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Download Circuit Theory By musicmarkup.info, musicmarkup.info – A Guideline for student to understand Basic Circuits Analysis, Network Reduction and Network Theorems. Circuit Theory By musicmarkup.info- musicmarkup.info - Download as PDF File .pdf) or read online. Circuit Theory ISBN ·All rights reserved wHh Techn;col Publ; cciijons. No pon of!his book should be reproduced in any form, Electronic.

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Godse d A. Godse free with p. Godse, U A. Analog And Digital Electronics by U. Bakshi and A. Godse All www. Godse in English and here it is Torrent. Keywords: digital electronic circuits,analog communication. Electronic Circuits - I book : A. Bakshi, www. download Electronic Circuits - I online, free home delivery.

Basic Electrical and by U. Topic includes basic circuits ,laws ,basic. Author: U. Bakshi And A. Godse, U. Consider the two sources as shown in the Fig. Thus the voltage source is as shown in the Fig.

It must be noted that at the terminals open circuit voltage provided by each source must be equal as the sources are in parall2l.

The technique can be used to reduce the series combination of more than two voltage sources connected in series. The voltage sources to be connected in series must have s. The equivalent single source has a value same as 11 and These laws are of tremendous importance from network simplification point of view.

Tire cu"o't S urces to be conntcled in parallel must l! The law can be stated as. Tlu algebraic 61lm of all the currmt meeting 11t a junction point Is always zero. Another way to state the law is. The total change in potential along a closed path is always zero.

Refer to Fig. Applying KCL. Now while tracing a closed path. Once all such polarities are marked in the given circuit. The polarity of this voltage drop always depends on direction of the cu: This is called potential rise. This is called potential drop. If the branch is traced from ll! Due to these currents the various voltage drops taken place across various resistances are marked as shown in the Fig. While tracing a closed path. For example. I R while writing the equations.

As the loop is assumed to be a part of complex network. I R while writing the equation. Similarly in the Fig. Mark aU the branch currents with some assumed directions using KCL at various nodes and junction points. Across R 3 there is a drop! So while applying KVL. Battery Ez is getting traced from. Step 2: Assumed directions may be wrong.

Step 1: Draw the circuit diagram from the given information and insert all the values of sources with appropriate polarities and all the resistances. Let us trace this clo. Kept the number of unknown currents minimum as far as possible to limit the mathematical calculations required to solve them later on. We can write an equation by using KVL around this closed path as.

Then KVL must be applied to the loop bcdcb. The loop abefa should not be used for KVL application. Mark all the polarities of voltage drops and rises as per directions of the assumed branch cu. For example.. Its effect is already considered at the time of current distribution. Step 4: Apply KVL to different closed paths in the network and obtain the corresponding equations Step 5: Solve the simultaneous equations for the unknown currents.

What to do if current source exists? If tlrert is current source in tire ndwork thtn compltlt tht currtnt distribution considering tire current source But whik applying KVl.

From these currents unknown voltages and power consumption in different resistances can be calculated.. In such case. Then Cramer's rule says that form a system determinant 6 or D as Let us assume that set of simultaneous equations obtained is.

This is combined with step 2 shown in the network below in Fig.. Step 3: Mark all the polarities for different voltages across the resistances. Basic Cln: Application of Kirchhoff's law: Step 1 and 2: Draw the d rcuit with all the values which are same as the given network.

Write down lk equations refilling lk various lmmch curretlls. SOV ': Calculating 0 1. Calcula ling 0 2. Generated o. Amplitude Practically a purely sinusoidal waveform is accepted as a standard alternating waveform for alternating voltages and currents.

For I1 answer is 0. For I2 answer is 2. Let us consider the waveform of an alternating quantity which is purely sinusoidal as shown in the Fig. Fundamentals Let us discuss in brief. An alternating quantity is the one which changes periodically both in magnitude and direction. It is denoted by f and it is measured in cycles I second which is known as Hertz. A cycle can also be defined as that interval of time during which a complete set of non-repeating events or wave form variations occur containing positive as wcll as negative loops.

Such repetition occurs at regular interval of time. Such a waveform which exhibits variations that reoccur after a regular time interval is called periodic waveform. As time period T is time for one cycle i. One such cycle of the alternating quantity is shown in the Fig. After every T seconds. I In our nation.

