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Skip to main content. Next, we write the two required nodal equations: Using nodal analysis is less desirable in this case, as there will be a large number of nodal equations needed.

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. A general difference amplifier: Log In Sign Up. So, we obtain v1 by KVL: This also makes vP easier to find, as it will be a nodal voltage.

In order to find the voltage across the 2-mA source we will need to write a KVL equation, however. Now look at the range of possible loads.

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This should resit in a rms current amplitude entineering Diameter of a dime: For LF, a current of You dismissed this ad. V4 V2 V1 V3 Ref.

Writing KVL for this instant in time, 16 — 10 1. We first number the nodes as 1, 2, 3, 4, and 5 moving left to right. Engineering circuit-analysis-solutions-7ed-hayt i hope this will help you.

Thus, we need a total of 30 W from the new network of lamps. We might also choose to obtain an expression for vdep in terms of mesh currents using KVL around mesh 2 or 3. The following comparator setup would give a soolution 0 for voltages below 1.

The bottom left mesh current is labeled i1, the bottom right mesh current is labeled i3, and the fdition mesh current is labeled i2. Thus, mesh analysis has an edge here. We next designate the bottom node as the reference terminal, and define VA and VB as shown: Thus, we need to solve the quadratic equation 0. This is confirmed by the PSpice simulation shown below.


With no source present, we expect a new response but with the same general form: The first step is to perform a simple source transformation, so that a 0. This op amp circuit is an open-loop circuit; there is no external feedback path from the output terminal to either input. KCL provides us with the means to find this current: This circuit does not require the supermesh technique, as it does not contain any current sources.


The top mesh current is labeled i4.

Engineering Circuit Analysis () :: Homework Help and Answers :: Slader

Reading from the graph, this corresponds to roughly 0. For a 20 V Zener diode, three 9 V batteries giving a voltage of 27 V would be needed.

We now have two dependent current sources in parallel, which may be combined to yield a single —0.