eNote.Page

  Originally published on 4–May–2013   FARADAY and ELECHEM are the two portable computer programs, find useful for electrochemical researchers, electroplating units, battery as well as supercapacitor industries, process engineers, materials scientists & engineers, electroplaters, hobbyists, and students to evaluate various electrochemical deposition parameters.   These computational tools can be used to simulate various electrochemical deposition data or electroplating parameters for metals, alloys, composites, and some of them are listed below:   ·        cathode / cathodic current efficiency on electro plating ·        energy required for element or metal electrodeposition ·        current density calculations ·        estimation of energy efficiency ·        Hull cell data for the given current ·        throwing power of the chemical bath based on Harring Blum cell ·        applied current required for plating of metals and alloys ·      

Brief note on definite integrals Definite integrals are used to find the area under the given limits ‘a’ and ‘b’ for a differential derivative function. Though many rules are there to determine definite integrals for every specific function, basic algebraic function is given.  For example, if ‘y’ as a function of ‘x’ or f(x) is given as, where ‘a’ is lower limit, ‘b’ is upper limit and n ≠ –1.   Find the definite integral between the limits 1 to 4 for the function: y = 2x 2 + 3x. On solving this, the value is 64.5.   Now applying this concept to charging a battery... a battery is charged under constant voltage by varying the current (C) in Amperes. The current passed with reference to time (t) in hours is governed by the equation, C = 0.5t 2 + 0.5t. Estimate the charge (Ah) given from 2.5 hours to 4.5 hours. (1 Ah is One Ampere given in one hour.)   The problem is represented by the following figure shown below. This figure shows the curre

BOX UP puzzle game: Classic, yet interesting mind game for logical thinking. How to play? > Goal: Put the small red box inside the big blue box. > For this, move the boxes by pushing from inside. > Small box (black or red) inside a large box can be pushed together. > Play in full screen mode. Solution to Level 4 Solution to Level 5 Solution to Level 6 Note: Number of moves may be reduced.... Play (Desktop version only)

What is Haircut in stock market? Haircut is the amount of margin money deducted while pledging the Equites/ Mutual Funds / Bonds for trading. A haircut value of 13% means that, if a share worth of Rs 100 is pledged for trading, then Rs. 13 will be deducted and Rs. 87 will be given as collateral margin for trading. It is based on previous closing price.   This program is used to identify the real worth or money value of an equity / share / sovereign Gold Bonds (SGB) at the selected date.   This program fetches the historical Haircut data (%) and Price of Equities listed in NSE, India at the specified date.   Enter the Date (in the format: dd-mm-yyyy) to get Haircut data for equities listed in NSE. It lists the haircut value (%) and the respective price of the equity at the given date.   Enter the symbol for equity listed in NSE. For example, Infosys should be entered as INFY.   Note: If the date falls on holiday, no results will be displaye

Formula to calculate the compound interest from fixed deposit is: Final value = Principal × [1+(r/n)]^(n×t) Principal refers the invested amount. ‘n’ is compounding frequency, generally the value is 4 (quarterly for banks). ‘t’ is total period in years. ‘r’ is the annual interest rate.   Rate of return (in %) is calculated as: [(Final value – Initial value) / Initial value] × 100   Compound Annual Growth Rate (CAGR)is used to measure the rate of return in a given period from an investment. It is a measure of the average yearly growth rate. Formula to estimate CAGR is: CAGR (%) = {[(Final value/Principal)^(1/t)] - 1} × 100 t = Number of years (Period).   Find the return and CAGR, if $100 is invested in fixed deposit at 7% interest rate for a period of 5 years.   n = 4((quarterly basis) t = 5 years r = 7/100 = 0.07 Principal = 100 Maturity (Final) value is calculated as: = 100 × [1+(0.07/4)]^(4×5) = 141.47 So, the return after 5

Brief note on definite integrals Definite integrals are used to find the area under the given limits ‘a’ and ‘b’ for a differential derivative function. Though many rules are there to determine definite integrals for every specific function, basic algebraic function is given.  For example, if ‘y’ as a function of ‘x’ or f(x) is given as, where ‘a’ is lower limit, ‘b’ is upper limit and n ≠ –1.   Find the definite integral between the limits 1 to 4 for the function: y = 2x 2 + 3x. On solving this, the value is 64.5.   Now applying this concept to find the area under a curve. The given data (x and y) and the corresponding curve is given below:   x      y 0.0    0.000 0.1    0.146 0.2    0.284 0.3    0.414 0.4    0.536 0.5    0.650 0.6    0.756 0.7    0.854 0.8    0.944 0.9    1.026 1.0    1.100 1.1    1.166 1.2    1.224 1.3    1.274 1.4    1.316 1.5    1.350 1.6    1.376 1.7    1.394 1.8    1.404 1.9    1.

Quadratic equation is given in the form: ax 2 + bx + c = 0, where ‘a’, ‘b’ and ‘c’ are constants and a ≠ 0. The solution to this quadratic equation is given by the following quadratic formula: Here b 2 – 4ac is called as discriminant. If b 2 – 4ac > 0, then the equation has two solutions. If b 2 – 4ac = 0, then the equation has one solution. If b 2 – 4ac < 0, then the equation has no real solutions. Using quadratic equation, it is possible to find the value of ‘x’, if ‘a’, ‘b’ and ‘c’ are known. For example, consider a 2 nd order polynomial equation: y = ax 2 + bx + c. By using quadratic formula, it is possible to determine the value of ‘x’ data when, y = 0. If y = 0, then 2 nd order polynomial equation becomes a quadratic equation as, 0 = ax 2 + bx + c when a ≠ 0. Example #1 In an experimental analysis, for a set of ‘x’ and ‘y’ data, the derived 2 nd order polynomial curve fit equation is:  y = 2x 2 – 9x + 10. Determine t