Why when I reverse one of the three batteries in a torch light, the bulb does not light up? The three batteries (one of them is reversed) and the bulb suppose to form a closed loop. According to Kirchhoff's 2nd law,

$IR = E_1 + E_2 - E_3$

If all the batteries are identical ($1.5$ V for example), then IR$ = 1.5$V. That means there is still an electric current flows through the bulb. Is there any reason why the bulb does not light up?

  • $\begingroup$ Have you thought about the current not being strong enough? $\endgroup$ – Aaron Stevens Feb 20 at 5:24
  • $\begingroup$ Is it an incandescent bulb, or an LED? $\endgroup$ – PM 2Ring Feb 20 at 6:02
  • 1
    $\begingroup$ Re, "That means there is still an electric current flows through the bulb." Not true if the "bulb" actually is an LED. An LED is a diode, not a resistor. It does not obey Ohm's Law. (see electronics-tutorials.ws/blog/i-v-characteristic-curves.html). A "white" LED might not glow at all if you supply it with only 1.5V. $\endgroup$ – Solomon Slow Feb 20 at 15:19

Yes you are correct that the net voltage in the circuit is 1.5 V. You can be assured of it using a multi meter. The bulb of your flashlight is designed to use 4.5 V but you say that you are just powering it with a 1.5 V battery as the other 2 batteries are no use. This way there is just $\frac{1}{3}$rd the voltage it was designed to work with .

Simply said, that small voltage might not be enough to produce a considerable current through your bulb in the circuit.

  • $\begingroup$ Thank you for your response. But, I have tried with a 1.1 V bulb and it still couldn't light up. Is there any other possibility? $\endgroup$ – Hussaini Hashim Feb 21 at 5:05
  • $\begingroup$ was it a filament bulb or an led? There is a possibility of the battery being dead or the bulb being broken $\endgroup$ – user8718165 Feb 21 at 6:20

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