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gnarlyskim

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OK so I am new to these forums so please forgive me for any mistakes. I have a prelab for my electromagnetism class. We just began this chapter in class and the book is harder to understand than the problems! Here's the problems and my shot at them:

1. This prelab explores the characteristics of wires in DC circuits. Keep in mind that an ordinary wire has low but finite resistance. For these questions assume the wire resistance is 1.0 Ohms, the power supply is an ideal 6V voltage source, and the light bulb resistance is 30 Ohms.

For the circuit below, one such wire is connected directly across the power supply. What is the voltage drop (in V) across the wire?

http://img156.imageshack.us/img156/3593/phy1sy6.png

2. Calculate the current (in A) in the wire.

http://img156.imageshack.us/img156/7119/phy2ks5.png

3. In the circuit below, a lightbulb is connected to the power supply using two such wires. Calculate the current (in A) in the circuit.

http://img26.imageshack.us/img26/6919/phy3st0.png

4. Calculate the voltage drop (in V) across each wire.

http://img120.imageshack.us/img120/7647/phy4yt2.png

5. Calculate the voltage drop (in V) across the lightbulb.

http://img26.imageshack.us/img26/2457/phy5sl4.png

6. Now assume the wires in have zero resistance (0 Ohms). Calculate the current (in A) through the lightbulb.

http://img26.imageshack.us/img26/2165/phy6gd1.png

7. Again, assuming the wires have zero resistance, calculate the voltage drop (in V) across the lightbulb.

http://img120.imageshack.us/img120/4402/phy7go6.png

8. A current flows in the resistor below. Which of the following is true?

http://img159.imageshack.us/img159/3627/phy8jc0.jpg

A. V1 = V2 but I1 ≠ I2

B. I1 = I2 but V1 ≠ V2

C. V1 = V2 and I1 = I2

D. V1 ≠ V2 and I1 ≠ I2

9. Positive currents flow in the directions shown in the resistors below. Which of the following is true?

http://img159.imageshack.us/img159/5792/phy9rh8.jpg

A. I1 = I2 + I3

B. I2 = I1 + I3

C. I3 = I1 + I2

D. I1 + I2 + I3 = 0

V=IR

R=V/I

I=V/R

Ok so here is my attempt for each:

So the question is asking for voltage drop (V). V=IR, but what is the I? I took a guess and did V=(6A)(1[tex]\Omega[/tex]). The 1[tex]\Omega[/tex] coming from the wire.

Thus, voltage drop=6V

This problem is how I found 6A for the I in question 1. I used I=V/R.

Thus, I=6V/1[tex]\Omega[/tex] .;. I=6A

This one is asking for current I (A) for the circuit including two wires (contributing 1[tex]\Omega[/tex] each?) and the lightbulb (contributing 30[tex]\Omega[/tex]) powered by 6V.

So, I=V/R, I=(6V)/(30[tex]\Omega[/tex]+2[tex]\Omega[/tex])=.1875A

I don't know where to go with this one. I assume that voltage remained constant in these...so V=6?

Same situation as above...but this started making me feel a little suspicious that my inclinations were not as correct as I had hoped.

No contribution to resistance from the wires, so only the 30[tex]\Omega[/tex] from the lightbulb should be factored in.

I=V/R, I=(6V)/(30[tex]\Omega[/tex])=.2A

Same situation as the voltage drop questions above.

A. V1 = V2 but I1 ≠ I2

B. I1 = I2 but V1 ≠ V2

C. V1 = V2 and I1 = I2

D. V1 ≠ V2 and I1 ≠ I2

I was following my instinct as said before on this problem, so my guess was A. Voltage remained constant, but the current changed as indicated through the problems above.

A. I1 = I2 + I3

B. I2 = I1 + I3

C. I3 = I1 + I2

D. I1 + I2 + I3 = 0

This one I based off human instinct (so it's probably wrong ha). My guess was B, where I2=I3+I1. It makes sense that this should be so, but I have some underlying feeling that there is a physics property that is going to come into play here.

THANK YOU for anyone who can look at these and give me some help!

**1. Homework Statement**1. This prelab explores the characteristics of wires in DC circuits. Keep in mind that an ordinary wire has low but finite resistance. For these questions assume the wire resistance is 1.0 Ohms, the power supply is an ideal 6V voltage source, and the light bulb resistance is 30 Ohms.

