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Skip to main content. Log In Sign Up. Engineering Circuit Analysis 7ed solution manual-by William Hayt.
Limited distribution permitted only to teachers and educators for course preparation. If you are a student using this Manual, you are using it without permission. The mJ pulse lasts anapysis ns. The 1-mJ pulse lasts 75 fs.
The power drawn from the battery is not quite drawn to scale: It takes 2 pJ to move —1. Reading from the egnineering, this corresponds to approximately 3.
Reading from the graph, this corresponds to roughly 0. The maximum power point, therefore, occurs somewhere between these two points.
By trial and error, Pmax is roughly mV 2. We will compute absorbed power by using the current flowing into the positive reference terminal of the appropriate voltage passive sign conventionahalysis we will compute supplied power by using the current flowing out of the positive reference terminal of the appropriate voltage.
We are asked to determine the voltage vs, which is identical to the voltage labeled v1. The only remaining reference to v1 is in the expression for the current flowing through the dependent source, 5v1. This current is equal to —i2. The battery soluiton an energy of The supply voltage is V, and the maximum dissipated power is W. The fuses are specified in terms of current, so we need to determine soluion maximum current that can flow through the fuse.
This exceeds the specified maximum power. If we choose the 4. This leads to a maximum power of 4. Although W is less than the maximum power allowed, this fuse will provide adequate protection for the application circuitry. If a fault occurs and the application circuitry attempts to draw too much power, W for example, the fuse will blow, no current will flow, and the application circuitry will be protected.
However, if the application circuitry tries to draw its maximum rated power Wthe fuse will also blow. In practice, most equipment will not draw its maximum rated power continuously—although to be safe, we typically assume that it will. Since we are informed that the same current must flow through each component, we begin by defining a current I flowing out of the positive reference terminal of the voltage source.
The power supplied by the voltage source is Vs I. The power absorbed by resistor R1 is I2R1. The power absorbed by resistor R2 is I2R2. Since we know that the total power supplied is equal to the total power absorbed, we may write: Thus, we choose to estimate the resistance using the two extreme points: Referring to Table 2. This seems like a lot of wire to be washing up on shore. In that case, perhaps we should take our cue from Eq. We have stated that deition can affect resistance—in other words, if the temperature changes during operation, the resistance will not remain constant and hence nonlinear behavior will be observed.
Most discrete resistors are rated for up to a specific power in order to ensure that temperature variation during operation will not significantly change the resistance value.
7tth bulbs, however, become rather circyit when operating and can experience a significant change in resistance.
The resistor is unable to store the energy itself. Copper wire Rotating short Leads to wire deter- connect to mines length circuit of long wire used in circuit.
But this is somewhat impractical, as the leads may turn out to have almost the same resistance unless we have a very long wire, which can also be impractical. One improvement would be to replace the copper wire shown with a coil of insulated copper wire.
A small amount of insulation would then need to be removed from where the moveable wire touches the coil so that electrical connection could be made. We have 10, ft of each of the gauges listed in Table 2. Quick inspection of the values listed eliminates 2, 4 and 6 AWG wire as their respective resistances are too low for only 10, ft of wire. Would the AWG wire weight less? Again referring to Table 2. The best but not the only choice for a portable application is clear: Thus, we see that the lower doping level clearly provides material with higher resistivity, requiring less of the available area on the silicon wafer.
By KCL, we may write: We obtain the current each bulb draws by dividing its power rating by the operating voltage V: The DMM is connected in parallel with the 3 load resistors, across which develops the voltage we wish to measure.
If the DMM appears as a short, then all 5 A flows through the DMM, and none through the resistors, resulting in a false reading of 0 V for the circuit undergoing testing. If, instead, the DMM has an infinite internal resistance, then no current is shunted away from the load resistors of the circuit, and a true voltage reading results. In either case, a bulb failure will adversely affect the sign.
Still, in the parallel-connected case, at least 10 up to 11 of the other characters will be lit, so the sign could be read and customers will know the restaurant is open for business. We begin by making use of the information given regarding the power generated by the 5-A and the V sources.
The 5-A source supplies W, so it must therefore have a terminal voltage of 20 V. The V source supplies W, so it must therefore provide a current IX of KCL provides us with the means to find this current: Begin by defining a clockwise current i. Calculate Vout by writing two KVL equations.
Refer to the labeled diagram below. A quick check assures us that these power quantities sum to zero. Therefore, none of the conditions specified in a to d can be met by this circuit. Thus, we need to solve the quadratic equation 0.
In order to determine which of these two values should be used, we must recall that the idea is to charge the battery, implying that it is absorbing power, or that i as defined is positive. Drawing the circuit described, we also define a clockwise current i.
Performing the necessary substition, Eq. We begin by defining a clockwise current i.
Note that in fact vx appears across each of the four elements. It engineerinf a simple matter now to compute the power absorbed by each element: It is now a straightforward matter to compute the power absorbed by each element: Despite the way it may appear at first glance, this is actually a simple node-pair circuit.
The power supplied by each source is then computed as: Thus, 7h sum of the supplied power of each circuit element is zero, as it should be.
Engineering Circuit Analysis () :: Homework Help and Answers :: Slader
On the left, the two 1-A sources in series reduce to a single 1-A source. The new 1-A source and the 3-A source combine to yield a 4-A source in series with the unknown current source which, by KCL, must be a 4-A current source. The result is two V sources in parallel, which is permitted by KVL.
We see that to determine the voltage v we will need vx due to the presence of the dependent current soruce. Returning amalysis the left-hand side of the circuit, and summing currents into the top node, we find that 12 — 3. We begin by noting several things: The mS and mS conductances are also in series, equivalent to Redrawing for clarity, These combinations are in series.
The voltage dropped across the 2. We first note that the 4. One possible solution of many: We may combine the A and 5-A current sources into a single 7-A current source with its arrow oriented upwards. Voltage division only applies to resistors connected in series, meaning that the same current must flow through each resistor.
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 solutiion the solution. We next designate the bottom node as the reference terminal, and define VA and VB as shown: So, we obtain v1 by KVL: 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 cirucit be a nodal voltage.