...TAX ON SHORT-TERM CAPITAL GAINS Introduction Gain arising on transfer of capital asset is charged to tax under the head “Capital Gains”. Income from capital gains is classified as “Short Term Capital Gains” and “Long Term Capital Gains”. In this part you can gain knowledge about the provisions relating to tax on Short Term Capital Gains. Meaning of Capital Gains Profits or gains arising from transfer of a capital asset are called “Capital Gains” and are charged to tax under the head “Capital Gains”. Meaning of Capital Asset Capital asset is defined to include: (a) Any kind of property held by an assesse, whether or not connected with business or profession of the assesse. (b) Any securities held by a FII which has invested in such securities in accordance with the regulations made under the SEBI Act, 1992. However, the following items are excluded from the definition of “capital asset”: i. any stock-in-trade (other than securities referred to in (b) above), consumable stores or raw materials held for the purposes of his business or profession ; ii. personal effects, that is, movable property (including wearing apparel and furniture) held for personal use by the taxpayer or any member of his family dependent on him, but excludes— (a) jewellery; (b) archaeological collections; (c) drawings; (d) paintings; (e) sculptures; or (f) any work of art. “Jewellery" includes— (a) ornaments made of gold, silver, platinum or any other precious metal or any...
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...CAPITAL GAINS TAX (CGT) PFP 14/15 Introduced in 1965 to tax assets which were not bought and sold as part of a trade. A CGT liability may arise when a chargeable person makes a chargeable disposal of a chargeable asset. CHARGEABLE PERSON • individuals resident in UK • partnerships - assessed on individual partners • companies - pay corporation tax on chargeable gains CHARGEABLE DISPOSAL • sale or gift of part or all of an asset • receipt of insurance claim following loss of chargeable asset • loss or destruction of a chargeable asset Certain disposals are not subject to CGT because of the nature of the recipient • disposals by gift to charity • disposal by gift to institutions which exist for public benefit e.g. museums • disposals between spouses: Disposals between a husband and wife (or civil partners) who live together at any time during the tax year in which the disposal occurs are chargeable disposals. However, such disposals are deemed to occur at a disposal value such that neither a chargeable gain nor an allowable loss arises on the disposal. • disposals of chargeable assets caused by death: Net capital losses incurred in the tax year in which the taxpayer dies obviously cannot be carried forward Therefore such net losses may instead be carried back and set off against net gains of the previous three tax years (later years first). CHARGEABLE...
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...SEVEN & EIGHT - CAPITAL GAINS: AN INTRODUCTION (Div B, Subdiv c , S38-55) HISTORY Prior to 1972 capital gains were not taxed in Canada and capital losses could not be claimed. The portion (inclusion rate) of a capital gain or loss which is taxable/deductible has changed since then. [calendar 1972 to 1987 => 50%; calendar 1988 to 1989 => 66.67%; calendar 1990 to Feb 27, 2000 => 75%; Feb 28 to Oct 17, 2000 => 66.67%; Oct 18, 2000 to present => 50%] Since 1972, each disposition of capital property requires a separate calculation of taxable capital gain or allowable capital loss. Section 3(b) requires that allowable capital losses be offset against taxable capital gains, except as discussed below for PUP (including LPP) and ABILs. If the net result of all current year capital dispositions is a taxable capital gain, this amount is included in Division B income. If the net result is an allowable capital loss, this amount is not deductible currently since allowable capital losses can ONLY BE DEDUCTED AGAINST taxable capital gains. (A net allowable capital loss for the current year can be carried over to other years under Division C, as will be discussed in chapter 10.) ABILs Allowable Business Investment Losses are effectively allowable capital losses, resulting from the disposition of shares or debt of Small Business Corporations (CCPCs carrying on Active Business), which are deductible against any type of income. (ie not restricted to capital gains.) As such they are identified...