As one cycle of an alternating quantity corresponds to 2 It radians. This Is shown in the Fig. Hence alternating voltage is expressed as. Thus E10 is called peak value of the voltage while I m is called pca.. The sinusoidally varying alternating quantity can be represented graphically by a straight line with an arrow in this method. If the projectio. Such a representation is called phasor representation of an alternating quantity.

In all the above equations. Made of the calculator should be converted to radillns. One complete cycle of a sine wave is represented by one complete rotation of a phasor. I The r. The job of adding. The relation between r. The anticlockwise direction of rotation is purely a conventional direction which has been universally adopted. The length of the line rep. This is similar to a vector representation. J2 Ir. Consider a phasor.

At point 'b'. Ialltities camrot be represented on the same plulsor diagram. Phasors are always assumed to be rotated in antlcloc: The ins tan taneous value of the current is aJso zero.

The length of the phasor is equal to the maximum value of an alternating quantity. The angular velocity ' w ' in an anticlockwise direction of the phasor should be such that it completes one revolution in the same time as taken by the alternating quantity to complete one cycle i. Consider the va. Two allemating quantities of same frequencies can be represented on same phasor diagram.

At point 'c'. At point 'e'. If frequmcies of the two qwmlities are differmt. In practice. At point 'a'. Consider alternating quantity represented in the Fig. The phase of an alternating quantity at any particula: As per above definition It is represented in tenns of angle e in radians or degrees..

In genera I. In terms of phase the equation of alternating quantity can be modified as. Let us consider three cases. Dse 3: Negative phase Q When phase of an a! This is shown in the Pig. Case 1: Dse 2: Circuit Theory a. The phase measured in anliclockwise direction is positive while the phase measun.

The phase is measured with respect to reference direction i. Kay Point: When s11ch phase diflnence bet Such two quantities arc said to have phase difference between them. Consider an e.. The two alternating quantities having same frequency. If there is difference between the phases angles of the two quantities..

This is called lagging phase difference. Their amplitudes may be different In the a. The equations of th It is possible that if one is achicv.. H ence. At all instants.

To clear this point. The current I achieves its maximum.. But in this case. X and Y rucis are not included in it. The phasor diagram drawn a t different instants are shown in the Fig. Important Points Regarding Phasor Diagram: This is just from convenience point of view. The phase of current i 1 is ' while the phase of the current i 2 is. Some of them may be current and some may be voltages or any other illtcmating quantities like flux.

The phase is always with respect the x-axis as shown in the Fig. Hence the phase difference between the two is. Two sinusoidal cu"cnts arc given by. The current i above is represented as. Instead of using above rtlations. Write the polAr fonn of tire voltage given by.

In rectangular system. So current i above. Very Important: Tire polar fonll of eqomtion or pllasc as.. The answer by analytical method is a. Rectangular system As the gt'aphical method is time consuming which includes plottin.

Such rectangular to polar and polar to rectangular conversion is often required in phasor mathematical operations like addition.. The r. To obtain phRse.

To express a voltage in polar form expres. A wltagt is defined as. Converting it to polar form. In general.. Express it in polAr form. Tlrtu to find r. Em cos Oll. Similarly for rectangular to polar conversion. But the rectangular form is not suitable to perform multiplication and division of phasors.!: An inductive reactance is the ohmic representation of an inductance denoted as XL and given by. Thus itl multiplication of complex mbers i11 polar fonn..

Hence multiplication and division must be performed using polar form of the phasors. Qz Key Point: Tlws in division of compltr 11umbers in polar form. For converting polar to rectangular form.. Let P and Q be the two phasors such that. The inductances are represented by inductive reactances in a. Now consider division of the phasors P and Q. While mu1tiplication and division.

C series circuit.. The Impedance is denoted by letter Z. But the behaviour of R.

L and C is different from each other in a. For R. Let us summarize the behaviour of R L and C in the tabular form. A capacitive reactance is the ohmic representation of a capacitance denoted as Xc and given by.

XL and Xc present in the circuit is called an Impedance of the circuit. Xc in the impedance. Copynghtcd materia It is also defined as the ratio of resistance to the impedance. For other combinations. VL and Vc is the supply voltage V. Hence power also can be obtained as. Xc v Fig. The network shown in the Fig. C circuit is shown in the Fig. Only resistance consumes power. R L c The impedance is given by.

