# Homework Help: Is a 120 Watt or 60 Watt bulb brighter?

1. Dec 5, 2003

### sk8rlindz

I had to do some homeowrk problems and i was hoping someone could just double check my answers... much appreciated.

When you connect a 120 watt bulb and a 60 watt bulb in a parallel circcuit across a standard 120 volt outlet, which is brighter?
the 120 watt bulb is brighter- right? because the current flowing through the circuit connects to both bulbs, and both get the 120 volts but the 120 watt bulb can actually use that amount of power.

If you connect the same bulbs in a series circuit, which bulb is brighter? and why?

The 60 watt bulb is brighter. The current is now divided among the two bulbs and the 60 watt bulb can utilize the entire 60 volts it gets ,while the 120 watt bulb can only light half of it's potential. or does it have to do with the 60 watt bulb being the path of least resistance?

2. Dec 5, 2003

### chroot

Staff Emeritus
First, determine the resistance of each bulb. Remember, power is voltage * current.

$$\begin{gather*} V = I \cdot R\\ P = IV = V^2 / R = I^2 R\\ 120\ W = (120\ V)^2 / R \Rightarrow R_{120\ W} = 120\ \Omega\\ 60\ W = (120\ V)^2 / R \Rightarrow R_{60\ W} = 240\ \Omega \end{gather*}$$

The 120W bulb has a resistance of 120 ohms. The 60W bulb has a resistance of 240 ohms.

So how do a 120 ohm and 240 ohm resistance act in parallel? Each one experiences the same voltage (120 V), so the power dissipated, $P = V^2/R$, is 120W for the 120 ohm bulb, and 60W for the 240 ohm bulb.

How about in series? The resistances in series add together, for a total of 360 ohms. 120 volts applied to 360 ohms is 1/3 A current.

How much power is dissipated by each bulb? $P = I^2 R$, or 10.8 W for the 120 ohm bulb and 21.6 W for the 240 ohm bulb.

Your suspicion is current -- the 120 ohm bulb is dimmer than the 240 ohm bumb when wired in series.

- Warren

Last edited: Dec 5, 2003
3. Dec 5, 2003

### Staff: Mentor

Re: Electricity

Study chroot's answer for a full explanation. I just want to comment on some of your statements.

For bulbs connected in parallel across a voltage supply, the same voltage connects to each bulb, not current. Each bulb has its own independent current (which you can calculate as chroot explains). Since each bulb gets the full voltage (120V) they can each operate at their full power. Of course, a 120W bulb (when operating at its full 120 Watts) is brighter than a 60W bulb.
For bulbs in series, the same (but smaller!) current flows through each. The voltage is divided (but not equally!) between the bulbs. Since the same current flows through each, the one with the most resistance will have the biggest share of the voltage and power.

4. Dec 6, 2003

### ShawnD

Re: Re: Electricity

Yep, that's right. P = I^2R. Since both have the same current, the variable is resistance.