The attached circuit is an interesting variation. It's a passive 9V battery tester which is a nice way of demonstrating that bipolar transistors don't actually need a 0.6V base-emitter voltage to turn on.
If you build the circuit (or simulate it in Spice), a 9V supply will light all five LEDs with about 6-7mA (the differing resistor values balance the currents). The base input to Q5 will be about 3.9V, so each transistor has about 0.6V Vbe as expected.
However, if you attach a flat battery (about 5V), the last LED (D1) will still light up with about 3mA of current, and D2 may be just visible with about 0.3mA. The base input to Q5 will only be about 2.2V, yet for Q1 to be on, all the other transistors must also be "on".
If you measure them, Q5 will have a Vbe of about 0.2V, Q4 about 0.3V, Q3 about 0.45V. Q2 about 0.6V and Q1 about 0.65V, giving a total of about 2.2V.
Technically, all the LEDs are actually "on" as well, but the currents are too small to make them visible.
If you want to build the circuit for real, the capacitor just creates a small delay so the LEDs light up one-by-one. It makes a really nice battery tester, especially with green, yellow, orange and red LEDs.