Question

At t_0 the door opens, and K_1 closes at t_0 = 0 s. The initial voltage on capacitor C_1 is 0V ( C_1 is discharged). If the proximity sensor is still active after some time at t_1, meaning that the door is still open, the circuit:

- Turns on the LED (at the same time t_1)
- Current for LED light to be visible (limit low) = 1 mA
- Current for LED to operate (full bright) = 10 mA

**Assumptions:**

- Assume constant if the LED is forward-biased based on the offset model.

Answer 1

When a capacitor is uncharged and connected to a resistor and battery, its voltage will rise rapidly. With a proximity sensor indicating the door is open, the LED can be turned on, and the circuit's resistor value should be chosen to achieve desired LED currents.

Step-by-step explanation:

When a capacitor is initially uncharged and connected to a resistor and a battery, the voltage on the capacitor will rise rapidly at first. The voltage across the capacitor can be described by the equation V = emf(1 - e-t/RC) during the charging process. After some time, when the proximity sensor indicates that the door is still open, an LED can be turned on in the circuit.

To limit the current for the LED light to be visible, a current of at least 1 mA is required. To operate the LED at its full brightness, a current of 10 mA is needed. These current values should be achieved by appropriately choosing the resistor value in the circuit.

It's important to note that if the LED is forward-biased based on the offset model, the LED current is mainly determined by the resistor value in the circuit.