More Voltage Woes: Solving Electrical Circuit Problems

[scorpa]

More Voltage ... :(

Hi again,

Once again I am hopelessly stuck on some voltage questions.

#1) Three identical lamps are connected in parallel and then connected to a 6V battery. What is the voltage drop across each lamp?

Voltage stays constant throughout the entire parallel circuit so shouldn't each lamp continue to have a voltage of 6V?

#2) During a lab exercise you are supplied with a battery of potential difference V, two heating elements of low resistance that can be placed in water, an ammeter of negligable resistance, a voltmeter of very high resistance, wires of negligable resistance, a beaker that is well insulated and has negligible heat capacity, as well as 100g of water at 25 degrees celsius.
If the voltmeter reading holds steady at 50V and the ammeter holds steady at 5A, estimate time in seconds required to vaporize the water in the beaker. Use 4200 J/kg*degree celsius as the specific heat capacity of the water and 2.3 x 10^6 J/Kg as the heat of vaporization of the water.

I am totally lost on this question, seems more like chem one than a physics question, we haven't even done any questions like this, but it has showed up as a question in our assignment. Please help!

#3) A power saw is operated by an electric motor. When first turned on these motors have very low resistance. Suppose a kitchen light and a power saw hooked up in parallel with it are both turned on. The saw , light, and lead lines have an initial total resistance of 6 ohms.

Calculate the equivalent resistance of the light-saw parallel circuit. What current flows to the light. What is the total voltage drop across the two leads to the light? What voltage remains to operate the light?

I don't know if you can figure this out without seeing the picture. I can try and describe it. There is a switch box of 120 V with two lead lines hooked up in series to a 240 ohm kitchen light which is hooked up in parallel with the power saw. The two lead lines have a resistance of .25 ohms.


Thanks again.

 

[dextercioby]

The first problem is really simple.It should come up to 2V per each lamp.Can u see why?HINT:Those lamps have identical electrical resistence & are connected in series.What does that tell u about the current?

Daniel.

EDIT:Yeah,Warren,i know.I misread the question... :yuck:

 

[chroot]

#1) Three identical lamps are connected in parallel and then connected to a 6V battery. What is the voltage drop across each lamp?

Voltage stays constant throughout the entire parallel circuit so shouldn't each lamp continue to have a voltage of 6V?

You're not hopelessly stuck, you're right.

In general, just follow the wires. Wires have constant potentials all throughout. The entire top of the circuit is connected with one continuous wire network to the battery, so its potential is everywhere 6V. Similarly, the bottom of the circuit is all ground. Thus, each lamp has 6V across it.

#2) During a lab exercise you are supplied with a battery of potential difference V, two heating elements of low resistance that can be placed in water, an ammeter of negligable resistance, a voltmeter of very high resistance, wires of negligable resistance, a beaker that is well insulated and has negligible heat capacity, as well as 100g of water at 25 degrees celsius.
If the voltmeter reading holds steady at 50V and the ammeter holds steady at 5A, estimate time in seconds required to vaporize the water in the beaker. Use 4200 J/kg*degree celsius as the specific heat capacity of the water and 2.3 x 10^6 J/Kg as the heat of vaporization of the water.

If the heating elements are carrying 5 amps at 50V, how much power are they dissipating?

Next, look at the units on the constant you're given. You don't even need to worry about remembering an equation -- the units give you everything you need. The heat capacity of water is 4200 joules, per kilogram, per degree celsius. It takes 4200 joules of energy to raise the temperature of one kilogram one degree celsius. You should be able to figure out how much energy is required to heat up 0.1 kg of water 75 degrees, yes?

The heat of vaporization is even easier! It takes 2.3 x 10^6 joules of energy to vaporize one kilogram of water.

#3) A power saw is operated by an electric motor. When first turned on these motors have very low resistance. Suppose a kitchen light and a power saw hooked up in parallel with it are both turned on. The saw , light, and lead lines have an initial total resistance of 6 ohms.

Calculate the equivalent resistance of the light-saw parallel circuit. What current flows to the light. What is the total voltage drop across the two leads to the light? What voltage remains to operate the light?

I don't know if you can figure this out without seeing the picture. I can try and describe it. There is a switch box of 120 V with two lead lines hooked up in series to a 240 ohm kitchen light which is hooked up in parallel with the power saw. The two lead lines have a resistance of .25 ohms.

Do you know how to figure out the total resistance of two resistor in parallel? You know that 240 ohms in parallel with an unknown resistance = 6 ohms.

- Warren

 

[chroot]

The first problem is really simple.It should come up to 2V per each lamp.Can u see why?HINT:Those lamps have identical electrical resistence & are connected in series.What does that tell u about the current?

Daniel.

Read more carefully.

- Warren

 

[asrodan]

Question 1: Correct

Question 2: Knowing the specific heat capacity of water and the heat of vaporization you can find the amount of energy required to vaporize the water. You will also need to find the power dissapated by the elements in the water.

Question 3: Assuming the 6 ohms is the equvialent resistance, and knowing the resitance of the lead lines and the kitchen light, you should be able to find the resistance of the power saw.

Once you've found this it should be easy to find the resistnce of the light-saw parallel circuit.

For the current flow and voltage drop remember V = IR

Voltage remaining to operate light = 120 - D where D is the sum of the potential drops on the way to the light.

 

[scorpa]

So for the time to vaporize 0.1 kg of water this is what I am trying to do:

p = IV = 5A * 50V = 250 W

Later when I find work I should be be able to use P = W/T

So...

q = mct = (0.1KG)(4200J/Kg*celsius)(75 degrees) = 31 500 J

If it takes 2.3 x 10^6 J to vaporize 1 KG it should take 2.3 x 10^5 to vaporize 0.1 kg

From there I think I should add the two together because it takes energy to get up to to boiling point and to vaporize so...

2.3 x 10^5 + 31500 = 261500J of energy or work

so t = W/P 261500J/250W = 866 seconds?

 

[scorpa]

For the one about the power saw and lead lines and light. I was having trouble because the lead lines look like they are in series but the light and saw are in parallel.
I know R in series = R1+R2+R3 and in parallel (1/R) = (1/R1) + (1/R2) = (1/R3)

so to combine them do i have to go like this

6ohms = 0.25 + 0.25 + (1/240) + (1/R) ? This just doesn't seem right to me.

 

[scorpa]

Should I be figuring them out separately somehow? I am so lost :(

 

[scorpa]

anyone? I am getting really desperate, I've been working on them all day and have to hand them in tomorrow morning...im so stressed out right now :(