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Faraday's Law

24 MHz clock ...

Question

24 MHz clock frequency is applied to a cascaded counter of MOD-3, MOD-4 and MOD-5 counter. The lowest output frequency and the MOD value of the cascaded counter are

A

6000 kHz, 60

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B

400 kHz, 60

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C

400 kHz, 120

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D

600 kHz, 120

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Solution

The correct option is B 400 kHz, 60
Overall MOD = $3 \times 4 \times 5 = 60$
$f_{o u t} = \frac{24 \times 10^{6}}{60} = 400 k H z$

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