I need the formula for calculating resistor wattage when the resistors are in series. For example, say I have 3 1/4 watt 10 ohm resistors in series. What is the potential power dissipation for the entire circuit?


Since resistors are in series, the current will be the same throughout the chain. Assuming that the voltage V across the string of resistors is constant, you can calculate the current \$I={V \over R_1+R_2 +R_3+ ...}\$. Power dissipated on each of the resistors: \$P_i=I^2R_i\$. If \$P_i\$ is less than the power rating of the resistor, it should dissipate the power without burning.

A string of N identical resistors in series each rated for P watts can dissipate \$NP\$ watts.

  • \$\begingroup\$ Nice and simple. Great explanation! \$\endgroup\$
    – Matt Ruwe
    Mar 19 '12 at 2:34
  • 6
    \$\begingroup\$ Twinkle, twinkle little star, power = i^2 r :-) \$\endgroup\$
    – JonnyBoats
    Mar 19 '12 at 3:04
  • \$\begingroup\$ Nice... I don't think I'll be forgetting that one any time soon. =) \$\endgroup\$
    – Matt Ruwe
    Mar 19 '12 at 3:37
  • \$\begingroup\$ A uniform series or parallel string of identical resistors can dissipate NP watts. The word identical is important, though. Each resistor's contribution to dissipation will be proportional to its resistance, so a 1K 1W in series with a 2K 1W would only dissipate 1½W total (though one could use 1K ½W in series with a 2K 1W and get total power dissipation equal to the sum of the individual ones). \$\endgroup\$
    – supercat
    Mar 19 '12 at 14:44
  • 3
    \$\begingroup\$ This simple formula is valid when the distance between the resistors is not too small. But a very close packaging of a lot of resistors may be very hot without derating. \$\endgroup\$
    – Uwe
    Dec 29 '16 at 9:36

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