It is my first time to encounter an experiment with parallels. That's why I'm not sure in what part I got mistake. I believe IN5 and IN6 should be lit on.

Here's the diagram:

And here's the circuit:

 

The diagram doesn't show the diodes (are they diodes?).

The LEDs don't have current limiting resistors.

Pin 8 is not connected to IN6.

The lower power supply for the LEDs is not connected.

 

Diodes aren't on the materials list and we're not told to use diodes.

I'm using Circuit Wizard for that and it works even there's no resistor. I'll change the wires into resistors later in the actual experiment.

Thanks for pointing out the pin8 to IN6.

The lower power supply is connected with the upper, else all LEDS won't lit.

Now, the only problem left is the IN5..

 

I take it the parts labeled D1, D2, and D3 are SPDT switches?

Get those resistors in there and make sure that each of the LEDs is still working when you connect it, through the resistor, to the supply.

Without the resistors, the chip may be getting so heavily loaded that it just can't push any more current.

I can't tell whether you are trying to turn the LEDs one when the output is HI or when it is LO. Check the current sourcing capability of the LS family to see if it can meet your needs.

 

The reason why I don't put resistors in Circuit Wizard is because I don't know which of these is the 220 ohms resistor:

Here's the actual 220 ohm resistor which (I actually have and the one that will replace the wires):

The last time I try put resistor in Circuit Wizard some LEDS didn't lit.

EDIT: As far as I know, based on the diagram, if the output of D1 or D2 is HI then IN5 will lit.

 

You probably have to edit an attribute once you place the symbol and type in the desired resistance.

 

Here, I put the resistors but IN4 doesn't lit anymore:

 

The LEDs are limiting the voltage that the inputs on pins and they very likely can't get high enough to be seen as a HI.

You need to have the resistors in series with the LEDs, meaning that whatever current flows in one must flow in the other. You are tapping of from the junction and going to the IC input pin.

Try placing the resistors between the other side of the LED and the power supply.

 

Here's the actual 220 ohm resistor which (I actually have and the one that will replace the wires):

If the colours on that resistor are RED-RED-BLACK then that is a 22Ω resistor.
220Ω would be RED-RED-BROWN.

 

Yay! It worked! Thank you so much! ^^

 

I didn't even look at the resistor values.

I think I see BROWN-BLACK-RED...GOLD, which would make it a 1kΩ resistor.

And what is it we are looking at? Is this an actual picture of a breadboarded circuit? I'm having a hard time believing that. Look at the top edge of the four resistors, for instance, they are perfectly aligned. The same with all the LEDs. And there is no trace of solder on their legs. And no shadows. A number of other discrepancies.

What is this? Some virtual solderless breadboard program?

 

Whoa! Either your eyesight is really bad or your monitor is crappy.
This is definitely not the real thing but a simulation.

The resistor I am referring to is the one OP posted on post #5.

 

Also I don't see a GROUND connection on the LEFT side power rail to the RIGHT side of the breadboard.

 

Whoa! Either your eyesight is really bad or your monitor is crappy.

Given that my eyes are currently dialated and that I was diagnosed with both a fast-growing cataract and a detaching retina this afternoon, I am leaning toward the former.

The resistor I am referring to is the one OP posted on post #5.

Yep, I agree. That looks like a 22 ohmer.