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How to Calculate the Carrying Capacity of Silver

The electrical circuits in your cellphone, refrigerator, desk lamp and everything else electronic have been designed according to certain rules. The rules govern things like what wattage of light bulb can be put in a socket, how thick the wires need to be and how long the wires can be. The rules are designed to protect the wires from getting too hot.

The amount a wire heats up depends upon the amount of current running through it and the resistance of the wire. The resistance of the wire depends upon its geometry: longer wires have higher resistance, while larger diameter wires have lower resistance. The resistance of a wire also depends upon the material it's made of: the same shape wire made of aluminum or made of silver will have a different resistance. To determine the current carrying capacity of silver, you need to figure out how much current silver can carry before it gets too hot.

Instructions

- 1
  Collect information about the properties of silver. You will need the resistivity, which indicates how much a material resists electrical current; the mass density, which is a measure of how heavy a certain amount of material is; and the specific heat, which measures how easily the material changes temperature when it absorbs energy.
  
  According to Periodictable.com, determine the resistivity of silver by calculating: rho_r, is 1.6 x 10^-8 meter-ohms; the mass density, rho_m, is 10.49 g/cm^3; the specific heat, C, is 235 joules/kg-kelvin.
- 2
  Define the geometry of your silver wire. The resistance depends upon the geometry.
  
  For example, you could have a 30 cm piece of 20 gauge wire, which has an area of 0.5 square millimeters.
- 3
  Calculate the resistance of your wire. The resistance is given by the resistivity times the length/area. The resistance determines how much energy will be absorbed by the wire when current travels through. For the example, R = rho_r x l / area = 1.6 x 10^-8 m-ohms x 0.3 m / (5 x 10^-7 m^2) = 9.6 x 10^-4 ohms.
- 4
  Calculate the maximum permissible temperature. This temperature could be the melting point of the wire insulation, the ignition temperature of a printed circuit board or --- in the most extreme case possible --- the melting point of the material. For this example, assume the maximum temperature is 200 degrees C, a common temperature rating for insulation.
- 5
  Determine how the temperature changes as your wire absorbs energy. You will be calculating the energy the wire absorbs in Joules, a unit of energy. Calculate the temperature rise per Joule of energy in the wire. The temperature rise is given by the inverse of the heat capacity times the mass of the wire, which is rho_m x length x area.
  
  For the example, the wire has a mass of rho_m x length x area = .0105 kg/cm^3 x 30 cm x 0.005 cm^2 = .0016 kg. Multiply this by the heat capacity and invert the result to get the temperature coefficient, Tc: 1/(235 J/kg-K x .0016 kg) = 2.6 K/J. That is, the wire will rise in temperature by 2.6 kelvin (which is the same as 2.6 degrees C) for every joule of energy dissipated in the wire.
- 6
  
  Heat is the enemy of electrical circuits: an overheated wire can damage the circuit or even start a fire.
  
  Calculate the amount of current the wire can take before reaching the maximum temperature. The power dissipated in a resistor is given by the current squared times the resistance, so the maximum total current is given by I_total = sqrt ((maximum temperature - starting temperature)/(Tc x R)).
  
  You found the maximum temperature in Step 4, you found Tc in Step 5, and you calculated R in Step 3. For the example, assume the circuit starts at 25 degrees C, so the maximum current is given by I_total = sqrt ((200 -25)/(2.6 x 9.6 x 10^-4)) = 265 amps.
- 7
  Decide how much time the wire needs to carry current. Divide the total current by the amount of time the wire will be carrying current. For example, you may have a circuit that will send current through your silver wire for two minutes --- 120 seconds. Then the maximum average current it could carry would be 265 amps/120 seconds: 2.2 amps.

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