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...Income Tax Law Assessment Two Question One - Capital Gains Tax The CGT consequences for the year 2006/2007 for Cynthia are as follows: Firstly, in January 2007 Cynthia incurs a CGT event A1, by selling her house. As the original acquisition of the house was contracted in November 1999, (s. 104-10(3b) if there is no contract, then when the change of ownership occurs.). The discount method can be used to form the cost base and the indexation method will be disallowed as this is only available for assets acquired before 21 September 1999. With this in mind the gain/loss will be calculated by: Capital proceeds from this event: Sale of house 600,000 Take over of loan 10,000 610,000 Cost base: Acquisition cost 450,000 Incidental acquisition costs 20,000 Capital enhancement expenditure 10,000 Council rates 1,200 481,200 Capital Gain 128,800 Then you take 50% of this nominal gain of 128,800 to calculate the capital gain figure that will be included in your tax, which becomes: 128,800/2 = 64,400 The $10,000 loan which the purchaser has agreed to take on becomes part of the capital proceeds under s116-55. Acquisition costs are at market value, of the time the acquisition took place in this case it was $450,000 as in s116-30(1) which states if you are given a CGT asset that you are taken to have received it at market value. The stamp...
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...Based on the success of the company in that time period we can assume that this strategy was working. However by the mid-eighties, when Phil Satre became the CEO, he believed that the company was losing its competitiveness. He specifically believed it happened because of the atmosphere of complacency that had been inadvertently fostered by the tradition of employees’ loyalty. In a way, this loyalty lead to stagnation, when long-standing with the company people, especially executives and managers, were holding tight to the old “tricks” and could not appreciate the need for change. As Loveman, the COO of the company, felt, the ideals of excellence and customer service were replaced with priorities of long tenure and employee happiness. This was how the company was losing to its competition. The building out of new properties was calculated to be not as beneficial as usually thought, because the effect of “newness” was not financially appreciable after just two years. What could be a real differentiating factor in re-gaining and keeping bigger market share was customer satisfaction. The goal of increasing employees’ motivation to provide outstanding customer service was the reason for introducing the gainsharing program. This program showed its effectiveness in increasing both customer satisfaction and employees loyalty. It also fostered the team building spirit within departments and on broader organizational level, because the results were based not on the individual...
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...Operational amplifiers-OPAMP Ideal- characteristics Real- you buy LM741C Characteristics 1. AOL- Open loop gain – infinite (typical value 1000000 for LM741) 2. input resistance Rin=infinite(2Mohms) 3. input current = 0 4. Rout=0(small 75ohms) 5. AOLdoes not change with the frequency of the input signal- the voltage gain is constant VCVS-voltage controlled voltage source Negative feedback Lowers the voltage gain, and also makes it constant for a bigger range frequency INVERTING AMPLIFIER USING AN OP AMP Positive Vin negative Vout Negative Vinpositive Vout (1800 out of phase) Op amp with negative feedback will have ACL(closed-loop gain)—the new circuit (op amp +Rf+R)has a new gain which is called closed-loop gain Acl=-Rf/R Vout=-(Rf*Vin)/R; Rf>R The gain of the inverting depends on two resistors, Rf and R,Rf is connected between the output pin and inverting pin and Rf must be grater than the value of R. To have an amplifier, R is connected between the input signal and the inverting pin. The negative sign in the formula for the closed loop gain means that the input signal and the ouput signal are 180 out of phase, meaning that +inputs produce-output and vice versa. The output voltage cannot go above Vcc and Vee . The output wave is clip off wave’ The output wave will be a saw wave(triangular), if the gain is too much big. IR=IF+Iin (Iin=0) IR=IF Ir=Vin/R If=Vout/ Rf Vout/Vin=Rf/R To find the formula for ACL of inverting op amp amplifiers: ...
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...For this paper I took a look at possible volunteer opportunities in my local community. I gave examples of what opportunities were available and how to find information on how to become a volunteer. I choose to use google as my research criterion and found a very resourceful website with the information I needed. As you read my paper you will find information about being a volunteer, what benefits it has, and what opportunities are available. Below I listed some great information about becoming a volunteer. As a volunteer As a volunteer, you can volunteer on your own or with your family or with a group. Volunteer your time, skills, services and talents. Opportunities are available with children, teens, adults, individuals with special needs and seniors. Volunteer your expertise and experience on a community board. Share your time as an instructor. Design a web page, troubleshoot computer problems, build a bookcase or serve as a mentor Volunteer by coordinating an event, project or publication. Volunteer by making an in-kind donation (new or used). Our partner agencies need everything from computers (hardware and software), office equipment and supplies, household goods to clothing and canned goods. There are so many ways an individual can volunteer these days. Finding one that best suits you is important. You want to be comfortable while volunteering and know your gaining a great deal of knowledge to help improve your education. Search criterion When researching for...
