# Chegg circuits

One of these outputs will be active High based on the combination of inputs present, when the decoder is enabled. That means decoder detects a particular code. The block diagram of 2 to 4 decoder is shown in the following figure. The Truth table of 2 to 4 decoder is shown below. Each output is having one product term.

So, there are four product terms in total. The circuit diagram of 2 to 4 decoder is shown in the following figure. If enable, E is zero, then all the outputs of decoder will be equal to zero. In this section, let us implement 3 to 8 decoder using 2 to 4 decoders.

We can find the number of lower order decoders required for implementing higher order decoder using the following formula. Therefore, we require two 2 to 4 decoders for implementing one 3 to 8 decoder.

The block diagram of 3 to 8 decoder using 2 to 4 decoders is shown in the following figure. The complement of input A 2 is connected to Enable, E of lower 2 to 4 decoder in order to get the outputs, Y 3 to Y 0.

These are the lower four min terms. The input, A 2 is directly connected to Enable, E of upper 2 to 4 decoder in order to get the outputs, Y 7 to Y 4. These are the higher four min terms.

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In this section, let us implement 4 to 16 decoder using 3 to 8 decoders. Therefore, we require two 3 to 8 decoders for implementing one 4 to 16 decoder. The block diagram of 4 to 16 decoder using 3 to 8 decoders is shown in the following figure.

The complement of input, A3 is connected to Enable, E of lower 3 to 8 decoder in order to get the outputs, Y 7 to Y 0. These are the lower eight min terms. The input, A 3 is directly connected to Enable, E of upper 3 to 8 decoder in order to get the outputs, Y 15 to Y 8. These are the higher eight min terms. Digital Circuits - Decoders Advertisements.Free Chegg Account equips learners with an array of digital academic services.

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They are often drawn in such a way that makes it difficult to follow which components are in series and which are in parallel with each other. The purpose of this section is to show you a method useful for redrawing circuit schematics in a neat and orderly fashion. Like the stage-reduction strategy for solving series-parallel combination circuits, it is a method easier demonstrated than described.

Perhaps this diagram was originally drawn this way by a technician or engineer. Perhaps it was sketched as someone traced the wires and connections of a real circuit. In any case, here it is in all its ugliness:. With electric circuits and circuit diagrams, the length and routing of wire connecting components in a circuit matters little. Actually, in some AC circuits it becomes critical, and very long wire lengths can contribute unwanted resistance to both AC and DC circuits, but in most cases wire length is irrelevant.

The strategy I have found easiest to apply is to start by tracing the current from one terminal of the battery around to the other terminal, following the loop of components closest to the battery and ignoring all other wires and components for the time being.

While tracing the path of the loop, mark each resistor with the appropriate polarity for voltage drop. When tracing this direction, I will mark each resistor with positive polarity on the entering side and negative polarity on the exiting side, for that is how the actual polarity will be as current according to the conventional-flow model enters and exits a resistor:.

Now, proceed to trace any loops of components connected around components that were just traced. Tracing those loops, I draw R 2 and R 4 in parallel with R 1 and R 3 respectively on the vertical diagram.

### Digital Circuits - Decoders

Now we have a circuit that is very easily understood and analyzed. In this case, it is identical to the four-resistor series-parallel configuration we examined earlier in the chapter. Re-drawing vertically and keeping track of voltage drop polarities along the way, our equivalent circuit starts out looking like this:. Next, we can proceed to follow the next loop around one of the traced resistors R 6in this case, the loop formed by R 5 and R 7.

As before, we start at the positive end of R 6 and proceed to the negative end of R 6marking voltage drop polarities across R 5 and R 7 as we go:. Now we add the R 5 —R 7 loop to the vertical drawing. Notice how the voltage drop polarities across R 7 and R 5 correspond with that of R 6and how this is the same as what we found tracing R 7 and R 5 in the original circuit:.

We repeat the process again, identifying and tracing another loop around an already-traced resistor. In this case, the R 3 —R 4 loop around R 5 looks like a good loop to trace next:. Adding the R 3 —R 4 loop to the vertical drawing, marking the correct polarities as well:. With only one remaining resistor left to trace, then next step is obvious: trace the loop formed by R 2 around R 3 :.

