Circuit Lab B/C

Cathy-TJ
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Re: Circuit Lab B/C

Postby Cathy-TJ » February 26th, 2019, 6:58 pm

Calculate the resistance between terminals A and B in the infinite chain of resistors where all resistors are 1 ohm.
https://drive.google.com/open?id=1SuAGn ... q1QjmSB_6H
Answer
Since the network repeats, if its resistance is R, it consists of a 1-ohm resistor in series with a unit consisting of a parallel 1-ohm and R-ohm resistor. Solving for R, we see that [math]R=\frac{1+\sqrt5}2\Omega[/math], or assuming sufficient sig figs, about [math]1.618\Omega[/math].
Nice!
Fun Fact
The answer is also the golden ratio!
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Re: Circuit Lab B/C

Postby mjcox2000 » February 26th, 2019, 7:53 pm

Positive and negative charge carriers behave the same in almost all circuits. However, there are some circuits in which it actually makes a difference whether charge carriers are positive or negative. What is an example of a circuit in which this matters, and why does the distinction matter for this circuit?
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Cathy-TJ
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Re: Circuit Lab B/C

Postby Cathy-TJ » February 26th, 2019, 8:04 pm

Positive and negative charge carriers behave the same in almost all circuits. However, there are some circuits in which it actually makes a difference whether charge carriers are positive or negative. What is an example of a circuit in which this matters, and why does the distinction matter for this circuit?
A guess
N-gate and P-gate transistors? P gates allow flow when the signal is low (logic 0), while N-gates allow flow when the signal is high (logic 1). Might be misinterpreting this question.
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mjcox2000
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Re: Circuit Lab B/C

Postby mjcox2000 » February 27th, 2019, 4:14 am

Positive and negative charge carriers behave the same in almost all circuits. However, there are some circuits in which it actually makes a difference whether charge carriers are positive or negative. What is an example of a circuit in which this matters, and why does the distinction matter for this circuit?
A guess
N-gate and P-gate transistors? P gates allow flow when the signal is low (logic 0), while N-gates allow flow when the signal is high (logic 1). Might be misinterpreting this question.
That’s not what I was going for.

Let me try rephrasing the question: Describe a circuit in which macro-level measurements of the circuit would differ depending on whether charge carriers are electrons flowing opposite conventional current or positive charge carriers flowing in the direction of conventional current.
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Re: Circuit Lab B/C

Postby Cathy-TJ » February 27th, 2019, 12:07 pm

Positive and negative charge carriers behave the same in almost all circuits. However, there are some circuits in which it actually makes a difference whether charge carriers are positive or negative. What is an example of a circuit in which this matters, and why does the distinction matter for this circuit?
A guess
N-gate and P-gate transistors? P gates allow flow when the signal is low (logic 0), while N-gates allow flow when the signal is high (logic 1). Might be misinterpreting this question.
That’s not what I was going for.

Let me try rephrasing the question: Describe a circuit in which macro-level measurements of the circuit would differ depending on whether charge carriers are electrons flowing opposite conventional current or positive charge carriers flowing in the direction of conventional current.
Answer
A circuit that uses Hall effect, since changing the charge carrier would reverse the voltage produced.
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mjcox2000
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Re: Circuit Lab B/C

Postby mjcox2000 » February 27th, 2019, 12:33 pm

A guess
N-gate and P-gate transistors? P gates allow flow when the signal is low (logic 0), while N-gates allow flow when the signal is high (logic 1). Might be misinterpreting this question.
That’s not what I was going for.

