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At t =0, the switch is closed. Because the voltage across the capacitor cannot change instantaneously, the value of the voltage across the resistor immediately after the closing of the swith ( v (0 + ) will be 10 V. After the switch has been closed, current will flow throughout the circuit and the voltage across the resistor will diminish exponentially. Because the decay can be described by the use of an exponential function, this is an example of exponential decay .

We can write an expression for the transient response, that is, the voltage across the resistor for positive time

v ( t ) = 10 e t / RC size 12{v \( t \) ="10"`e rSup { size 8{ - t/ ital "RC"} } } {}

Here, the units of v ( t ) are Volts.

The product of the resistance and the capacitance is called the time constant for the circuit and is often denoted as (τ). The time constant is measured in seconds. For this example, τ = ( 100 × 10 3 ) ( 8 × 10 6 ) = 800 × 10 3 = 0 . 800 s size 12{τ= \( "100" times "10" rSup { size 8{3} } \) ` \( 8 times "10" rSup { size 8{ - 6} } \) ="800" times "10" rSup { size 8{ - 3} } =0 "." "800"`s} {}

So the transient response of the circuit becomes

Let us now work a problem about this circuit that involves logarithms.

Question : At what instant of time will the transent response be equal to 5 Volts?

Solution : We have been asked to find the value of t that satisfies the following equation.

5 = 10 e t / 0 . 8 size 12{5="10"`e rSup { size 8{ - t/0 "." 8} } } {}

Let us divide each side of the equation by (10) and interchange the sides

e t / 0 . 8 = 0 . 5 size 12{e rSup { size 8{ - t/0 "." 8} } =0 "." 5} {}

We now take the natural logarithm of each side

t 0 . 8 = 0 . 693 size 12{ { { - t} over {0 "." 8} } = - 0 "." "693"} {}
t = ( 0 . 8 ) ( 0 . 693 ) = 0 . 555 size 12{t= \( 0 "." 8 \) ` \( 0 "." "693" \) =0 "." "555"} {}

So we conclude that at the time 0.555 seconds after the switch closes, the value of the voltage across the resistor will be 5.0 volts.

Compound interest

It is always essential to consider the financial aspects of an engineering project. The field of industrial engineering provides us with the analytical tools necessary to weigh the merits of competing designs.

Most situations surrounding the financial aspects of an engineering project involve the determination of what is most economical in the long run. That is, engineers must be aware of the costs and benefits of a project over a considerable period of time. In situations as these, it is important to consider the time value of money. Because of the existence of interest, the current value of a dollar is worth more than the value of a dollar some time in the future.

Interest can be defined as money that is paid for the use of money that has been borrowed. The rate of interest is the ratio between the interest chargeable or payable at the end of a period of time. This period of time is typically yearly, quarterly or monthly. In this module, we will restrict our attention to interest that is paid yearly or per annum .

As an example, a sum of money is invested at an annual rate of interest of 4%. One year later, the interest that would be paid on the investment would be $40 (4% of $1,000). So after one year, the initial sum would grow to a value of $1,040 one year later.

Suppose that the $1,040 were invested for a second year at the end of the first year. At the end of the second year, the amount of interest that would be payable would be 4% of $1,040 or $41.60. The amount of interest earned in the second year exceeds the amount earned in the first year because of a phenomenon known as compound interest .

Questions & Answers

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Source:  OpenStax, Math 1508 (laboratory) engineering applications of precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11337/1.3
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