The pulse width of the output of a monostable multivibrator using IC 555 if the external components are R = 1000 KΩ and C = 10 µF is _____.

Question

Question

This question was previously asked in

RRB JE CBT 2 (Electronics Engineering) ReExam Official Paper (Held On: 04 Jun, 2025)

Answer 1

Concept:

The pulse width of a monostable multivibrator using IC 555 is given by:

\( T = 1.1 \times R \times C \)

where R is in ohms and C is in farads.

Calculation:

Given:

\(R =1000~k\Omega = 1,000,000~\Omega, C = 10~\mu F = 10 \times 10^{-6}~F\)

\( T = 1.1 \times 1,000,000 \times 10 \times 10^{-6} \)

\( T = 11~seconds \)

Answer:

The pulse width of the output of the monostable multivibrator is: 11 s