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...Instructions: Complete the three parts—Part 1, Part 2, and Part 3 of Lab 1. When you have completed each part, answer the questions and transcribe/transfer the test results recorded in the lab manual’s tables to the tables provided. ------------------------------------------------- Part 1—Common Emitter (CE) Amplifier Theory: 1. What type of bias is used in the common-emitter amplifier shown in Figure 3-1(a)? * Voltage Divider 2. Explain the reason(s) why the AC equivalent circuit appears as it does in Figure 3-1(b). * When an AC input signal is applied to the amplifier, the capacitors will act as shorts. This is why we call this Figure the “AC Equivalent”. 3. What values must be calculated to determine the DC operating parameters? * VB * VE * IE * VC * VCE * VCC 4. What information can be obtained by determining the DC load line of an amplifier? * We use the DC load line to graph all possible current and voltage characteristics of a biased transistor. This includes any measurements that are between saturation and cutoff, including the Q-point. 5. What values must be calculated to determine the AC operating parameters? * Vin = Vb * r’ e * Ve * Av * Vout = Vc * Rin(total) * Rac 6. What information can be obtained by determining the AC load line of an amplifier? * We use the AC load line to graph the maximum/minimum current...
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...There are five phases that a hacker will go thru when trying to attack your system. Each one is different and requires different ways to limit the hacker’s ability to gain information about your system. The first phase is reconnaissance, this can be passive or active. One of the things that a hacker might try is social engineering to gain information on the system. The best way to combat this, is by training and more training of the employees on the various ways that a hacker will attempt to get information. There is also dumpster diving, the only way to combat this is to make sure that the information that is being of disposed is of such a nature that it is useless to them. Information that they could use to gain access to the system, should be destroyed in such a way that it cannot be reconstructed in any way. Also a hacker could try sniffing the network, this is where system hardening will assist in preventing the hacker from gaining information. The second phase is scanning, in this phase the hacker will try to scan the network to see what information he can obtain to assist him in determining what weakness exist. This scanning he can find out such information as to what type of OS is being used, the version of the OS, and many other things about the network. To help prevent him from getting this information, system hardening is the best defense. This will include but not limited to disabling all ports but those that are needed, turning off certain ICMP features which...
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...“Class A Power Amplifiers” ITT Technical Institute• 3rd Quarter/December2014 •3-6-2015 Introduction The common-emitter (CE) amplifier provides high voltage gain with moderate input resistance whereas the common collector (CC) amplifier provides current gain and low output resistance. Combining the two amplifiers gives the advantages of each, allowing the amplifier to drive a relatively low resistance load such as a speaker. In all amplifiers, some power from the supply is wasted – that is, it does not show up as signal power in the load. In class A power amplifiers, the transistor is biased on at all times, causing power to be dissipated in the transistor, even when no signal is present. Because of this, class A amplifiers are not as efficient as class B designs. For low power applications, this reduced efficiency is not a major problem. Further, the power dissipated in the transistor is highest when no signal is present, so it is simple to compute the worst case power dissipated in the transistor − it is simply VCEQICQ. In this experiment, you will combine a CE and CC amplifier to form an amplifier that will be used for driving a small speaker. Most speakers are low resistance devices, requiring the driving amplifier to have a low output resistance. Because the CE amplifier typically has relatively high output resistance, a Darlington CC amplifier is selected to minimize the loading effect. In the For Further Investigation section, you can complete the amplifier...