This simplified layout greatly eases the task of determining where to start and how to proceed in reducing the circuit down to a single equivalent total resistance. In this particular case, we would start with the simple parallel combination of R 2 and R 3reducing it to a single resistance.

Sergio Franco. Don't have an AAC account?Use superposition to analyze circuits that have lots of voltage and current sources. Superposition helps you to break down complex linear circuits composed of multiple independent sources into simpler circuits that have just one independent source. The total output, then, is the algebraic sum of individual outputs from each independent source. With the help of superposition, you can break down the complex circuit shown here into two simpler circuits that have just one voltage source each.

To turn off a voltage source, you replace it with a short circuit. The next diagram shows the same circuit with one voltage source turned off: Circuit B contains one voltage source, with v s 2 turned off and replaced by a short circuit. The output voltage due to v s 1 is v o 1. Similarly, Circuit C is Circuit A with the other voltage source turned off. Circuit C contains one voltage source, with v s 1 replaced by a short circuit.

The output voltage due to voltage source v s 2 is v o 2. Summing up the two outputs due to each voltage source, you wind up with the following output voltage:. To find the output voltages for Circuits B and C, you use voltage divider techniques.

That is, you use the idea that a circuit with a voltage source connected in series with resistors divides its source voltage proportionally according to the ratio of a resistor value to the total resistance. In Circuit B, you simply find the output voltage v o 1 due to v s 1 with a voltage divider equation:. In Circuit C, finding the output voltage v o 2 due to v s 2 also requires a voltage divider equation, with the polarities of v o 2 opposite v s 2.

Using the voltage divider method produces the output voltage v o 2 as follows:. The plan in this section is to reduce the circuit shown here to two simpler circuits, each one having a single current source, and add the outputs using superposition.

You consider the outputs from the current sources one at a time, turning off a current source by replacing it with an open circuit. Circuit A consists of two current sources, i s 1 and i s 2and you want to find the output current i o flowing through resistor R 2.

Circuit B is the same circuit with one current source turned off: Circuit B contains one current source, with i s 2 replaced by an open circuit.

The output voltage due to i s 1 is i o 1.

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Similarly, Circuit C is Circuit A with only current source, with i s 1 replaced by an open circuit. The output current due to current source i s 2 is i o 2. Adding up the two current outputs due to each source, you wind up with the following net output current through R 2 :.

To find the output currents for Circuits B and C, you use current divider techniques. That is, you use the idea that for a parallel circuit, the current source connected in parallel with resistors divides its supplied current proportionally according to the ratio of the value of the conductance to the total conductance.

For Circuit B, you find the output current i o 1 due to i s 1 using a current divider equation. In Circuit C, the output current i o 2 due to i s 2 also requires a current divider equation. Note the current direction between i o 2 and i s 2 : i s 2 is opposite in sign to i o 2. Adding up i o 1 and i o 2you wind up with the following total output current:. You can use superposition when a circuit has a mixture of two independent sources, with one voltage source and one current source.

## Analyze an RLC Second-Order Parallel Circuit Using Duality

You need to turn off the independent sources one at a time. To do so, replace the current source with an open circuit and the voltage source with a short circuit. Circuit A of the sample circuit shown here has an independent voltage source and an independent current source. How do you find the output voltage v o as the voltage across resistor R 2? Circuit A with its two independent sources breaks up into two simpler circuits, B and C, which have just one source each.

Circuit B has one voltage source because the current source was replaced with an open circuit.You can then use the voltage divider technique for a series circuit to obtain the load voltage, or you can use the current divider technique for a parallel circuit to obtain the load current. For example, consider Circuit A shown here.

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In this circuit, the voltage source v s and resistors R 1 and R 2 are connected in series. Because independent current sources are in parallel and point in the same direction, you can add up the two source currents, which produces the equivalent Norton current, i N :.

Circuit B shows the combination of the two current sources. When you combine the two current sources into one single current source connected in parallel with one resistor, you have the Norton equivalent.

You can convert the current source i N in parallel with R T to a voltage source in series with R T using the following source transformation equation:. John M. Santiago Jr. During that time, he held a variety of leadership positions in technical program management, acquisition development, and operation research support. About the Book Author John M.In the last section, we learned what would happen in simple resistor-only and capacitor-only AC circuits.

