Consider the circuit shown in Figure.All diodes are ideal. Knowing that the maximum electrical specifications of the lamps are VL = 6 V and PL = 120 mW, determine:

1)The state of operation of each of the diodes.

2 ) The lamps do not light up and they get damaged. Justify each case.


3)The lamps that light up without exceeding the maximum specifications. Justify each case.


4)Equivalent resistance of the circuit (RIn)


1)Considering that Vs=12 V and Vl=6 V i would say:

D1 ON,D2 OFF,D3 ON,D4 ON, D5 OFF

2)L1,L5 dont light up.

For analyze what lamps got damage i did:

\( Ilamp=(120 mW)/(6 V)=0.02 A

Rlamp=(120 mW)/(0.02 A)^2=300 Ohm

\)

Then i calculated the current that goes through:

L2

\( IL2=(12-0.7-0.7)/(300 Ohm)=35 mA \) which is > 20 mA so L2 gets damaged(too much power dissipated on L2)

For L4 and L5,

\(
IL4=IL5=(12-0.7-0.7)/(600)=17,6 mA \) so these lamps dont get damaged.

4) \( Rin=((300 *300)(300+300)) + 600 =750 Ohm \)

Is this correct?

Thanks

 

Anyone to check this calculations?


Thanks

 

Ok,

Answer 1 is correct. D2, D5 are reversed biased. D1, D3, D4 are forward biased.

Answer 2 is Lamp 1 and Lamp 3 are not energized. Lamp 2 was possibly destroyed due to excessive current. Lamp 4 and Lamp 5 are energized.

You correctly identified the current through the Lamps.

You correctly identified the equivalent resistance of the lamps.

Answer 3. L2 current calculation is correct and is the reason of the failure.

Lamp 4 and Lamp 5 current calculation is correct.

Answer 4. Rin Calculation is not correct.

Disregarding the forward biased diodes, Lamp 2 is in parallel with Lamp 4 and Lamp 5. Your formula does not reflect that reality. Revisit and redraw if necessary to do the correct calculation.

Of course your theoretical diodes will be zero ohms when they are forward biased for the resistance calculations. I doubt you would get near that calculation in real life.