Let me try rephrasing the question: Describe a circuit in which macro-level measurements of the circuit would differ depending on whether charge carriers are electrons flowing opposite conventional current or positive charge carriers flowing in the direction of conventional current.
Answer
A circuit that uses Hall effect, since changing the charge carrier would reverse the voltage produced.
Right! Your turn.
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Re: Circuit Lab B/C

Postby Cathy-TJ » March 1st, 2019, 3:12 pm

When and why would you use a Kelvin Double Bridge?
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Re: Circuit Lab B/C

Postby Things2do » March 1st, 2019, 3:18 pm

To measure unknown resistors under 1 Ohm?
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Cathy-TJ
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Re: Circuit Lab B/C

Postby Cathy-TJ » March 1st, 2019, 8:29 pm

To measure unknown resistors under 1 Ohm?
That's right!
This is because
the resistance of the wires becomes significant
You have the next question!
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Re: Circuit Lab B/C

Postby mjcox2000 » March 13th, 2019, 8:17 am

Since no one has posted a question yet:

An LED with voltage drop and internal resistance is in series with a voltage and resistance . The LED’s maximum rated current is , and the minimum current at which its light is visible is .

1. What is the minimum series voltage for the LED to emit visible light?
2. Given the value of , write equations for the minimum and maximum value of for the LED to emit visible light without burning out.
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Cathy-TJ
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Re: Circuit Lab B/C

Postby Cathy-TJ » March 13th, 2019, 3:37 pm

Since no one has posted a question yet:

An LED with voltage drop and internal resistance is in series with a voltage and resistance . The LED’s maximum rated current is , and the minimum current at which its light is visible is .

1. What is the minimum series voltage for the LED to emit visible light?
2. Given the value of , write equations for the minimum and maximum value of for the LED to emit visible light without burning out.
Answer?
Voltage = [math]10^-^4 (R+2)+3.2[/math] Volts Min R = [math][(V-3.2)/10^-^2]-2[/math] Ohms Max R = [math][(V-3.2)/10^-^4]-2[/math] Ohms
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mjcox2000
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Re: Circuit Lab B/C

Postby mjcox2000 » March 13th, 2019, 4:58 pm

Since no one has posted a question yet:

An LED with voltage drop and internal resistance is in series with a voltage and resistance . The LED’s maximum rated current is , and the minimum current at which its light is visible is .

1. What is the minimum series voltage for the LED to emit visible light?
2. Given the value of , write equations for the minimum and maximum value of for the LED to emit visible light without burning out.
Answer?
Voltage = [math]10^-^4 (R+2)+3.2[/math] Volts Min R = [math][(V-3.2)/10^-^2]-2[/math] Ohms Max R = [math][(V-3.2)/10^-^4]-2[/math] Ohms
Looks good. Your turn!
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Re: Circuit Lab B/C

Postby Cathy-TJ » March 14th, 2019, 6:17 pm

What are the values of R3 and R4 if you want the battery to supply 83.3 mA to the circuit while the ammeter reads 0.00mA?

Image:https://drive.google.com/open?id=1_cVQz ... GLcoC9kaRg
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Re: Circuit Lab B/C

Postby mjcox2000 » March 14th, 2019, 6:45 pm

What are the values of R3 and R4 if you want the battery to supply 83.3 mA to the circuit while the ammeter reads 0.00mA?

Image:https://drive.google.com/open?id=1_cVQz ... GLcoC9kaRg
Answer
The 0 current condition requires that [math]\frac{R_3}{R_4}=\frac{20}{280}[/math], and the 83.3mA current condition requires that the total parallel resistance be [math]180\Omega[/math], so [math]R_3+R_4=450\Omega[/math], so [math]\boxed{R_3=30\Omega,\ R_4=420\Omega}[/math].
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Re: Circuit Lab B/C

Postby Cathy-TJ » March 14th, 2019, 6:54 pm

What are the values of R3 and R4 if you want the battery to supply 83.3 mA to the circuit while the ammeter reads 0.00mA?

Image:https://drive.google.com/open?id=1_cVQz ... GLcoC9kaRg
Answer
The 0 current condition requires that [math]\frac{R_3}{R_4}=\frac{20}{280}[/math], and the 83.3mA current condition requires that the total parallel resistance be [math]180\Omega[/math], so [math]R_3+R_4=450\Omega[/math], so [math]\boxed{R_3=30\Omega,\ R_4=420\Omega}[/math].
You're correct!
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