A zener diode exhibits a constant voltage of 5.6 V for currents greater than five times the knee current. IZK is specified to be
1 answer:
Answer:
The maximum power dissipation of the zener diode 112mV.
Explanation:
The minimum zener current should be:
5 * Iza= 5 * 1= 5 mA.
Since the load current can be at maximum 15 mA, we should select R so that, IL= 15 mA.
A zener current of 5 mA is available, Thus the current should be 20 mA, which leads to,
R = = 470 Ω.
Maximum power dissipated in the diode occours when, IL=0 is
Pmax = 20 * * 5.6 = 112mV.
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Answer:
the torque capacity is 30316.369 lb-in
Explanation:
Given data
OD = 9 in
ID = 7 in
coefficient of friction = 0.2
maximum pressure = 1.5 in-kip = 1500 lb
To find out
the torque capacity using the uniform-pressure assumption.
Solution
We know the the torque formula for uniform pressure theory is
torque = 2/3 × × coefficient of friction × maximum pressure ( R³ - r³ ) .....................................1
here R = OD/2 = 4.5 in and r = ID/2 = 3.5 in
now put all these value R, r, coefficient of friction and maximum pressure in equation 1 and we will get here torque
torque = 2/3 × × 0.2 × 1500 ( 4.5³ - 3.5³ )
so the torque = 30316.369 lb-in