{"id":17675,"date":"2022-09-13T04:42:11","date_gmt":"2022-09-13T04:42:11","guid":{"rendered":"https:\/\/www.booksofall.com\/tw\/?post_type=product&p=17675"},"modified":"2022-09-13T04:42:11","modified_gmt":"2022-09-13T04:42:11","slug":"electric-circuits-v-references-kuphaldt","status":"publish","type":"product","link":"https:\/\/www.booksofall.com\/tw\/electric-circuits-v-references-kuphaldt\/","title":{"rendered":"Electric Circuits V – References (Kuphaldt)"},"content":{"rendered":"

If you are involved with electronics<\/a> and\/or electricity, you will find that despite the many complex circuits and equations that exist, you will frequently use the few most common equations and circuit configurations<\/a>. For example, V = IR is arguably the most important equation in electronics, followed by Kirchhoff\u2019s current and voltage laws. But there will occasionally be situations where you find yourself needing an uncommon component (such as a gas discharge tube) or equation (e.g., the relationship between inductance and the physical characteristics of a coiled wire). Maybe you, like many others, just need a conversion chart for a resistor value or wire colour codes for power distribution cables. Here you will find many references for situations such as these, ranging from DC circuit equations and unit conversions to derivatives and troubleshooting techniques.<\/p>\n

Wheeler\u2019s formulas for inductance of air core coils<\/a> which follow are useful for radio frequency inductors. The following formula for the inductance of a single layer air core solenoid coil is accurate to approximately 1% for 2r\/l < 3. The thick coil formula is 1% accurate when the denominator terms are approximately equal. Wheeler\u2019s spiral formula is 1% accurate for c>0.2r. While this is a \u201cround wire\u201d formula, it may still be applicable to printed circuit spiral inductors at reduced accuracy.<\/p>\n

Locate the row corresponding to known unit of torque along the left of the table. Multiply by the factor under the column for the desired units. For example, to convert 2 oz-in torque to n-m, locate oz-in row at table left. Locate 7.062 x 10 at intersection of desired n-m units column. Multiply 2 oz-in x (7.062 x 10 ) = 14.12 x 10 n-m.<\/p>\n

Converting between units is easy if you have a set of equivalencies to work with. Suppose we wanted to convert an energy quantity of 2500 calories into watt-hours. What we would need to do is find a set of equivalent figures for those units. In our reference here, we see that 251.996 calories is physically equal to 0.293071 watt hour. To convert from calories into watt-hours, we must form a \u201cunity fraction\u201d with these physically equal figures (a fraction composed of different figures and different units, the numerator and denominator being physically equal to one another), placing the desired unit in the numerator and the initial unit in the denominator<\/a>, and then multiply our initial value of calories by that fraction.<\/p>\n

Since both terms of the \u201cunity fraction\u201d are physically equal to one another, the fraction as a whole has a physical value of 1, and so does not change the true value of any figure when multiplied by it. When units are canceled, however, there will be a change in units. For example, 2500 calories multiplied by the unity fraction of (0.293071 w-hr \/ 251.996 cal) = 2.9075 watt-hours.<\/p>\n

The \u201cunity fraction\u201d approach to unit conversion may be extended beyond single steps. Suppose we wanted to convert a fluid flow measurement of 175 gallons per hour into liters per day. We have two units to convert here: gallons into liters, and hours into days. Remember that the word \u201cper\u201d in mathematics means \u201cdivided by,\u201d so our initial figure of 175 gallons per hour means 175 gallons divided by hours. Expressing our original figure as such a fraction, we multiply it by the necessary unity fractions to convert gallons to liters (3.7854 liters = 1 gallon), and hours to days (1 day = 24 hours). The units must be arranged in the unity fraction in such a way that undesired units cancel each other out above and below fraction bars. For this problem it means using a gallons-toliters unity fraction of (3.7854 liters \/ 1 gallon) and a hours-to-days unity fraction of (24 hours \/ 1 day): Our final (converted) answer is 15898.68 liters per day.<\/p>\n","protected":false},"excerpt":{"rendered":"