Question

Turn off time of astable multivibrator is if Ra=1 kohm, Rb-2kohm and C=0.1 uF​

Answer 1

Answer:

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Answer 2

Answer:

The turn-off time of astable multivibrator is $T = 0.20\times 10^{-3} s$.

Explanation:

Given,

The resistance, $R_{a} =1 k\Omega = 10^{3} \Omega$

The resistance, $R_{b} = 2 k\Omega= 2\times10^{3} \Omega$

The capacitance, $C = 0.1\mu F = 0.1\times 10^{-6}F$

The turn-off time of astable multivibrator, T =?

As we know,

• The turn-off time is also known as the discharging time.
• The turn-off time or discharging time can be calculated by the equation given below:
• $T = 0.693(R_{a} + 2R_{b}) C$

Here,

• T = The turn-off time
• $R_{a}$ and $R_{b}$ = Resistances

After putting the given values in the equation, we get:

• $T = 0.693(10^{3} + 2\times 10^{3} ) 0.1\times 10^{-6}$
• $T = 0.693\times10^{3}(1 + 2 ) 0.1\times 10^{-6}$
• $T = 0.693\times10^{3}\times3\times 0.1\times 10^{-6}$
• $T = 0.20\times 10^{-3} s$

Hence, the turn-off time, $T = 0.20\times 10^{-3} s$.