KCL equations at the essential nodes labeled 1 and 2. From Eq. The power absorbed by the resistors is Therefore, the dependent source is developing 46, W. Therefore, the independent source is developing W, all other elements are absorbing power, and the total power developed is thus W.
Therefore, the 5 A source delivers W. Therefore, only the current source delivers power and the total power delivered is W. Thus the total power dissipated is W. Use the mesh current method to minimize the number of simultaneous equations. The node voltage method has the advantage of having to solve the three simultaneous equations for one unknown voltage provided the connection at either the top or bottom of the circuit is used as the reference node.
Therefore recommend the node voltage method. If the connection at the bottom of the circuit is used as the reference node, then the voltages controlling the dependent sources are node voltages.
This makes it easy to formulate the constraint equations. The current in the 20 V source is obtained by summing the currents at either terminal of the source. The mesh current method requires summing the voltages around the two meshes not containing current sources in terms of four mesh currents. In addition the voltages controlling the dependent sources must be expressed in terms of the mesh currents.
Thus the constraint equations are more complicated, and the reduction to two equations and two unknowns involves more algebraic manipulation. The current in the 20 V source is found by subtracting two mesh currents.
Because the constraint equations are easier to formulate in the node. Instead, use the node-voltage method and choose the reference node so that a node voltage is identical to the voltage across the 40 mA source. Since the 40 mA source is developing 0 W, v1 must be 0 V.
This shortcut will simplify the set of KVL equations. The node-voltage method has no equivalent simplifying shortcut, so the mesh-current method is preferred. Write the mesh current equations. Now use a source transformation on each voltage source, thus.
This follows because they are in parallel with an ideal voltage source. Hence our circuit can be simplified to. Short-circuit current:. Thus we need only find RTh. Use voltage division to find the voltage drop across this load resistor, and use the voltage to find the power delivered to it: v2. Calculate power absorbed by the 2. After making a couple of source transformations the circuit simplifies to.
Thus our problem is reduced to analyzing the circuit shown below. Open circuit voltage:. Open circuit voltage. Further study shows that by replacing the parallel resistors with their equivalent values the circuit reduces to four meshes and four essential nodes as shown in the following diagram. The node-voltage approach will require solving three node voltage equations along with equations involving vx, vy , and ix.
The mesh-current approach will require writing one mesh equation and one supermesh equation plus five constraint equations involving the five sources. Thus at the outset we know the supermesh equation can be reduced to a single unknown current. Since we are interested in the power developed by the 50 V source, we will retain the mesh current ib and eliminate the mesh currents ia, ic and id.
The supermesh is denoted by the dashed line in the following figure. The mesh equations are: i1 These values are in agreement with our predicted values. Note our predicted values are within a fraction of a volt of the actual values. Use the negative power supply value to determine one limit on the value of Rx :. The Operational Amplifier. Since we cannot have negative resistor values, the lower limit for Rx is 0. Now use the positive power supply value to determine the upper limit on the value of Rx :.
Therefore the voltmeter reads 7. If the gain of the inverting amplifier is to be 2. There are many possible designs that use a resistor value chosen from Appendix H. We present one here that uses 3. Use a single 3.
Then construct a network of 3. The resulting circuit is shown here:. Multiply through by ,, plug in the values of input voltages, and rearrange to solve for vo:! This is an inverting summing amplifier, so each input voltage is amplified by a gain that is the ratio of the feedback resistance to the resistance in the forward path for the input voltage. Solve for each input resistance value to yield the desired gain:.
Now create the 5 resistor values needed from the realistic resistor values in Appendix H. Of course there are many other acceptable possibilities. The final circuit is shown here:. The resulting non-inverting amplifier circuit is shown here:. This instructor solution manual provide all answers that you will need to know as teacher or instructor. Please bear in mind that we do not own copyrights to these books.
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