But the veclor addition of VR.. Find voltage across capacitor.. While calculating the equivalent impedance. VL and Vc is not V as these three voltages are not in phase. Expressing it as. The applied voltage is 10 cos t. J 10 cos SOOt rv Fig. This result is applicable for 'n' such impedances connected in parallel.

Parallel Circuit A parallel circuit is one in which two or more impedances are connected in parallel across the supply voltage. Remark Xc! Circuit Theory Sr. Pure L XL L 90'! Pute R R tt. The Fig. L and C series circuits 1. Each impedance may be a separate series circuit. Applying Kirchhoff's law.

Each impedance is called branch of the parallel circuit.. Is Fig. The three impedances connected in parallel can be replaced by an equivalent circuit.

It is denoted by Y and is measured in unit siemens or mho. The sign convention for the reactance and the susceptance are opposite to each other. It is measured in the unit siemens. Impedances in parallel get 1xmoerled to admilla1JCts i11 series while impedances in series gel converted to admillar1ces in parallel. Let z. II Zz " " Z2 in redangular form 5L Hmce carculate the pawn..

Draw a neat phasor ditlgram showing voltage. The power consumed can be calculated as. The circuit is shown in the Fig. The phasor diagram is shown in the Fig. This reduces the number of unknowns and hence network can be analysed very quickly for the required result.. In s uch a case application of Star-Delta or Delta-Star transformation.

Both the connections Fig. Kirchhoffs laws give us complex set of simultaneous equations. Let us see what is star connection? U the three resistances are connected in such a manner that one end of each is connected together to form a junction point called St. It is time consuming to solve such set of simultaneous equations involving large number of unknowns.

The star point is indicated a. The Flig. These transformations allow us to replace three star connected resistances of the network. If the three resistances are connected in such a manner that one end of the first is connected to first end of second.

Star Transformation Consider the three resistances Rt2. The 1 terminals between w! R3 between the same terminals 1. Such a Star is shown inside the Delta in the Fig. Reduction Techniques ParaHel. L' a Given Delta Fig.. Is I and 2 as shown in the Fig. Between 1 and 2 the resistance is. Now terminal '3' we are not considering.. And hence we can red. Let us find equivalent resistance between 1 and 2.

We can red. R 1 and R 3 divided by the third star resistance which is R2. U all three resistances in a Star connection are of same magnitude say R. We can wr1te. Tltt tquivalf. So if we want equivalent delta resistance between terminals 3 and 1.

Result for equal resistances In star and delta: U all resistances in a Delta connection have same magnitude say R. Convert the given Dtlta in the Fig. Its equivalent star is as shown in the Fig. Star Star-Delta Equ. For star to delta conversion. Use rectangular system for addition or subtraction while polar for multiplication or division. The equivalent delta is shown in the Fig. Cm1vertthe given star into an equivala1t delta A or Find resi.

Redraw the circuit..

B B Fig. The resistances 2. Example Circuit becomes after simplification as shown in the Fig. The various branch currents are shown in the Fig. This is the current through branch AB. Determine the equiwlent resistance the circuit si10W11 in Fig. It -l: Circuit Theoty Bulc: N to find cumnts I1 and Consider the various branch currents and node voltages Fig. Converting all the voltage sources to the current sources.

Current division rule. Using source transformation. Hl 20 10 40 10 Fig. For tire circuit shown in tM Fig. Redrawing the given network we get. Apply KVL to various loops. Find the cu"ent through glllvanomder by Kircluaoffs laws. Step 2 and 3: Mark all the branch currents applying KCL at various codes and then mark the polarities of various voltages due to these assumed branch cu.. So lution: Circuit diagram from given inJormation.

I Loop 2: A Wheatstone bridge consists of the folluwing resistanas. Cu rrent through galvanometer is 7. A We can use branch current method. Applying Cramer's rul! Two batttries A and 8 luzving e. Show the branch currents and polarities.. The circuit diagram is as shown in Fig. If for th. Converting internal star to delta. Circuit Theory 1-U i. Find tire tquilllllent resislllnct betwten 8 Rrrd C. Copynghted rna. Use Kirchhoff's laws. KCL at node A. Circuit Theory R I.

Converting 10 V voltage source to c: Mark all the branch currents starting from positive of 10 V source Power supplied by 50 V source..

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