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I was trying to heat up a nicrome wire (32gauge) using a 21650 lithium battery, and I have charged it. I have used jumper wires to connect all and joined the nicrome using wire connecting joint. The wire is 6cm and the voltage is 3.7v. I'm trying to make a heating coil.

I don't know what its called, but it is tightened by screws. The wire is heating up but very slowly and isn't getting very hot. I'm a complete beginner and have very little knowledge about electronics, and I was looking for some guidance.

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  • \$\begingroup\$ How long is the nichrome wire? Edit your question with the answer. As for what you see, it's what you should expect. To get more heat, you need a bigger voltage source. \$\endgroup\$ Commented Feb 28 at 6:41
  • \$\begingroup\$ Get a variable voltage supply with current limit capability. They are not particularly expensive. You probably don't need a lot of current compliance so focus on buying higher voltage range. Gradually increase the voltage until you see what you need. Let is know what you find from this. \$\endgroup\$ Commented Feb 28 at 13:40
  • \$\begingroup\$ Are there any specific battery types that I should use ? \$\endgroup\$
    – Hanreet
    Commented Feb 29 at 3:46

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You can try increasing the voltage (multiple cells in series), doubling the wire back so you have 1/2 the length with two parallel strands, or getting a thicker wire such as AWG30. Doubling the wire back (two strands of half the length will give you double the power in each strand x 2 strands so 4x the power.

In general the power delivered to each strand is \$P = \frac{V^2A}{\rho L}\$ where L is the length of the wire, V is the voltage, A is the cross-sectional area of the wire and \$\rho\$ is a constant that depends on the alloy (actually it's not quite constant- it generally increases with temperature for metals and alloys).

The final temperature the wire will reach is a (often quite nonlinear) function of the power, and depends on the surface area of the wire, air currents, if it's in contact with anything, if it's wound in a coil so it heats itself etc.

In any case, you can see that increasing the voltage has a large effect on the power - it's squared so going from 3.6 to 11.1V will increase the power by almost 10:1. Decreasing the length has a reciprocal effect. And fattening up the wire increases the power (but also increases the surface area).

Also make sure you're not losing too much power in the connecting copper wires.

Of course you need to take care not to draw excessive current from then battery (use a protected battery to avoid possible fires or worse from the battery) and be careful not to overload or short the battery anyway and the wire itself could cause burns or start a fire. With proper precautions a bit of judicious experimentation can help get a feel for how it behaves.

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  • \$\begingroup\$ Thanks for the help I will try it. Also if I use 2 same batteries so is the voltage going to increase or be the same. \$\endgroup\$
    – Hanreet
    Commented Feb 29 at 16:06
  • \$\begingroup\$ If you use two in series, the voltage will be 2x and the power will be 4x as much, all other things being equal. 4x is a big change so you should see a significant difference. \$\endgroup\$ Commented Feb 29 at 16:18

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