I have a chandelier with 10 LED lights in parallel. I replaced one with an incandescent lamp and now the 9 LEDs glow much brighter. Why does happen?

  • 2
    $\begingroup$ Welcome to Physics Stack Exchange. All circuit questions should come with a circuit diagram. Please include one. $\endgroup$ – DanielSank Feb 13 '18 at 18:07
  • $\begingroup$ Is this a theoretical or a practical question? i.e. are we assuming perfect voltage supplies or otherwise? $\endgroup$ – Dancrumb Feb 13 '18 at 18:09
  • $\begingroup$ Also, was the resistor that was in series with the replaced LED taken out of the circuit? $\endgroup$ – Dancrumb Feb 13 '18 at 18:11

There is a lot of information missing from this question, but there is enough to discuss how to address the physics.

First, LED brightness is (generally) proportional to the current flowing through them. Thus, if the LEDs are glowing more brightly, then the current through them must have increased.

If the current flowing through each LED has increased, then it means that the voltage over the LED-resistor sub-circuit has also increased (I'm assuming that each LED has a resistor in series with it).

Assuming that the power supply is relatively stable, the only way for this to have happened is if the voltage over the terminals of the power supply has gone up.

For this to have happened, it's likely that this power supply is not a perfect one and has some internal resistance $R_{SUPPLY}$.

The chandelier represents a loading resistance $R_{LOAD}$ on the supply. The voltage over the terminals of the supply ($V_{OUT}$) is given by:

$$ V_{OUT} = V_{SUPPLY-RATING}\cdot\frac{R_{LOAD}}{R_{LOAD}+R_{SUPPLY}} $$

Since this is a parallel circuit, if LED was just replaced with an incandescent bulb without removing the resistor in series, then the resistance of that leg will have gone up. Thus, the net load ($R_{LOAD}$) will have gone down, resulting in $V_{OUT}$ going up, which increases the current through the LEDs and thus they glow more brightly.


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