Question

A monostable multivibrator has r = 120k and the time delay t = 1000ms, calculate the value of c

Answer 1

Time delay for a monostable multivibrator, T = 1.1RC

=> C = T/(1.1R) = 1000ms/(1.1×120kΩ) = 7.57µF.

Answer 2

Answer:

From the given question we should calculate the capacitor(c) of the monostable multi vibrator with the help of given resistance(r) and time delay(t).Final solution of the capacitor(c) $=7.57$μ$F$

Step-by-step explanation:

The Output Time Delay for monostable multi vibrator $T=1.1$ × $R$ × $C$

here, $T$ is the time delay and the unit is milliseconds($ms$) or second($s$) or minute($m$) or hour($H$).

        $R$ is the resistance and the unit is ohms(Ω) or Kilo Ohms($k$Ω) or Mega Ohms($m$Ω).

        $C$ is the capacitor and the unit is farads($F$) or Nano Farad(η$F$) or Micro Farad(μ$F$).

From the given data, we have resistor value($r$) and time delay($t$). So the formula for the capacitor($c$) is,

              $C=T/1.1$ × $R$

Already we have the following values $T = 1000ms$ and $R=120k$Ω

So that,    $C = \frac{1000 * 10^{-3} }{1.1 * 120 *10^{3}}$

here milliseconds denoted as $10^{-3}$.kilo ohms denoted as $10^{3}$.

Now we get the following values,

                $C = \frac{1000}{1.1 * 120}$  ×  $10^{-6}$

$1000$ in the numerator can be written as $1$×$10^{3}$.

                $C=\frac{1*10^{3} }{1.1*120}$  × $10^{-6}$

Now we add $10^{3}$ and  $10^{-6}$ and we will get $10^{-3}$.

                  $C= \frac{1}{1.1 *120}$  × $10^{-3}$

                  $C= \frac{1}{132}$ × $10^{-3}$

                  $C=0.00757$ × $10^{-3}$

                  $C=7.57$ ×$10^{-3}$ × $10^{-3}$

                  $C=7.57$ × $10^{-6}$ $F$

capacitor   $C=7.57$μ$F$