Here is the question word by word: In the Circuit in Figure A, what would happen

Question

Here is the question word by word:
In the Circuit in Figure A, what would happen if the load resistorwere shorted? what would happen if the load resistor wereremoved?Hint: think in terms of power ratings

circuit:
Vin(12V)0----------Resistor (.25W,100ohm)--------------------------------
                                                                      |                                             |
                                                                      |                                             |     
                                                                   Zenerdiode                             RL
                                                                   (.25W,5.1V)                            |
                                                                      |                                             |
                                                                      |                                             |
     ---------------------------------------------------------------------
               |
             ground
I guess if RL is removed then it is just a circuit with zener diodeand a resistor but the power rating for the zener diode=IzVz=((Vin-Vz)/R)*Vz=.3519W which exceeds the power rating for thediode.

Can some one please help me with this question

Explanation / Answer

1. When the load resistance is shorted All the current from the supply, flows through theshort circuit and no current flows through the zener diode. The current through series resistor R is I = Vin / R = 12 / 100 = 0.12A Power dissipation across R will be P = I2 R = (0.12)2 (100) =1.44W This exceeds the power rating of the resistor R which is0.25W. Hence the resistor will be damaged 2. When the load resistance is open The current from the supply flows through R and zenerdiode The current is given by I = (Vin - VZ) / 100 = (12-5.1) /100 = 6.9/100 = 0.069A The power dissipated across R will be P = (0.069)2 (100) = 0.476W The power dissipated across zener dioide will be Pz = (0.069)(5.1) = 0.352W The power dissipated exceeds the power rating of thezener diode which is 0.25W and hence the diode will be over heatedand can damage it

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