Ohm's Law, Resistance, and Power

  • Thread starter jperentosh
  • Start date
  • #1

Homework Statement



Please look at the attachment.


Homework Equations



ΔV=I*R
R=ΔV/I
Power=I*ΔV


The Attempt at a Solution



I said that ΔV of A and ΔV of B are the same (each 3/2 V). Using this, I got that the current at A = 3/20 Amps. However, this means that the current at B and C is each 3/40 Amps. However, then the 1.5 V * 3/40 does not equal 10 ohms, which is given. How am I supposed to go about this problem to find the power dissipated in each bulb and find what changes in terms of bulb brightnesses in A and B if bulb C is unscrewed?
 

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Answers and Replies

  • #2
SteamKing
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You have one resistor in series with two others which are in parallel. You have not found the correct current in the circuit.
 
  • #3
Simon Bridge
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I said that ΔV of A and ΔV of B are the same (each 3/2 V)
The resistance of B and C in parallel is less than the resistance of A, so the volt-drop will be less.
 
  • #4
How so? They're each 10 ohms.
 
  • #5
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J, they are in parallel. There are 2 possible paths for current so there is less total resistance for electrons being pushed through this path. You can think of this sort of like having 2 checkout lanes open at a grocery store. More shoppers can move through faster than if there were one lane open.

Resistors in series (in a row) are added as R1+R2. Resistors in parallel (multiple paths) are not added this way. Hint: If there are two parallel paths, both must have the same voltage across them.
 
  • #6
CWatters
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How so? They're each 10 ohms.
Apply KCL to the point where A is connected to B and C. You will find the current through B (or C) will be half the current through A. So the voltage drop oved B will be half that of A.
 
  • #7
Simon Bridge
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How so? They're each 10 ohms.
J, they are in parallel.
Apply KCL to the point where A is connected to B and C.
Thanks folks - jperentosh: yes they are correct. By now you should have seen the equations for resistors in series and parallel. You can always check with an ohmmeter if you don't believe it.
 

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