Question

There are n identical light bulbs, each designed to draw power P from main supply. These all are joined in series across the main. what is the total power drawn by this combination

Answer 1

Let the mains voltage be V and let the resistance of one light bulb be R. Then the voltage across each light bulb will be V/n. The power dissipated in each light bulb is:
\[\frac{(\frac{V}{n})^{2}}{R}\]
and the power dissipated in n light bulbs will be:
\[\frac{(\frac{V}{n})^{2}}{R} \times n=\frac{V ^{2}}{nR}\]\
But V^2/R = P
Therefore the total power drawn by the combination is P/n watts.

Answer 2

Just curious as to look at from the standpoint of current I. Let R equal resistance of one lamp, then total R would be nR. Power equals current squared times R, in a series arrangement current is the same everywhere in the circuit. The total power would then be:\[(nR)\times I ^{2}\] but:\[P=I ^{2}R\] Total power is nP Where P is the power dissapated by one lamp.

Answer 3

@radar Not so! Looked at from the standpoint of current, where the current drawn by one lamp connected across the mains is I, the current through the n series connected light bulbs is I/n. The reason for this reduction in current being the increase in resistance across the mains from R (for one lamp) to nR for n lamps.
The total power would then be:
\[(nR) \times (\frac{I}{n})^{2}=\frac{I ^{2}R}{n}\]
but:
\[P=I ^{2}R\]
Total power is P/n where P is the power dissipated by one lamp.

Answer 4

@kropot72 I was stating that I (current) is the total current the results of E/nr, where E is the voltage of the "mains" and n is the number of lamps, and R is the resistance of a single lamp. It is clear that the total power is the sum of the power dissipated in each individual lamp. And stating that the total power is equal to the power dissipated in a single lamp divided by the number of lamps is confusing. That is like saying the total power of the series string is less than the power consumed by a single lamp.

Answer 5

@radar. Lets put some numbers to the question. Say we have 4 identical light bulbs, each designed to draw 100 watts from a 120 volts mains supply. The current drawn by one lamp connected across the mains is I =100/120 A. The resistance of one lamp is therefore:
\[R=\frac{P}{I ^{2}}=\frac{100}{0.694}=144.1\ \Omega\]
The resistance of 4 lamps in series is:
\[R _{s}=144.1\times4=576.4\ \Omega\]
The current drawn by the 4 lamps in series across 120 V is:
\[I _{s}=\frac{120}{576.4}\ A\]
\[I _{s}^{2}R _{s}=(\frac{120}{576.4})^{2}576.4=25\ W\]

Answer 6

@kropot72 Now continue, what is the power dissipated by a single lamp? Using your numbers, R equals 144.1 Ohm, the current is equal to (120/576.4) Amperes. The power drawn by a single lamp is equal ti: (usine P=I^2 R) \[P=(120/576.4)^{2}\times 144.1=6.24 \]watts. Now the total power is the number of lamps n (4) and that is:\[4\times6.24=25 W\] this is as I posted in my previous post. Please note that in your circuit the lamps do not draw the 100 watts as they were designed. Can't help it, it is your circuit. Your should realize that from your numbers. If they were really dissipating 100 watts each, the total power would be 400 watts not 25.

Please review your numbers and after computing the total power, then compute the power dissipated in a single lamp, you will see that the total power dissipated is equal to the sum of the power dissipated in each lamp. If you fail to see this, you will soon see that in the real world, you don't pay for less power than you use.

Answer 7

@radar I can see where you are coming from. The question defines P as the rated power of each lamp when it is connected directly across the mains. My solution adheres to this definition, whereas your calculation redefines P as the actual power consumed by each lamp in the series circuit described in the question.

Answer 8

The problem is really not realistic in that the power consumed is specified on a "per lamp" basis, but then ask for total power with a unspecified number of these unique lamps connected in series. Your original method using voltage (E^2/R) was just fine. I tried using the current equation (I^2R) and my answer confused rather than clarify.

Answer 9

@radar Thank you for your reply and your input. I fully agree that the problem is unrealistic and I feel that it is liable to confuse students.