n the bistable multivibrator, RESET pin is kept fixed high. If the PIN 2 is always LOW, output will
Question
n the bistable multivibrator, RESET pin is kept fixed high. If the PIN 2 is always LOW, output will
Solution
In a bistable multivibrator, the RESET pin is typically used to control the state of the circuit. If the RESET pin is kept high, it means the circuit is not in a reset state.
The PIN 2 in a bistable multivibrator is usually the trigger pin. If this pin is always LOW, it means the trigger is constantly active.
In this scenario, the output of the bistable multivibrator will depend on the design of the circuit. However, in a typical design, if the RESET pin is high (not resetting the circuit) and the trigger pin (PIN 2) is always LOW (constantly triggering), the output will likely be HIGH. This is because the trigger being active will typically set the output to its active state.
Please note that this is a general explanation and the actual behavior could vary based on the specific design of the bistable multivibrator.
Similar Questions
In the bistable multivibrator, RESET pin is kept fixed high. If the PIN 2 is always LOW, output willa.Change to HIGH and won’t change furtherb.Change to LOW and won’t change furtherc.Change to HIGH for short durationd.Keep changing between LOW & HIGH by itself.
What starts a free-running multivibrator?ans.nothinga triggeran external circuitan input signal Previous Marked for Review Next
When a high is applied to the reset input of the NAND Gate SR Latch, the output is
Fault 2 open circuits the feed back loop of the first stage of detector voltage amplifier, so that the final amplifier output saturates.Select one:TrueFalse
What starts a free-running multivibrator
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.