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...1. Is there any evidence that the program pays out for Sainsbury? Elaborate. Sainsbury’s had been losing market share until partnering with Nectar and then became U.K.’s second largest supermarket chain after the first year of using the program. They also discovered weekly spending was 40% greater among people collecting points from several sponsors than only from Sainsbury’s. A study suggested that Sainsbury’s gained market share from non-Nectar retailers through gasoline sales due to the competition between BP and Sainsbury’s with Nectar. Sainsbury’s benefited from the Nectar program by: · Increasing the frequency of customer’s visits and spending more per transaction · Acquiring new customers · Retaining current customers · Up-selling to customers to buy higher margin products While Sainsbury’s was the dominant sponsor, there had been a steady decrease in its importance. In the launch month, four out of five collectors earned points from Sainsbury’s alone. Within six months, half had earned points from a second sponsor and at the anniversary 30% earned from one, 60% from two and 10% from three sponsors. Spending at Sainsbury’s was greater among collectors earning points from multiple sources than from Sainsbury’s alone. Weekly spending at Sainsbury’s was 40% greater when people collected from them and two other sponsors than from Sainsbury’s alone. It also became apparent that collectors who earned points from multiple sponsors...
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...-50dBm (g) log2 (64) = 6 (h) log2 (1000) =10 (i) log2 (496) = 9 3. A three-stage amplifier is shown with power gains. Calculate total power gain in decibels, and as a number. Ap = AP1 * AP2 * AP3 = 10 * 25 * 30 A = 7500 20log(Ap) = 77.501dB AP1 = 10 AP2 = 25 AP3 = 30 4. (a) (b) Determine the overall power gain. i. Ap = Ap1 * Ap2 * Ap3 = 10 * 10 * 20 Ap = 2000 Find the output power if Pin = 15 mW i. Ap = (Po/Pi) = 2000 ==> (Po/15*10-3) = 2000 ii. Answer: Po = 30W : Output power !2 AP1 = 10 AP2 = 10 AP3 = 20 5. If Pin = -20 dBm, determine the output power in dBm and watts. Pin = -20dB (Pin)dB = 20logPin = -20 Pin = 0.1 20log(Ap1) => Ap1 = 4.46 20log(Ap2) = 16dB => Ap2 = 6.309 20log(Ap3) = -6dB => Ap3 = +0.501 Ap = Ap1 * Ap2 * Ap3 = 4.46 * 6.309 *0.501 Ap = 14.102 Po = 1.4102 AP1 = 13 dB AP2 = 16 dB !3 AP3 = -6 dB 6. Determine: (a) Net gain (b) Net gain in dBs (c) Power output PS = 0.1W AP2 AP1 1 L = 2000 RL AP2 = 500 AP1 = 400 0 ! A. Net Gain = 100 B. Net gain in dBs = 20dB C. Power output = 10dbW or 10W 7. Determine: (a) Net gain (b) Net gain in dBs (c) Power output PS = -8 dBm AP2 AP1 1 L = 24 dB AP1 = 32 dB RL AP2 = 28 dB ! 0 !4 A. Net gain = 36dB = 3981 B. Net gain in dBs = 36dB C. Power Output = 631mW...
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...The Fruits of Legitimacy:Why Some New Ventures Gain More from Innovation Than Others Summary: New Ventures play a vital role in the development of any company.In addition,the actions of new ventures may even spur large, incumbent firms into action, thus accelerating the pace of technological change.The new ventures success depend upon their ability to produce new products.Sometimes even amongst the ventures producing the same good they have different fates.One of the difficulties faced by the new ventures is the liability of being new to the business. Potential stakeholders view firms in these industries with skepticism. An important way that new ventures can overcome the liability of newness and increase their gains from new products is by taking actions that provide them with legitimacy in the eyes of stakeholders. new ventures can gain legitimacy by creating associations with more established entities, either external or internal to the firm. Article Review: This article is written by Raghunath Singh Rao, Rajesh K. Chandy, & Jaideep C. Prabhu.In this study, the role of a variety of legitimizing actions have been highlighted and empirically tested. In order to legitimize the new ventures in the eyes of the stakeholders different actions can to undertaken.It has also been shown that these legitimizing actions may not always work together. Among internal means of gaining legitimacy, four types of actions has been proposed: historical, scientific, market, and locational...
