Expressing current through a circuit

In summary: I understand now how the order in which the resistors are combined affects the overall resistance and current. And for part D, I see now that when the switch is closed, the potential at the branch after A is zero, so the bulbs in that branch will not light up. Thank you both for your help!
  • #1
Ozmahn
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Homework Statement



Six identical light bulbs (A-F) are connected in the following circuit. The resistance of each bulb is R and the idealistic battery has an emf of E. The switch K remains open (not connected) for questions (a) through (c).

https://mail.google.com/mail/u/0/?ui=2&ik=bc505e9c61&view=fimg&th=14d778cdba4f2ea7&attid=0.1&disp=inline&realattid=1501801808870244352-local0&safe=1&attbid=ANGjdJ_jML1j_uxhOOejtQqrl6EizyxYtYjk-AO81psFg2w8clQTt8C-cGlvgZ2H049s4UYhNnWwWph-RWZ5dgCDVC0ZsduDmthKvAT4lWWMo8iGGe23ryb7qEsBK1E&ats=1432229831281&rm=14d778cdba4f2ea7&zw&sz=w1342-h523

(a) Express the current through the battery using E and R.

(b) solved

(c) Rank the brightness of the bulbs from the highest to the lowest.

(d) Now the switch K is closed (connected). Which bulbs go off as a result?

Homework Equations



Resistors in series
R1+R2+...=Req

Resistors in parallel
1/R1+1/R2+...=1/Req

V=IR

The Attempt at a Solution


(a)
C and D are in series
R+R=2R
CD parallel to B
1/2R+1/R=3/2R=>2R/3
BCD and A in series
2R/3+R=5R/3
ABCD and F are in series
5R/3+R=8R/3
ABCDF and E are parallel
3/8R+1/R=11/8R=>8R/11=Req

V=IR => I=V/R, where V is the emf (E)
I=E/8R/11 => 11E/8R

This is what I got as the answer, but the solution shows that instead of combining ABCD and F, you combine ABCD and E first, then ABCDE and F. I worked it out that way and got 8E/13R, but I'm not seeing any reason why you would combine the resistors in that order. Why would the order in which you combine these matter?

(c) Okay so I know that A must be the brightest since it is getting the full current (E/R), but the other ones I'm not as sure about. C and D should be getting the same, since they are in series in the same branch. And B should get twice as much as C or D, so IC+ID=IB. I'm assuming E is the second brightest because there are no resistors between it and A. So then F would be third, since the current flowing through the CD and B branches would be split between that and E...that's my rationale anyway. So the answer is:

A>E>F>B>C=D

It's a bit confusing to be honest, and I used the solution to come up with that explanation. Is there an easier way to determine this?

(d) The solution says that BCD are turned off when the switch is closed. Would this be because there is zero resistance along the wire that bypasses those branches, so the current follows the wire with least resistance? Just a wild guess on that one.

Any help/clarification for these problems is greatly appreciated. Thank you!

EDIT: The problem didn't show up in the threa, so I attached it as a jpg file.
 

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  • #2
Ozmahn said:
This is what I got as the answer, but the solution shows that instead of combining ABCD and F, you combine ABCD and E first, then ABCDE and F. I worked it out that way and got 8E/13R, but I'm not seeing any reason why you would combine the resistors in that order. Why would the order in which you combine these matter?

Try re-drawing the circuit, and you'll see that this does make sense. Perhaps when you understand part D it'll help
Ozmahn said:
(d) The solution says that BCD are turned off when the switch is closed. Would this be because there is zero resistance along the wire that bypasses those branches, so the current follows the wire with least resistance? Just a wild guess on that one.

Any help/clarification for these problems is greatly appreciated. Thank you!

If that switch is closed, what is the Potential at the branch after A and the potential on the wire that connects B to CD? Remember when doing these circuits if you can trace a path between 2 points with no resistors they'll be at the same potential
 
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  • #3
Ozmahn said:
(a)
C and D are in series
R+R=2R
CD parallel to B
1/2R+1/R=3/2R=>2R/3
BCD and A in series ( F is in series with BCD, you must find BCDF, then BCDFE, then ABCDFE )
2R/3+R=5R/3
ABCD and F are in series
5R/3+R=8R/3
ABCDF and E are parallel
3/8R+1/R=11/8R=>8R/11=Req

See the correction in the quote.
 
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  • #4
That makes a lot of sense, thank you both!
 

Related to Expressing current through a circuit

1. What is the purpose of expressing current through a circuit?

The purpose of expressing current through a circuit is to understand how electricity flows through a circuit and to calculate the amount of current that is present in a particular point in the circuit. This information is crucial in designing and troubleshooting electrical circuits.

2. How is current expressed through a circuit?

Current is expressed through a circuit using the unit of measurement called "amperes" or "amps." This unit represents the amount of electric charge passing through a point in the circuit per unit of time. It is typically denoted by the symbol "I" in equations.

3. What factors affect the amount of current in a circuit?

The amount of current in a circuit is affected by the voltage of the power source, the resistance of the components in the circuit, and the type of circuit (series or parallel). Changes in any of these factors can alter the amount of current flowing through the circuit.

4. How is current measured in a circuit?

Current is measured using a device called an ammeter. It is typically connected in series with the circuit and measures the flow of current through a particular point. Ammeters have a very low resistance to prevent altering the current in the circuit being measured.

5. What are the implications of expressing current through a circuit?

Expressing current through a circuit helps scientists and engineers to understand how electricity behaves and how to design and troubleshoot circuits. It also allows for the safe and efficient use of electrical power, as excessive current can cause damage to components or even start fires.

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