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Power dissipation of series-connected resistors

If you don't have a 1W resistor, and you still want to perform the experiment above, you may connect four 25ohms resistors in series. We've already learned that resistors in series act like a voltage devider: the voltage across each resistor is 9V/4 = 2.25V. The current is still 90mA since the total resistance is the same. So each resistor dissipates 2.25V ∙ 90mA = 0.20W. (Of couse we could also use one of our 'easy' equations: P = I2 ∙ R = (90mA)2 ∙ 25 = 0.20W.)

Be carefull: always take resistors with the same resistance. Of course you could also create a 100ohms resistor with three 33ohms resistors and one 1ohm resistor in series, but you're gonna smell some smoke! Which resistor(s) will blow? The 1ohm resistor because it's the smallest? Let's see. Since we know the current is 90mA, we use the equation P = I2 ∙ R = (90mA)2 ∙ 1 = 8.1mW. The 1ohm resistor will survive! The power dissipation of each 33ohms resistor will be (90mA)2 ∙ 33 = 0.27W. It may take some time, but you certainly will loose three resistors!

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