Now we will combine the two components together in series form and investigate the effects. The term for this complex opposition to current is impedanceits symbol is Z, and it is also expressed in the unit of ohms, just like resistance and reactance. In the above example, the total circuit impedance is:.

To calculate current in the above circuit, we first need to give a phase angle reference for the voltage source, which is generally assumed to be zero. As with the purely capacitive circuit, the current wave is leading the voltage wave of the sourcealthough this time the difference is Voltage lags current current leads voltage in a series R-C circuit.

Notice how the voltage across the resistor has the exact same phase angle as the current through it, telling us that E and I are in phase for the resistor only.

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The voltage across the capacitor has a phase angle of In this case, the Again, it must be emphasized that the calculated figures corresponding to real-life voltage and current measurements are those in polar form, not rectangular form!

For example, if we were to actually build this series resistor-capacitor circuit and measure voltage across the resistor, our voltmeter would indicate 1. Real instruments connected to real circuits provide indications corresponding to the vector length magnitude of the calculated figures.

While the rectangular form of complex number notation is useful for performing addition and subtraction, it is a more abstract form of notation than polar, which alone has direct correspondence to true measurements. Thus, the voltage phasor diagram can be replaced by a similar impedance diagram. Find the impedance at 60 hertz. Don't have an AAC account?

Create one now. Forgot your password? Click here. Latest Projects Education. Textbook Series Resistor-Capacitor Circuits. Home Textbook Vol. Current Calculation To calculate current in the above circuit, we first need to give a phase angle reference for the voltage source, which is generally assumed to be zero. Series: R-C circuit Impedance phasor diagram. Impedances Z are managed just like resistances R in series circuit analysis: series impedances add to form the total impedance.

Just be sure to perform all calculations in complex not scalar form! Z n Please note that impedances always add in series, regardless of what type of components comprise the impedances. That is, resistive impedance, inductive impedance, and capacitive impedance are to be treated the same way mathematically.

Published under the terms and conditions of the Design Science License. You May Also Like. Log in to comment. Sign In Stay logged in Or sign in with. Continue to site.The circulatory system is a major organ system of the body.

This system transports oxygen and nutrients in the blood to all of the cells in the body. In addition to transporting nutrients, the circulatory system also picks up waste products generated by metabolic processes and delivers them to other organs for disposal.

The circulatory system, sometimes called the cardiovascular systemconsists of the heartblood vesselsand blood. The heart provides the "muscle" needed to pump blood throughout the body. Blood vessels are the conduits through which blood is transported and blood contains the valuable nutrients and oxygen that are needed to sustain tissues and organs. The circulatory system circulates blood in two circuits: the pulmonary circuit and systemic circuit.

The circulatory system performs a number of vital functions in the body. This system works in conjunction with other systems to keep the body working properly.

### Analyze Circuits with Dependent Sources

This artery branches into left and right pulmonary arteries. In the lungs, carbon dioxide in the blood is exchanged for oxygen at lung alveoli. Alveoli are small air sacs that are coated with a moist film that dissolves air. The pulmonary circuit is completed when pulmonary veins return blood to the left atrium of the heart. When the heart contracts again, this blood is pumped from the left atrium to the left ventricle and later to systemic circulation.

Digital Circuits - Electrical Engineering - Chegg Tutors

The systemic circuit is the path of circulation between the heart and the rest of the body excluding the lungs. Blood flows from arteries to smaller arterioles and on to the capillaries. After passing through the capillaries or sinusoids, the blood is transported to venules, to veins, to the superior or inferior vena cavae, and back to the heart. During circulation, fluid gets lost from blood vessels at capillary beds and seeps into the surrounding tissues. Lymph nodes filter the fluid of germs and the fluid, or lymph, is eventually returned to blood circulation through veins located near the heart.

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This function of the lymphatic system helps to maintain blood pressure and blood volume. Share Flipboard Email. Regina Bailey. Biology Expert. Regina Bailey is a board-certified registered nurse, science writer and educator. Updated August 19, Respiratory System: The circulatory system and respiratory system make respiration possible.

Blood high in carbon dioxide is transported to the lungs where carbon dioxide is exchanged for oxygen.