Question
Download Solution PDFThe length of cable required for transmitting a data at the rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits and for signal speed 2,00,000 km/s is
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFBandwidth = 500 mbps
Frame size = 10,000 bits
Signal speed = 2,00,000 km/sec
Transmission delay = 2 × propagation delay
frame size/bandwidth = 2 × (length of the cable /signal speed)
Length of the cable \(= \frac{{200000 \times 10000}}{{2 \times 500 \times {{10}^6}}} = 2\;km\)
Last updated on Jun 23, 2025
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