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Three Electronics ProblemsDate: 10/31/2000 at 13:51:55 From: Don Waild Subject: Electronics 1. An RC series circuit has a voltage source of 10 Vac, a resistor of 10 ohms, and a capacitor with an impedance of -j10 ohms. What are the magnitude and phase angle of the current flow through the capacitor? 2. A 4:1 step-down transformer uses a 60 Vac source on the primary windings. 120 mA is flowing through the load resistance on the secondary windings. What is the resistive load? 3. The outputs of two "nand" gates are connected to the inputs of an "exclusive-or" gate. Each of the "nand" gates has one input at a logic high level, and the other input at a logic low level. What is the output of the "exclusive-or" gate? I would deeply appreciate the help. Thank you.
Date: 11/01/2000 at 10:51:46
From: Doctor TWE
Subject: Re: Electronics
Hi Don - thanks for writing to Dr. Math.
>1. An RC series circuit has a voltage source of 10 Vac, a resistor of
>10 ohms, and a capacitor with an impedance of -j10 ohms. What is the
>magnitude and phase angle of the current flow through the capacitor?
Think of the problem as a right triangle, with the length of the
triangle being R, the height being Xc, and the hypotenuse being Z (see
diagram).
R = 10
+----->
\a |
\ |
Z \ | Xc = --j10
\ |
\|
V
We can see that the magnitude of the impedance comes from the
Pythagorean theorem, and is:
Z^2 = R^2 + Xc^2
Z = sqrt(R^2 + Xc^2)
The magnitude of the current is found by Ohm's law:
I = E / Z
To find the phase angle (a), we can use any of the basic trigonometric
functions:
sin(a) = Opp / Hyp = Xc / Z
cos(a) = Adj / Hyp = R / Z
tan(a) = Opp / Adj = Xc / R
I'd probably use the tangent function, since R and Xc are given while
Z had to be calculated (one less chance of error in this calculation).
>2. A 4:1 step-down transformer uses a 60 Vac source on the primary
>windings. 120 mA is flowing through the load resistance on the
>secondary windings. What is the resistive load?
Remember that "step-down" means that Vs < Vp, and the ratio is 4-to-1.
From that, you can find the voltage on the secondary. (There's 1 volt
on the secondary for every 4 volts on the primary.) Now you have the
voltage and current (given) in the secondary, so use Ohm's law:
E = I * R
R = E / I
>3.) The outputs of two "nand" gates are connected to the inputs of
>an "exclusive-or" gate. Each of the "nand" gates has one input at a
>logic high level, and the other input at a logic low level. What is
>the output of the "exclusive or" gate?
The easiest way is to draw a schematic diagram.
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