Calculate the Resistance of the RC circuit so that LED just stays on

[rhiannynd]

In a simple series circuit there is a 12V supply battery, 12μ6 F capacitor, a Resistor, and a 2v drop LED with an imax of 30mA. frequency is 10Hz
Calculate the resistance of the resistor so that the LED just stays on.

Homework Equations

Resistance of Capacitor Xc = 1/2∏fC
P=VI
V=IR
Impedence = √R^2+Xc^2

The Attempt at a Solution

Xc = 1/2∏x10x12x10^6
=1326.3 Ω
P=vxI
P = 12 x 0.03A
=0.36w

i have no idea what else to do ! any help would be greatly appreciated please please please

 

[rude man]

A battery with 10 Hz? That's a new one on me.

 

[BvU]

So, is Rhian (PS: welcome to PF) out of trouble now ?

 

[rude man]

So, is Rhian (PS: welcome to PF) out of trouble now ?

He/she doesn't seem interested in pursuing this fascinating ac battery problem, all the 'please pleases' notwithstanding ...

 

[rezeew]

To calculate the resistance of the resistor, we can use the formula V=IR, where V is the voltage, I is the current, and R is the resistance. In this case, we know the voltage (12V) and the current (30mA), so we can rearrange the formula to solve for R:

R = V/I = 12V / 0.03A = 400 Ω

This means that the resistance of the resistor should be 400 Ω in order for the LED to just stay on. You can also double check this by calculating the power using the formula P=VI, where P is power, V is voltage, and I is current. In this case, the power should be equal to the power of the LED (0.36W) since the resistor and LED are in series.

We can also use the formula for impedance, which takes into account both the resistance and the reactance (Xc) of the circuit. The formula is Z = √(R^2 + Xc^2). In this case, the impedance should also be equal to the resistance of the resistor, since there is no other reactance in the circuit. So, Z = √(400^2 + 1326.3^2) = 1390 Ω, which is close to the value we calculated for the resistance of the resistor.

It is important to note that this calculation assumes ideal conditions and does not take into account any potential losses in the circuit. In a real-world scenario, the actual resistance of the resistor may need to be slightly higher to compensate for these losses and ensure that the LED stays on.