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In serial communication employing $8$ data bits, a parity bit and $2$ stop bits, the minimum band rate required to sustain a transfer rate of $300$ characters per second is

  1. $2400$ band
  2. $19200$ band
  3. $4800$ band
  4. $1200$ band
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Best answer
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Since stop bit is given it is asynchronous communication and $1$ start bit is implied. So, 

$\text{(8 + 2 + 1 + 1) * 300 = 3600 bps}$

Minimum band rate required would be $4800$ here. 

Correct Answer: $C$

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Transfer rate = Baud rate * (data bits/total bits)

300 * 8 = B * (8/12)

B = 3600

2400 < 3600 < 4800

4800 baud.

Answer:

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