555 Timer (Monostable) — Find R1 for Given Pulse Width A 555 timer operates as a monostable multivibrator with C1 = 100 µF. Determine the resistor R1 needed to obtain a pulse width of 500 ms using the standard relation t = 1.1 * R1 * C1.

Introduction / Context:
The 555 timer in monostable mode produces a single output pulse whose width depends on an external resistor R1 and capacitor C1. Correct dimensioning of R1 for a target pulse width ensures timing accuracy in applications like debouncing and pulse stretching.

Given Data / Assumptions:

• Pulsed width target t = 500 ms = 0.5 s.
• Capacitance C1 = 100 µF = 100 × 10^-6 F.
• Monostable formula: t = 1.1 * R1 * C1.

Concept / Approach:
Rearrange the monostable formula to solve for R1: R1 = t / (1.1 * C1). Substitute the given values and compute. Choose the closest standard value among the options.

Step-by-Step Solution:
Compute denominator: 1.1 * C1 = 1.1 * 100×10^-6 = 110×10^-6 = 0.00011.Compute R1: R1 = 0.5 / 0.00011 ≈ 4545.45 Ω.Express in kΩ: ≈ 4.545 kΩ, which rounds to 4.5 kΩ.Select the nearest offered value: 4.5 kΩ.

Verification / Alternative check:
Back-calculate t with R1 = 4.5 kΩ: t ≈ 1.1 * 4500 * 100×10^-6 ≈ 0.495 s, which is within 1% of 0.5 s and typical component tolerances dominate anyway.

Why Other Options Are Wrong:

• 45 Ω or 455 Ω: Yield pulse widths near 5 ms and 50 ms, far too short.
• 455 kΩ: Produces t near 50 s, far too long.

Common Pitfalls:
Confusing µF with nF, or forgetting the factor 1.1; either error produces pulse widths off by orders of magnitude.

Final Answer:
4.5 kΩ