For the circuit below, one such wire is connected directly across the power supply. What is the voltage drop (in V) across the wire?

http://img156.imageshack.us/img156/3593/phy1sy6.png

2. Calculate the current (in A) in the wire.

http://img156.imageshack.us/img156/7119/phy2ks5.png

3. In the circuit below, a lightbulb is connected to the power supply using two such wires. Calculate the current (in A) in the circuit.

http://img26.imageshack.us/img26/6919/phy3st0.png

4. Calculate the voltage drop (in V) across each wire.

http://img120.imageshack.us/img120/7647/phy4yt2.png

5. Calculate the voltage drop (in V) across the lightbulb.

http://img26.imageshack.us/img26/2457/phy5sl4.png

6. Now assume the wires in have zero resistance (0 Ohms). Calculate the current (in A) through the lightbulb.

http://img26.imageshack.us/img26/2165/phy6gd1.png

7. Again, assuming the wires have zero resistance, calculate the voltage drop (in V) across the lightbulb.

http://img120.imageshack.us/img120/4402/phy7go6.png

8. A current flows in the resistor below. Which of the following is true?

http://img159.imageshack.us/img159/3627/phy8jc0.jpg

A. V1 = V2 but I1 ≠ I2

B. I1 = I2 but V1 ≠ V2

C. V1 = V2 and I1 = I2

D. V1 ≠ V2 and I1 ≠ I2

9. Positive currents flow in the directions shown in the resistors below. Which of the following is true?

http://img159.imageshack.us/img159/5792/phy9rh8.jpg

A. I1 = I2 + I3

B. I2 = I1 + I3

C. I3 = I1 + I2

D. I1 + I2 + I3 = 0

**2. Homework Equations**V=IR

R=V/I

I=V/R

**3. The Attempt at a Solution**Ok so here is my attempt for each:

**1. For the circuit below, one such wire is connected directly across the power supply. What is the voltage drop (in V) across the wire?**So the question is asking for voltage drop (V). V=IR, but what is the I? I took a guess and did V=(6A)(1[tex]\Omega[/tex]). The 1[tex]\Omega[/tex] coming from the wire.

Thus, voltage drop=6V

**2. Calculate the current (in A) in the wire.**This problem is how I found 6A for the I in question 1. I used I=V/R.

Thus, I=6V/1[tex]\Omega[/tex] .;. I=6A

**3. In the circuit below, a lightbulb is connected to the power supply using two such wires. Calculate the current (in A) in the circuit.**This one is asking for current I (A) for the circuit including two wires (contributing 1[tex]\Omega[/tex] each?) and the lightbulb (contributing 30[tex]\Omega[/tex]) powered by 6V.

So, I=V/R, I=(6V)/(30[tex]\Omega[/tex]+2[tex]\Omega[/tex])=.1875A

**4. Calculate the voltage drop (in V) across each wire.**I don't know where to go with this one. I assume that voltage remained constant in these...so V=6?

**5. Calculate the voltage drop (in V) across the lightbulb.**Same situation as above...but this started making me feel a little suspicious that my inclinations were not as correct as I had hoped.

**6. Now assume the wires in have zero resistance (0 Ohms). Calculate the current (in A) through the lightbulb.**No contribution to resistance from the wires, so only the 30[tex]\Omega[/tex] from the lightbulb should be factored in.

I=V/R, I=(6V)/(30[tex]\Omega[/tex])=.2A

**7. Again, assuming the wires have zero resistance, calculate the voltage drop (in V) across the lightbulb.**Same situation as the voltage drop questions above.

**8. A current flows in the resistor below. Which of the following is true?**A. V1 = V2 but I1 ≠ I2

B. I1 = I2 but V1 ≠ V2

C. V1 = V2 and I1 = I2

D. V1 ≠ V2 and I1 ≠ I2

I was following my instinct as said before on this problem, so my guess was A. Voltage remained constant, but the current changed as indicated through the problems above.

**9. Positive currents flow in the directions shown in the resistors below. Which of the following is true?**A. I1 = I2 + I3

B. I2 = I1 + I3

C. I3 = I1 + I2

D. I1 + I2 + I3 = 0

This one I based off human instinct (so it's probably wrong ha). My guess was B, where I2=I3+I1. It makes sense that this should be so, but I have some underlying feeling that there is a physics property that is going to come into play here.

THANK YOU for anyone who can look at these and give me some help!

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