# Power dissipated by resistor -- Right answer, but need insight

You are absolutely correct. Everything you say is correct.

But your analysis below is not correct.

Your conclusion about the 60 W bulb consuming more power is correct, but your analysis in not. When resistors are connected in series the larger one has a dominant effect on the current, but you can't just assume that the effect of the other one disappears. You have to add their resistance to determine the total resistance. That total resistance predicts the correct current through both resistors.
Ok, so I'd add their currents to get their total currents? So $I_{total} = V(\frac{1}{R_1}+\frac{1}{R_2})$, and then use $I_{total}$ to calculate the power use of each "resistor"?

You are absolutely correct. Everything you say is correct.

But your analysis below is not correct.

Your conclusion about the 60 W bulb consuming more power is correct, but your analysis in not. When resistors are connected in series the larger one has a dominant effect on the current, but you can't just assume that the effect of the other one disappears. You have to add their resistance to determine the total resistance. That total resistance predicts the correct current through both resistors.
So, what specifically in that analysis of mine is incorrect? The part about the current needing to be the same everywhere in the circuit? My textbook quite clearly said that in an ideal circuit the current is the same everywhere, or else charge will accumulate.

haruspex
Homework Helper
Gold Member
So, what specifically in that analysis of mine is incorrect? The part about the current needing to be the same everywhere in the circuit? My textbook quite clearly said that in an ideal circuit the current is the same everywhere, or else charge will accumulate.
The part where you calculated the resulting current. You took a 120V supply and found the currents that would arise if that total voltage were applied to each of the two resistances. You then argued that with the two resistors in series the resulting current would be the lower of those two. That is quite wrong, and you know the equations to show it is wrong.
With the two resistors in series, neither would experience the full 120V. There would be a voltage drop across each, and the two voltages would add to 120V. To find the current, add the resistances: 384 Ohms. Divide that into 120V to get about 0.31A. That is the current in the circuit, so is the current through each resistor. The voltage drops are therefore roughly 0.31*144=45V and 0.31*240=74V.

• kostoglotov
Mister T
Gold Member
So, what specifically in that analysis of mine is incorrect? The part about the current needing to be the same everywhere in the circuit? My textbook quite clearly said that in an ideal circuit the current is the same everywhere, or else charge will accumulate.
Right, but the value of that current is not calculated in the way you calculated it. You assumed that the bulb with the higher resistance is all you need consider when calculating that current. You must instead consider both bulbs.

Mister T
Gold Member
To find the current, add the resistances: 384 Ohms.
I'll never forget embarrassing myself by assuming the bulbs have a resistance that stays the same when you hook them up in series. They don't, by a long shot. Last edited:
Right, but the value of that current is not calculated in the way you calculated it. You assumed that the bulb with the higher resistance is all you need consider when calculating that current. You must instead consider both bulbs.
Ok, so I'd add their currents to get their total currents? So $I_{total} = V(\frac{1}{R_1}+\frac{1}{R_2})$, and then use $I_{total}$ to calculate the power use of each "resistor"?

Mister T
Gold Member
Ok, so I'd add their currents to get their total currents? So $I_{total} = V(\frac{1}{R_1}+\frac{1}{R_2})$, and then use $I_{total}$ to calculate the power use of each "resistor"?
No, you add their voltage drops to get the total voltage drop. (The currents are the same, if you added them together you'd get 2##I##.).

$I = \frac{V}{R_1+R_2}$

haruspex
I'll never forget embarrassing myself by assuming the bulbs have a resistance that stays the same when you hook them up in series. They don't, by a long shot. 