In the monostable multivibrator, the frequency of the input triggered signal is 5 KHz with 75% duty cycle. If the value of C is 0.01 micro-Farads, what is the value of external resistance (in K Ohms)?a.45.45b.42.24c.35.23d.50.9

The monostable multivibrator is a one-shot timer. It is triggered by an input signal, and the duration of the output pulse is determined by the RC time constant.

The formula for the time period (T) of a monostable multivibrator is given by T = 1.1RC, where R is the resistance and C is the capacitance.

Given that the frequency of the input triggered signal is 5 KHz, the time period (T) is the reciprocal of the frequency, which is 1/5000 = 0.0002 seconds or 200 microseconds.

The duty cycle is the percentage of one period in which a signal is active. A duty cycle of 75% means that the signal is active for 75% of the period. Therefore, the duration of the output pulse (T) is 75% of 200 microseconds, which is 150 microseconds.

Substituting the given values into the formula, we get:

150 = 1.1R0.01

Solving for R, we get:

R = 150 / (1.1 * 0.01) = 13636.36 Ohms or approximately 13.64 K Ohms.

However, none of the given options match this result. It seems there might be a mistake in the question or the provided options.