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#### TokMor

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« on: August 22, 2003, 01:56:17 PM »

Since a knowledge of electronics is usefull when you deal with computers,hopefully this thread will give people a better understanding, and be fun. I will post questions dealing with electronics so that others can show what they know, and thoes who don't can learn from thier answers.
That being said, when you answer, please also describe how you arrived at your answer, so if I ask "If there is 1mA running through a 5Kohm resistor, what is the voltage across it" A correct answer would be "Since V=IR, V=1mA*5kOhm=5V" Rather than just saying 5.
The first person to correctly answer with suficient explination gets a point
(note: points will not get you anything more than the esteem and envy of others on GT, and I do not gaurentee that you will recieve anything other than that.)
Anyway, first question:
I need a 75 ohm resistor, but I only have 50 ohm resistors. How can I make a circuit with total resistance of 75 ohms with only 50 ohm resistors?
(assume connecting wires have no resistance and also provide a brief explination of your circuit) Logged

#### Styles

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« Reply #1 on: August 22, 2003, 02:49:46 PM »

ok i would put 2 50 ohm resistors in parallel, making a total resistance of 25 ohms>

then i would wire another 50 ohm resistor in series.

25 + 50=75  Logged

#### TokMor

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« Reply #2 on: August 22, 2003, 03:04:04 PM »

correct, one point to Styles.
Question 2:
If we put 7.5 volts across Style's circuit, how much current would be going through each resistor? Logged

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« Reply #3 on: August 22, 2003, 04:25:47 PM »

Since the total resistance it 75 ohms and the voltages is 7.5V, we use this formula I=V/R and get 0.1 amps for the total circuit.  In Styles ciruit, we have resisters in parallel, we will call these R1 and R2.  The resister in series will be called R3.  Since R3 is in series with the paralle resistors and the source it will recieve the full current of 0.1A and have a voltage drop of 5.0V.  That leaves the voltage drop on the R1/R2 resistors to be 2.5V (7.5-R1). Since R1/R2 are in parallel they have the same voltage drop across both of them.  This leaves the the currrent for the R1 and R2 resistors to be .05A, (I=V/R, I=2.5V/50 Ohm).

So in summery:
V= 7.5
I=7.5V/75Ohms = 0.1A
R3=0.1A
Vr3= 0.1A * 50 Ohm = 5.0V
Vr1r2= V-Vr3 = 7.5-5.0 = 2.5V
Vr1r2=Vr1=Vr2
Ir1=Vr1/R = 2.5V/50 Ohms = 0.05A
Ir2=Vr2/R = 2.5V/50 Ohms - 0.05A Logged

#### TokMor

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« Reply #4 on: August 22, 2003, 08:41:22 PM »

very detailed explanation, and very correct.  Shadowhawk gets the point for this one.
The next question in this sieries would be: what is the minimum power the resistors must be rated for in order for the circuit to operate correctly? Logged

#### TemplaraPheonix

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« Reply #5 on: August 22, 2003, 09:23:20 PM »

Power equals VxI. Since a different voltage will pass over the two types of resistors and thus produce different currents due to V=IR, we need to figure out both types seperately. Power is measured in Watts.

The ones in parallel need to be 2.5x0.05 or 0.125 Watt, and the one in series needs to be 7.5x0.1 or 0.75 Watt. Since they don't generally sell 0.125 or 0.75 Watt resistors, I'd say looking for 1 Watt. Logged

#### TokMor

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« Reply #6 on: August 22, 2003, 10:49:57 PM »

I belive you miscalculated something in the second part Logged

#### viridius

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« Reply #7 on: August 23, 2003, 12:49:36 AM »

One half-watt (5 volts x 0.1 amps) and two eighth-watt (2.5 volts x 0.05 amps). Logged
"I'm totally going to send ninjas after you." - BalefireX

#### TokMor

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« Reply #8 on: August 23, 2003, 08:06:15 PM »

correct, viridius gets the point.
TemplaraPheonix, you forgot that the first resistor does not get the full 7.5 volts because it is in sieries with the other two.

If I have a circuit that has a 9 volt battery in sieries with a 1Kohm resistor and a 1 mili Farad capacitor, after a long time what would the voltage across the capacitor be? Logged

#### Infindense

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« Reply #9 on: August 23, 2003, 09:04:08 PM »

It has been awhile since electronics class so I had to look through my old book.  I am not sure if this is right, I was never that good at electronics anyway, but maybe I'll learn something.

The sum of the dc voltages across the capacitor and the resistor equals the source voltage (Vs) at all times.
As the capacitor is charged, the voltage across the capacitor is series- opposing the source voltage.  Therefore, the voltage across the resistor at any given moment equals the source voltage minus the capacitor voltage.  That is, Vr = Vs - Vc.  Thus, as the capacitor charges to the Vs value, the resistor voltage decreases toward zero volts.  When the capacitor is fully charged, Vc equals Vs and zero current flows, and no IR drop occurs across the resistor.  When the switch is initially closed, the capacitor has zero charge and zero voltage.  Therefore, the source voltage is dropped by the resistor (R).  Since this is a series circuit, and the charging current is the same throughout a series circuit, Ohm's law reveals what the charging current is at the moment.
I = Vr/R

So I am guessing that the voltage across the capacitor will be eqaul to the source, 9v. Logged
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