Scioly.org

mjcox2000 wrote: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.

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

Cathy-TJ wrote:

mjcox2000 wrote: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? (Click to Show Hidden Text)

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

Cathy-TJ wrote: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

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].

mjcox2000 wrote:

Cathy-TJ wrote: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 (Click to Show Hidden Text)

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].

UTF-8 U+6211 U+662F wrote:Restarting... what force is exerted between two protons one meter apart?

krasabnk wrote:

UTF-8 U+6211 U+662F wrote:Restarting... what force is exerted between two protons one meter apart?

Using Coulomb's law and knowing protons are elementary charges.... F = (9x10^9)(1.6x10^-19)(1.6x10^-19) / (1)^2... Would the answer be 2.30x10-28 N?

krasabnk wrote:
Given this orientation of an operational amplifier, find the value of Rf in kΩ if the Vin = 20 mV, Rg = 3.8 kΩ, and Vout = 64 mV
(sorry for my bad drawing, haha)

8.36 kiloohms?