Linear Circuit Analysis: Basic Concepts - Work, Power,
Voltage, and Resistance
Determine the energy required to move 10
Coulombs of charge through 120 volts. Irwin. page 17 prob 1.7 Energy and Voltage are
related to Charge by the relationship.
Let's add this equation to the formula sheet
and solve the problem using the "with" operator.
Split the screen with the Text Editor on top and the Home
Screen Underneath.
Enter the equation on the next available line.
Press [F2] [ENTER] to add a Command character.
Press [F4] to transfer the equation to the Home Screen.
Press [2nd] [APPS] to select the Home Screen.
Scroll to the end of the Entry Line.
Press [2nd] [k] to add the "with" operator. Then enter q=10
and v=120.
Press [ENTER]
The Charge entering the positive terminal of an
element is given by the expression q = -30e-4t
mC. The Voltage to the element is defined as v =
120e-2t V. Find the Energy delivered to the
element in the time interval 0 < t < 1.5 msec. Irwin. page 18 prob 1.10
This problem requires two steps and the addition of another
formula. The new formula is the Integral form of Energy.
First lets add this to the formula Text.
Step one in solving this problem involves solving for I (The
Current) From the Charge.
On the Home Screen, enter the expression for Q and store it in
Q.
Press [2nd] [APPS] to go to the Text Editor and select
the differential form of Current.
Press [F4] to send the equation to the Home Screen and
evaluate it.
From the Home Screen, Store this expression in I.
Store the expression for the Voltage in V.
Press [2nd] [APPS] to go to the Equation Editor.
Select the Equation for the Integral form of Energy and press
[F4].
Press [2nd] [APPS] to go back to the Home Screen.
Add the limits to the Integral on the Entry Line.
Press [ENTER]
Two elements are connected in series. Element 1
supplies 24W of Power. Is element 2 supplying or absorbing
power? Irwin. page 18 prob 1.15
The newest equation to add to our formula text relates Power (in
Watts) to Current(in Amps) and Voltage.
Taking this equation and solving for the
current through Element 1, shows that the current is 4 Amps (from +
to - ) in the opposite direction of the arrow. Therefore we say that
the current is -4 Amps through Element 1. Since these elements are
in series, this is the same current through Element 2. The direction
of the current is going against the signs of Element 2 (from - to +
) We say that this element is absorbing or using Power.
Add the new formula to the Text Editor and press [2nd]
[APPS] to select the Home Screen.
Solve the new formula for I (Current).
Evaluate for the current through the first element using the
"with" operator ( [2nd] [k] ).
Note that the voltage of Element 1 is negative with respect to
the current on the circuit diagram.
Solve the equation for Element 2 using the current found from
Element 1 and the voltage given in the
problem.
The input current to the circuit is I(t)=20
cos(2 Pi t). Graph the Power delivered as a function of time on
the interval 0 < t < 3. DeCarlo. page 35 prob 37
The next formula to add to our text relates Power to Current and
Resistance.
Add the new equation to the formula text with a Command Mark.
Press [2nd] [APPS] to select the Home Screen.
Store the values of 10 to the variable R, And 20cos(2Pi*x) to
the variable I.
Press [2nd] [APPS] to select the Text Editor
Press [F4] To evaluate the expression. This value is
the instantaneous Power.
Press [2nd] [APPS] to select the Home Screen.
On the Entry line enter, Graph I^2*r,x to graph the function.
To Graph this function on the interval of 0 < t < 3 .
Press [Diamond] WINDOW to open the window variables
editor.
Set xmin=0
Set xmax=3
Press [F2] (Zoom) [a] (ZoomFit).
What is the Power absorbed by the 10K
resistor? Irwin. page 24 example
2.2b
This problem gives us the equation relating Power to Voltage and
Resistance.
After adding this equation to the formula text, we can solve for
the Power.
Enter the new equation to the Text Editor With a command line
and press [F4].
Press [2nd] [APPS] to select the Home Screen.
Enter the values for V and R.
Scroll back through the Home Screen history.
Select the new equation and press [ENTER]
twice.
Now we can find the Current for this circuit as well.
Solve the Power equation relating Voltage and Current, for
Current.
Press [2nd] [ANS] [2nd] [k] ("with" operator).
Scroll Back through the Home Screen history until the p=.0036
section is highlighted.
Press [ENTER] twice.
The Current could have been
solved for just as easy by using the well known equations for Ohm's
Law.
We will put the main one in the Text Editor
for safe keeping and consistancy.
