https://www.chegg.com/homework-help/questions-and-answers/consider-circuit-fig-1-square-wave-amplitudes-5-v-50-duty-cycle-means-pulse-width-half-per-q24321853
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Step 1

Consider the expression of time period:

` T=1/f `

` T=1/{100kHz} `

` T=0.01ms `

As Pulse width is half of period.

Pulse width ` ={T}/2 `

Pulse width ` ={0.01ms}/2 `

Pulse width ` =0.005ms~~5\mus `

Also the time constant of the given circuit is:

` tau=R_{eqv}C `

` tau={R+R_s}C `

` tau={50\Omega+R}\times 1nF `

Explanation:

Here the time constant, pulse width and time period is obtained.

Step 2

The value of resistor R so that the time constant of this RC circuit is less than 1/10 of the pulse width is obtained as follows:

` \text{Pulse width}-{1/10}\text{Pulse width}=tau `

` 5\mus-{1/10}\times 5mus={50\Omega+R}\times 1nF `

` {9/10}\times 5\times 10^{-6}s={50\Omega+R}\times 1\times 10^{-9}F `

` 45={50\Omega+R}times 10^{-2} `

` 4500={50\Omega+R} `

` R=(4500-50)Omega `

` R=4450\Omega `

Explanation:

Here the value of resistor R so that the time constant of this RC circuit is less than 1/10 of the pulse width is obtained.







Final Answer

Therefore the value of resistor R so that the time constant of this RC circuit is less than 1/10 of the pulse width is obtained as 4450ohms.