0
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With conductor:

enter image description here

Without conductor:

enter image description here

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    \$\begingroup\$ Can you edit your question to explain the purpose of your experiment and what you expected to happen? \$\endgroup\$
    – Transistor
    Apr 8, 2021 at 8:13

2 Answers 2

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You are short-circuiting your batteries, so the voltage output will be 0V when the conductor is connected and therefore not work and discharge your batteries

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ sorry im a rookie here I don't quite understand what short circuiting your batteries mean \$\endgroup\$ Apr 8, 2021 at 8:04
  • \$\begingroup\$ It means you are connecting you positive and negative through a very small resistance (your conductor) so all current will go through the conductor and almost none through the lights. Have added a schematic explaining this where the 100Ohm resistor is your light and the 1Ohm resistor your conductor. \$\endgroup\$
    – SolarTec
    Apr 8, 2021 at 8:13
  • \$\begingroup\$ @JamesOngkowijoyo The inductor has very low resistance, much lower than the lightbulb. The current from the battery will prefer to flow (mostly) through the path with the lowest resistance. When the inductor is connected, the path through that inductor has the lowest resistance (the 1 \$\Omega\$ resistor R3 above). So most of the current will flow through that inductor and not through the lamp (the 100 \$\Omega\$ resistor). \$\endgroup\$ Apr 8, 2021 at 8:13
  • \$\begingroup\$ Oh I get it now thank you very much for the help guys \$\endgroup\$ Apr 8, 2021 at 8:20
  • \$\begingroup\$ oh one more question the conductor doesn't effect a circuit that is not short circuited right ? \$\endgroup\$ Apr 8, 2021 at 8:29
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Let's take a look at your circuit as a schematic diagram:

schematic

simulate this circuit – Schematic created using CircuitLab

"Lamp circuit" is your circuit without your parallel connection. All the current goes to the light bulb, and it lights up.

"Lamp circuit 2" is your circuit with your parallel connection. All the current goes through the wire, leaving no current for the light bulb. The bulb doesn't light because it doesn't get any current.

Your circuit doesn't "turn off." There is still current flowing, even when the light is dark. There's actually more current flowing when the bulb is dark than when it is lit. More current flows through the second conductor than flows through the lamp when it is lit.

Electricity is "lazy." It flows most where there's the least resistance. In your parallel circuit, the least resistance is the wire, so nearly all of the current flows through the wire.


That is not a nice thing to do to your battery. The resistance of the wire is very low, so a very high current will flow. Batteries aren't made to do that. The battery will get hot, and it will discharge (become empty) very quickly.


A "short circuit" is when electricity takes a "short cut" and "goes around" a part of the circuit instead of through it.

It doesn't literally mean that the path is shorter. It just means that more current flows through an unwanted path.

Both of these are short circuits, even though the path is longer in one of them than in the other:

schematic

simulate this circuit

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    \$\begingroup\$ +1. Putting the short-circuit to the left of the lamp might reinforce the concept of short-circuit - as in a short-cut. \$\endgroup\$
    – Transistor
    Apr 8, 2021 at 8:20
  • \$\begingroup\$ @Transistor: Done. \$\endgroup\$
    – JRE
    Apr 8, 2021 at 8:28

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