We are given an input current Iin=100μA and a linear resistive load represented by the resistor RL . Our resistive load is not regulated for the high current provided by our current source, so we add a resistor RS in parallel to divide the current such that IL≈40μA. An additional requirement is that the Thevenin resistance as seen from the load terminals is between 60kΩ and 80kΩ. Assume first that the resistors have their nominal resistance. Come up with resistors RS and RL such that the divider ratio IL/Iin is within 10% of the requirement.

Of course, the resistances you chose are just nominal. Given that they are only guaranteed to have resistances within 10% of the nominal value, what is the largest and smallest value that IL may have?

ILmax and ILmin are wrong!

Enter your values below. (NOTE: Both resistor values are needed for a correct answer.)

RS (in Ohms):

RL (in Ohms):

ILmax (in Amps):

ILmin (in Amps):

RS (in Ohms): 100000

RL (in Ohms): 150000

ILmax (in Amps): 4.4e-5

ILmin (in Amps): 3.6e-5

Type all in at once, it does not accept one by one or errors!

Excelente respuesta, creo que podemos ayudarnos mediante esta pagina. La colocare en favoritos en mi navegador

Por favor ayudame con la respuesta en el segundo punto donde estan las resistencias optimas de 6.6 ohms. me falta la potencia, cual seria en ese caso. Hablo de la H2P2

Hi Spanish friend, can you write in English? I had to translate it online and could not understand it.

The answer is:
6.3
0.0137
6.3

Anyone for:

Consider a deuteron in a cyclotron with field strength 0.5T. The deuteron is accelerated twice per rotation by a potential of V=25 kV. (a) If the radius of the cyclotron is 2 meter, what is the maximum energy of the deuteron? Express your answer in Joules (the deuteron mass is 3.34×10−27kg) b)Starting from a negligibly small velocity, how many full rotations does the deuteron need before it reaches this maximum energy? c) What is the time it takes for the deuteron to make one complete rotation when its energy is about 500 keV and when it is about 5 MeV? Ignore possible relativistic effect

Consider a deuteron in a cyclotron with field strength 0.5T. The deuteron is accelerated twice per rotation by a potential of V=25 kV. (a) If the radius of the cyclotron is 2 meter, what is the maximum energy of the deuteron? Express your answer in Joules (the deuteron mass is 3.34×10−27kg) b)Starting from a negligibly small velocity, how many full rotations does the deuteron need before it reaches this maximum energy? c) What is the time it takes for the deuteron to make one complete rotation when its energy is about 500 keV and when it is about 5 MeV?

A current I travels counterclockwise through a closed copper wire loop which has the shape of a rectangle with sides a and b.

What is the magnitude of the magnetic field at the center, C , of the rectangle? Express your answer in terms of a, b, I and mu_0. (Enter mu_0)

To solve this problem, we need to find the values of resistors RS and RL that will give us a current divider ratio (IL/Iin) within 10% of the requirement, while also ensuring that the Thevenin resistance falls between 60kΩ and 80kΩ.

Let's start by finding the nominal values of RS and RL that will give us the desired current divider ratio. The current divider equation is given by:

IL/Iin = RS/(RS + RL)

We are given that IL should be approximately 40μA. Substituting the given values, we have:

40μA/100μA = RS/(RS + RL)

0.4 = RS/(RS + RL)

Solving for RL in terms of RS, we can rewrite the equation as:

RL = (RS/0.4) - RS

Now, let's consider the Thevenin resistance. The Thevenin resistance (RT) is given by the parallel combination of RS and RL, which is calculated as:

1/RT = 1/RS + 1/RL

We want the Thevenin resistance to be between 60kΩ and 80kΩ. Substituting the values, we have:

1/60kΩ ≤ 1/RS + 1/RL ≤ 1/80kΩ

Let's solve these equations:

For the largest value of IL:

Taking RS = 10kΩ (nominal), we can substitute RS = 10kΩ in the equation RL = (RS/0.4) - RS:

RL = (10kΩ/0.4) - 10kΩ = 15kΩ

Using RL = 15kΩ and RS = 10kΩ, we substitute the values in the Thevenin resistance equation:

1/RT = 1/10kΩ + 1/15kΩ

Solving this equation gives us:

1/RT = 0.1 + 0.0667

1/RT = 0.1667

RT = 6kΩ

Therefore, the largest value of IL will occur with RL = 15kΩ and RS = 10kΩ, where RT is equal to 6kΩ.

For the smallest value of IL:

Taking RS = 20kΩ (nominal), we substitute RS = 20kΩ in the equation RL = (RS/0.4) - RS:

RL = (20kΩ/0.4) - 20kΩ = 30kΩ

Using RL = 30kΩ and RS = 20kΩ, we substitute the values in the Thevenin resistance equation:

1/RT = 1/20kΩ + 1/30kΩ

Solving this equation gives us:

1/RT = 0.05 + 0.0333

1/RT = 0.0833

RT = 12kΩ

Therefore, the smallest value of IL will occur with RL = 30kΩ and RS = 20kΩ, where RT is equal to 12kΩ.

In summary, the largest value of IL will be when RL = 15kΩ and RS = 10kΩ, resulting in a Thevenin resistance (RT) of 6kΩ. The smallest value of IL will be when RL = 30kΩ and RS = 20kΩ, resulting in a Thevenin resistance (RT) of 12kΩ.