Lightbulb Puzzle 1 The Potential Difference Across The... | Course Hero

Each bulb dissipates an average power of 60 W. The circuit has a fuse F that burns out when the current Example: A 5-Ω resistor and 20-Ω resistor are connected in series and potential difference of 100 V QUICK QUIZ 18.4. You have a large supply of lightbulbs and a battery. You start with one...ramavatbharat ramavatbharat. Answer: 100 w bulb has a greater resistans.We have step-by-step solutions for your textbooks written by Bartleby experts! Given Information: Average power of the light bulb is 100 W , frequency of the power source is 60.0 Hz , maximum Therefore, the resistance of a 100 W light bulb is 144 Ω . Want to see more full solutions like this?Which has a larger resistance, a 60 W lightbulb or a 100 W lightbulb? Apply the junction rule to the junction labeled with the number 1 (at the bottom of the resistor of resistance R2). Answer in terms of given quantities, together with the meter readings I1 and I2 and the current I3.A 60W bulb means a bulb that when connected to it's rated supply voltage draws 60W. Unless otherwise stated one assumes that the rated supply voltage is the normal As we expect, the lower Wattage lightbulb has a higher resistance. The total resistance of the circuit, when you put the two...

33. Which has a greater resistance a 100 W bulb or a 60 W bulb?

A light bulb that produces more light and less heat is more efficient. Incandescent[change | change source]. The incandescent light bulb turns electricity into light by sending the electric current through a thin wire called a filament. Large LED light bulb. Edison Lightbulb Musée des Lettres et Manuscrits.The term "60-Watt equal" has little to do with any kind of wattage correlation; a 60W equal LED may use as little as 8 Watts to operate. If your fixture is rated to accept 60 Watts, you can safely use 75W, 100W Just know that 100W and 125W equal LEDs are typically made in a slightly larger A21 size...A 60 W light bulb and a 100 W light bulb are each connected to a 130 V outlet. Which bulb has more current in it? Use these flashcards to help memorize information. Look at the large card and try to recall what is on the other side. Then click the card to flip it.Lightbulb Puzzle #1. The figure shows two identical lightbulbs in a circuit. The current through both bulbs is exactly the same! It's not the current that the bulbs use up, it's energy. 4/7/2016 16. QuickCheck 31.8. Which has a larger resistance, a 60 W lightbulb or a 100 W lightbulb?

(a) What is the resistance of a lightbulb that uses an... | bartleby

Two lightbulbs, one rated 60 W at 120 V and the other rated 100 W at 120 V, are arranged in two different circuits. 1. Calculate the resistance of By the equation for power , the bulb with the lower resistance will dissipate more power. The 100 W bulb has a smaller resistance, so it will be the...Lets say that I have a 60 watt globe in my room. 1) When I switch on that light and keep it on for an hour I will use 60 watt of electricity, or does the light use extra electricity to switch on? I accept this, but this current spike will only last 50~100 milliseconds or something, and we should re-iterate that in...Problem 22 Easy Difficulty. $*$ Does a $60-$ W lightbulb have more or less resistance than a 100 - W bulb?Which bulb will glow brighter 40W or 60W when in series/parallel?In short, the 60 watt bulb has more resistance. Power can be calculated using the following formula: P = (V^2)/R The power rating for a light bulb is Because more energy goes through it, you pay more and you get more result.AnswerA 100-W lamp has a lower operating resistance than a 40-W lamp.

Answering the question from the title directly:

The bulb with lower power score has a upper resistance (at running temperature, but we need to think it to be roughly consistent), because at nominal voltage, less current must go with the flow thru it compared to the bulb with higher rating.

In a series circuit, the present is equal at all issues.

The following assumes consistent $R$ for every bulb.

The voltage drop at each and every bulb can also be calculated the usage of $V = RI$ (assuming constant $R$ regardless of changing temperature of the filament). Higher resistance leads to higer voltage drop at the bulb with lower wattage.

For 60W @ 120V we need 0.5A of current and with $R=V/I$ we get $R_60W = 240\Omega$. The other bulb has $R_100W= 144\Omega$.

Power is calculated using $P=VI$, and since $I$ is continuing for both bulbs, higher resistance leads to higer energy, more brightness.

The current is $I=V / (R_60W + R_100W) = 120V / 384 \Omega$

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