Power Source at 6.0V: Analyzing Current, Voltage, & Resistance

AI Thread Summary
The discussion focuses on analyzing a circuit powered by a 6.0V source, detailing voltage drops and current measurements across light bulbs in both parallel and series configurations. The voltage drop across the parallel branch and light bulb is 0.25V, while the series light bulb experiences a 0.5V drop. Current measurements indicate that the entire circuit has a current of 0.1A, with the parallel bulb drawing 0.05A, highlighting that current is lower in parallel than in series. Participants express uncertainty about determining resistance and the effects of adding cells or resistors on bulb brightness, while Ohm's Law (V=IR) is referenced as a potential method for calculating resistance. The discussion emphasizes the importance of ammeter placement for accurate current measurement in different circuit configurations.
Sowon
Messages
2
Reaction score
0
Power Source at 6.0V

a) Voltage drop across the parallel branch of the circuit: 0.25V
b) Voltage drop across the light bulb connected in parallel: 0.25V
c) Voltage drop across the light bulb connected in series: 0.5V

2) How would you measure the current through the entire circuit: By putting an ammeter in front of the power source.

3.a) Current through the entire circuit: 0.1A
b) Current through the light bulb connected in parallel: 0.05A
c) Current through the light bulb connected in series: 0.1A

Introduction:

1. Compare the values of the current flowing through the light bulbs: As I have thought, the current in parallel circuit is smaller than the series circuit.

2. Compare the values of the current flowing through the series and parallel section of the circuit: The values are lower in parallel circuit and in series circuit the current values are higher.

3. Describe how you could determine the resistance of each light bulbs: (I have no clue)

4. Determine the resistance of each of the light bulb: (Not sure about this one either)

5. Imagine that you are using cells instead of variable power source. Consider the following changes made to the circuit. What happens to the brightness of the light bulbs? Explain what caused changes in the brightness.
This part is the part that I have no clue on too

a) One more cell is added in series to the circuit:
b) One more cell is added in parallel to the circuit:
c) One more resister is added in series to the circuit:
d) One more resistor is added in parallel to the circuit:

This is the diagram of the circuit (the best I can provide sorry for bad quality):
http://puu.sh/iarE6/7639d4310f.jpg
 
Physics news on Phys.org
Sowon said:
Power Source at 6.0V

a) Voltage drop across the parallel branch of the circuit: 0.25V
b) Voltage drop across the light bulb connected in parallel: 0.25V
c) Voltage drop across the light bulb connected in series: 0.5V

2) How would you measure the current through the entire circuit: By putting an ammeter in front of the power source.

3.a) Current through the entire circuit: 0.1A
b) Current through the light bulb connected in parallel: 0.05A
c) Current through the light bulb connected in series: 0.1A

Introduction:

1. Compare the values of the current flowing through the light bulbs: As I have thought, the current in parallel circuit is smaller than the series circuit.

2. Compare the values of the current flowing through the series and parallel section of the circuit: The values are lower in parallel circuit and in series circuit the current values are higher.

3. Describe how you could determine the resistance of each light bulbs: (I have no clue)

4. Determine the resistance of each of the light bulb: (Not sure about this one either)

5. Imagine that you are using cells instead of variable power source. Consider the following changes made to the circuit. What happens to the brightness of the light bulbs? Explain what caused changes in the brightness.
This part is the part that I have no clue on too

a) One more cell is added in series to the circuit:
b) One more cell is added in parallel to the circuit:
c) One more resister is added in series to the circuit:
d) One more resistor is added in parallel to the circuit:

This is the diagram of the circuit (the best I can provide sorry for bad quality):
http://puu.sh/iarE6/7639d4310f.jpg
Post that as an image:
7639d4310f.jpg


For question 2) When you answer, " By putting an ammeter in front of the power source." Where exactly do you mean? Is that the only possible placement ?

Do you know the resistance of each resistor?

To answer 3 & 4: Do you know Ohm's Law ?
 
For #2 if you put the ammeter connecting directly with other light bulbs and resistors you'd be able to get the current from power source. Since in a series circuit all the current must be the same therefore I guess in series circuit it doesn't matter but the upper part of the diagram which is a parallel circuit will matter where you place the ammeter.
#3, #4 Yes I know, Ohm's Law which is V = IR but I'm not sure how I can explain it in words... Which is #3. Also, I do not know the resistance of each resistor that's why it's in the question I think.
 
Sowon said:
For #2 if you put the ammeter connecting directly with other light bulbs and resistors you'd be able to get the current from power source. Since in a series circuit all the current must be the same therefore I guess in series circuit it doesn't matter but the upper part of the diagram which is a parallel circuit will matter where you place the ammeter.
with what other light bulbs and resistors?

If you mean bulb #1 and the resistor that's between bulb #1 and the power source, that's right. Actually, anywhere from where it's shown in the circuit diagram to just above the switch is fine.
#3, #4 Yes I know, Ohm's Law which is V = IR but I'm not sure how I can explain it in words... Which is #3. Also, I do not know the resistance of each resistor that's why it's in the question I think.
The #3 and #4 which you had no clue about.

Use your measurements.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top