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I can sort of understand the one on the right. Without the resistor, the circuit would be shorted to ground and Vout spoiled.
But the one on the left... if light shines on the phototransistor, the circuit grounds out. If it doesn't, however, you get... some voltage less than Vcc on Vout. There's something very fundamental I don't get.
Remember, use small words. I had to google 'pull down resistor.'
And what makes the right one better? That Vout gets all Vcc, undiluted by a resistor?
That would mean there's no resistance between VCC and VOUT so VOUT would always equal VCC minus the loading effect of the photo transistor through the resistor, which would be very low. It would essential do nothing.
If you moved the resistor to the emitter lead as with the circuit on the right, and kept the left circuit output in the same place than you would have a constant volt output of Vcc no matter what the input is.
If you kept the resistor at the output and added a second resistor to the emitter lead than you would have an output voltage determined by the amount of current flow through the circuit which depends on the value of resistors as well as the amount of voltage drop across the transistor ect...
When there is no light shining on the one at the left, the transistor is off and it acts like an open switch, no current flowing, so all of Vcc is across the transistor and the output is high. When light shines on the transistor, it is on and acts like a closed switch, the output is low.