1.

If current in diode is five times that in R_(1). Breakdown voltage of diode is 6 volt. Find R= ?

Answer»

Solution :Current in zero diode is 5 times.
So total current DRAWN from battery `=6 mA+30 mA=36 mA`
POTENTIAL difference Across `R=24` volt
So V= IR `24=36xx10^(-3)R,R=2000//3 Omega`


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