I have a simulated circuit below, where a PNP determines when LED 1 or LED 2 is on or off. I don't understand why LED 2 is on when the PNP is active and LED 1 doesn't turn on but stays off. Is this because when the PNP is active there is less resistance and thus the current flows through LED 2 and not through LED 1? I need help to understand the circuit below.
The base current \$I_B\$ will be roughly \$17\text{ mA}\$ in your circuit, most likely putting Q1 into deep saturation, hence \$U_{CE} \approx 0\$. It follows that
$$ U_{10} = U_{D1} + U_{\text{LED}1} \approx U_{\text{LED}2} = U_{\text{LED,Forward}} \,. $$
The upper branch will behave approximately like a diode with forward voltage
$$U_{D1\text{,Forward}} + U_{LED\text{,Forward}} \approx 0.7\text{ V} + U_{\text{LED,Forward}} > U_{\text{LED,Forward}} = U_{10},$$
so there simply is not enough voltage applied to put LED1 and D1 into forward conduction. LED1 stays off.
(Of course, there is going to be some minuscule current flowing through LED1 owing to the exponential diode I/V curve, just not nearly enough to consider it "on".)
