# Brightness of bulbs

by ana111790
Tags: bulbs, circuits, power, resistance
 P: 43 1. The problem statement, all variables and given/known data There are two bulbs, one of 100W and one of 50W and a 120V battery. First, connect them in parallel, find the resistance and current of each and determine which will shine brighter and by what factor. Next connect them in series, again find the resistance and current of each and determine which will shine brighter and by what factor. 2. Relevant equations Parallel circuits have the voltage drop is the same on each loop. A circuit in series has equal current. V=IR P=IV 3. The attempt at a solution In parallel: I found the current across each resistor, I=P/V so that current across 100W bulb is .833 A and current across 50W bulb is .417. I then found the resistance of each by R=V/I so resistance of 100W bulb is 144 Ohms and across 50W is 288 Ohms. Because the resistance of 50W bulb is larger it glows brighter, by a factor of 2. In series: I set (Voltage across 100W + Voltage across 50W)= 120 V. Substituting for V=P/I, with I being the same I got: 100/I + 50/I = 120, so I=150/120=1.25 A. I then proceeded to find resistance, R=V/I, so Resistance across the 100W bulb is 100/1.25^2 = 64 Ohms and resistance across 50W bulb is 50/1.25^2 = 32 Ohms. Because it has a larger resistance 100W bulb will shine brighter by a factor of 2. Can anyone verify this is correct? I want to make sure my concepts are right. Thanks!
 HW Helper Thanks P: 10,618 Resistance of a bulb does not change (it depends a bit of the temperature but this is relevant only at very high temperature differences.) If a bulb is designed for a special voltage and power (the nominal values are written on the bulb) the resistance is R=U2/P. If the voltage across the bulb is different from the nominal value the power will be different. When the power is lower than nominal the bulb shines dimmer. If the power is higher than nominal, the bulb will be destroyed because of overheating. ehild
P: 462
 Quote by ana111790 [b] I found the current across each resistor, I=P/V so that current across 100W bulb is .833 A and current across 50W bulb is .417. I then found the resistance of each by R=V/I so resistance of 100W bulb is 144 Ohms and across 50W is 288 Ohms. Because the resistance of 50W bulb is larger it glows brighter, by a factor of 2. In series: I set (Voltage across 100W + Voltage across 50W)= 120 V. Substituting for V=P/I, with I being the same I got: 100/I + 50/I = 120, so I=150/120=1.25 A. I then proceeded to find resistance, R=V/I, so Resistance across the 100W bulb is 100/1.25^2 = 64 Ohms and resistance across 50W bulb is 50/1.25^2 = 32 Ohms. Because it has a larger resistance 100W bulb will shine brighter by a factor of 2. Can anyone verify this is correct? I want to make sure my concepts are right. Thanks!
In parallel the two get whatever they need to work from the battery, that mean one has 100W the other 50W. The battery delivers more energy in parallel. So the 100 W will be brighter, and by a factor of 2 (as no other variables are given that is right)

In series they share the same Intensity of current that means that the 50 W bulb has more Power than the 100W, by a factor of 2 as it has double the resistance.

 P: 43 Brightness of bulbs I am having trouble seeing this conceptually, can someone tell me which steps in the process (mathematically solving it) I am doing wrong? In parallel: (voltages are the same, currents are not) a) Find the resistance on each bulb R=V2/P Resistance of 100W bulb= 120V2/100W = 144 Ohms Resistance of 50W bulb= 120V2/50W = 288 Ohms b) Find the current on each bulb I=P/V Current on 100W bulb = 100W/120V = .833A Current on 50W bulb = 50W/120V = .417A c) Which bulb glows brighter? By what factor? The 50W bulb because it has a higher resistance, by a factor of 2. In series: (voltages are not the same, currents are) a) Find the current of each bulb I set the sum of the voltages (V1=voltage across 100W bulb, V2=voltage across 50W bulb) equal to 120: V1+V2= 120 (substituting for V=P/I:) P1/I + P2/I = 120 (common denominator:) 100/I + 50/I = 120 150/I = 120 I=120/150 = 1.25 A (The same across each bulb) b) Find the resistance across each bulb R=P/I2 Resistance on 100W bulb=100W/(1.25A)2 = 64 Ohms Resistance on 50W bulb=50W/(1.25A)2 = 32 Ohms c) Which bulb glows brighter? By what factor? The 100W bulb because it has a higher resistance, by a factor of 2. Thanks
 P: 43 Can someone check the above work please? I pretty much cannot do the rest of my homework since they are problems similar to this one... :(
 P: 462 What do you understand by glowing brighter? I mean what physical concept?

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