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...= 316.228 mW (g) –30 dBm into dBrn dBrn = dBm + 90 = -30dBm+90 = 60 dBrn (h) 15 dBrn into dBm AP3 = 30 AP2 = 25 AP1 = 10 2. A three-stage amplifier is shown with power gains. Calculate total power gain in decibels, and as a number. Ap(tot) = (10)(25)(30) = 7500 Ap(tot)(dB) = 10 log(10) + 10 log (25) + 10 log (30) = 38.75 dB AP3 = 20 AP2 = 10 AP1 = 10 3. (a) Determine the overall power gain. Ap(tot) = (10)(10)(20) = 2000 (b) Find the output power if Pin = 15 mW Po = (Pin)(AP(T)) = (15mW)(2000) = 30 W AP3 = -6 dB AP2 = 16 dB AP1 = 13 dB 4. If Pin = -20 dBm, determine the output power in dBm and watts. Ap(tot) = 13dB + 16dB+ -6dB = 23 dB Ap(tot) = log-1(23/10) = 199.526 Pin = log-1(-20 dBm/10) = 0.01 mW Po = (Pin)(AP(T)) = (0.01mW)(199.526) = 1.995 mW Po(dBm) = 10 log (Po)/1mW) = 10 log (1.995mW/1mW) = 2.999 dBm PS = 0.1W AP1 AP2 AP1 = 400 L = 2000 AP2 = 500 RL 0 1 5. Determine: (a) Net gain (b) Net gain in dBs (c) Power output AP(T) = (AP1)(1/L)(AP2) = (400)(1/2000)(500) = 100 AP(T)(dB) = 10log(100) = 20 dB Po = (Pin)(AP(T)) = 0.1W*100 = 10W PS = -8 dBm AP1 AP2 AP2 = 28 dB AP1 = 32 dB L = 24 dB RL 0 1 6. Determine: (b) Net gain (b) Net gain in dBs (c) Power output AP(T) = AP1 - L + AP2 = 32dB - 24dB + 28dB = 36dB AP(T) = log-1(36/10) = 3981.07 Pin = (0.001)(log-1(-8/10) = 2551.89...
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...Homework WK4 ECET 220 Chapter 4 10. Assume a JFET has the transconductance curve shown in Figure 4–76. (a) What is IDSS? (b) What is VGS(off)? (c) What is the transconductance at a drain current of 2.0 mA? ID= 2.3 mA VGs = -2 v VD= Vpp – ID( RS+RD) 15 – 2.3m (1k +3.3 K) 15 – 2.3 (4.3 K) VDS= 5.11 2.3 m * 1 k Vs= 2.3 V 11. Assume the JFET with the transconductance curve shown in Figure 4–76 is connected in the circuit shown in Figure 4–77. (a) What is VS? (b) What is ID? (c) What is VDS? VD = VDD - ID*RD = 15V - (2.1mA) (3.3k Ω) = 8.07V ID = IS 2.1mA VS = IS * RS = 2.1mA * 1k Ω 2.1V ID = IDSS ( 1- (VGS/VGSoff))2 6.4 ( 1 - ( -2.1 / -5 )2 2.1 mA VDS = VDD - ID(RD + RS) 15V - (2.1MA) ( 4.3k Ω ) 15V - 9.03 = 5.97V 13. For each circuit in Figure 4–78, determine VDS and VGS. VD = VDD – ID times RD VS = IS times RS VDS = VDD - ID(RD + RS) VGS = VG - VS a) VD = 12V - (1mA) (4.7k Ω) = 7.3V VS = (1mA) (1k Ω) = 1V VDS = 12V - (1mA)(5.7k Ω) = 6.3V VGS = 0V - 1V = -1V b) VD = 9V - (3mA) (470 Ω) = 7.59V VS = (3mA) (100 Ω) = .3V VDS = 9V - (3mA)(570k Ω) = 7.29V VGS = 0V - .3V = -.3V c) VD = -15V - (5mA)...
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