# Direct current electric circuit

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?

• Have you thought about the current not being strong enough? – Aaron Stevens Feb 20 at 5:24
• Is it an incandescent bulb, or an LED? – PM 2Ring Feb 20 at 6:02
• 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. – 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 .