I have a bulb rated 110 V, 60 W and is in series with another bulb which is 110 V, 110 W. It's being powered with a 220 V source. Now, what would be the resistance of the resistor to be added in parallel to the first bulb so that each bulb will get the rated power?
The way I approached this is first get the resistance of each bulb, then get the voltage drop of each bulb via current divider principle. After that, I'm stuck.
As Steven states, this is only true when the bulbs act like ordinary resistors.
The solution is easy. The voltage across the 'divider' will be evenly distributed when power at the top half and power at the bottom half are equal.
To have equal power both at top and at bottom halves, you have to add an extra \$110W - 60W = 50W\$ in parallel to the existing top bulb.
Ohms law:
\$R = \dfrac{U}{I}\$
and
\$I = \dfrac{P}{U}\$
Substituting the second equation into the first, gives us the familiar: \$R = \dfrac{U^2}{P}\$
Now fill in the details:
\$R = \dfrac{U^2}{P} = \dfrac{(110 V)^2}{50W} = 242 \Omega\$
I'll presume your bulbs are ordinary resistors first. The 60 W bulb is R1, the other one R2.
Resistance can be calculated as
\$ R = \dfrac{V^2}{W} \$
For R1 and R2 that's
\$ R1 = \dfrac{(110 V)^2}{60 W} = 202 \Omega \$
\$ R2 = \dfrac{(110 V)^2}{110 W} = 110 \Omega \$
Then Rp || R1 = R2, or
\$\dfrac{Rp \cdot R1}{Rp + R1} = R2 \$
Filling in the values gives
\$\dfrac{Rp \cdot 202 \Omega}{Rp + 202 \Omega} = 110 \Omega \$
Solving for Rp gives us 242 Ω.
That would it be if the bulbs were ordinary resistors. They're not they're PTC resistors with a very low resistance at room temperature, and higher when the filament is heated by the current. Operating them at lower voltages by placing two of them in series will decrease the resistance, but you'll have to measure the voltage across each of them ,and the current to know what the resistance at that voltage is. So for the bulbs there's no easy answer.
This looks like a homework problem, so I'll give some hints and let you do the math.
You want the voltage to be equal on both bulbs. Therefore, neglecting the thermal influence on the bulbs' resistances1), you need to decrease the resistance of the bulb with the lower power rating (60 W) such that it is equal to the resistance to the bulb with the higher power rating (110 W). This way, you get a voltage divider with two equal resistances R1 and R2, equally sharing the provided voltage of 220 V, with R1 being the parallel resistance of the 60 W bulb and your additional resistor, R2 being the resistance of the 110 W bulb.
Step 1: Using the ratings of the "brighter bulb", solve for its resistance.
Step 2: Do the same for the not-so-bright bulb.
Step 3: Using the formula for parallel resistances, determine the resistor needed in parallel with the other bulb.
Step 4: Upon successful accomplishment, you'll look like a very bright bulb.
1) Thermal variation of the bulbs' resistances is often neglected in homework problems, so I guess this is a safe simplification to do.
There is a problem with the assumptions all the other answers that series connected bulbs will draw the same current. This is incorrect and both bulbs must be equal rating for this to to work.
As "@stevenh" reports correctly. tungsten light bulbs are PTCs with a cold surge current ratio of 10:1 worst case which corresponds to the resistance ratio when hot.
Ref: R ratio for tungsten R/R300= 10.30 with temp rise = 1800 C so using 10.3 below with corrected resistance values for 100W bulb by user in latter remark at 110V we get;
\$ R = \dfrac{V^2}{W} \$
for 60W, R = 202 Ω hot, R/10.3 = 19.6 Ω at room temp. where currents are 0.54A, 5.6A, respectively.
for 100W, R = 121 Ω hot, R/10.3 = 11.7 Ω at room temp, where currents are 0.91A, 9.4A, respectively.
The series voltage divider ratio at room temp is,
For 60W bulb 19.6 / ( 19.6+10.3) * 220Vrms = 144Vrms or 144^2/19.6Ω = 1,058 W Current is then 144Vrms/ 19.6Ω = 7.3A at power on @ room temp. ( averaged over 1 cycle)
Note that this is 7.3/5.6 = 1.30 or 30% over voltage and significantly over-power rating.
This will lead to **P.O.O.F.** or "Premature Optical Ocular Failure" ™ j/k ;)
More proof is elementary. Running in parallel at 220V would be even fast Poof and runing in parallel at 110 is light how light bulbs are normally used.
Is there a point to this futile experiment? Unless you are trying to design a constant current load with PTC on a 50V DC circuit or similar, Please advise at your earliest convenience. Tony.
