I am providing a 5V to my circuit.

There is 10 µF capacitor in series with a 100 kΩ resistor and they are in parallel with a NPN transistor and LED.

I am trying to understand why the base of thetransistor only gets current when C1 is full.

Why does no/some current flow to Q1 (base), via R1, even though C1 is charging.

Current chooses the path of least resistance.

Current flows to base of Q1 only when C1 is fully charged which turns the transistor on and then LED lights.


simulate this circuit – Schematic created using CircuitLab

  • 3
    \$\begingroup\$ ”Current choses the path of least resistance.” No it does not. Place two resistors in parallel, one 1000 ohm and one 1001 ohm. Apply voltage across them. According to your theory no current would flow in the 1001 ohm one. C will absorb current according to I=dV/dt and the base will only sink current when you reach Vb threshold. \$\endgroup\$
    – winny
    Commented Jan 29, 2022 at 9:03
  • 1
    \$\begingroup\$ @winny it does take the path of least resistance, that's why it's a common saying, even if you misinterpret it. A highway with a higher speed limit than side-streets can be slower than the side-street during traffic jams. Unlike cars, charge instantly re-routes to find the fastest (easiest) route. If your 1000 ohm resistor were busy, the least resistance from the point of view of each charge is found by roughly splitting the flow between the 1000 and 10001 ohm paths. \$\endgroup\$
    – dandavis
    Commented Jan 29, 2022 at 9:14
  • 1
    \$\begingroup\$ @dandavis Without the per charge statement, the “rule” falls apart and people misunderstand. Current flows in whatever way minimizes the total losses in the system. Set up a simulation of the example I gave you and try it yourself. \$\endgroup\$
    – winny
    Commented Jan 29, 2022 at 11:28
  • \$\begingroup\$ Dang it, no C in my I = C x dV/dt above, or proportional sign instead of equal. \$\endgroup\$
    – winny
    Commented Jan 29, 2022 at 11:30
  • 3
    \$\begingroup\$ The LED is not a resistor, hence the division of currents according to resistance does not apply to it. An LED clamps the voltage. Please stop spread misinformation. \$\endgroup\$
    – winny
    Commented Jan 29, 2022 at 11:47

1 Answer 1


There is no priority, just standard circuit. The transistor will be off until there is enough voltage in the capacitor.

The base needs about 0.7V of Vbe to turn on the transistor. And the LED at the emitter also requires some voltage, let's assume 2V.

The capacitor charges slowly due to the resistor.

Therefore it takes time before capacitor voltage which equals Vb rises from 0V to about 0.7V plus the LED voltage and only then LED turns on.


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