text: # Homework Help: Circuit Related Q
  
  1. Jul 24, 2006
  
  ### MD2000
  
  Determine the power dissipated in the 6.0 resistor in the circuit shown in the drawing. (R1 = 4.0 , R2 = 6.0 and V1 = 15 V.)
  
  Am I supposed to be using P = V^2/R..is the power flowing through each resistor gonna be diff?
  
  Anyone wanna help me out where to start?
  
  2. Jul 24, 2006
  
  You can solve this many different ways. First note that the current through the 2ohm and the 1ohm resistor is the same, and that the current through R2 and the 1ohm resistor is the same.
  
  I would first solve for the currents, and then you know that:
  
  $$P=IV$$
  
  $$V=IR$$
  
  Thus,
  
  $$P=I(IR)=I^2R$$
  
  Do you know KVL, KCL?
  
  3. Jul 25, 2006
  
  ### sdekivit
  
  4. Jul 25, 2006
  
  ### MD2000
  
  How exactly can I get the currents for each individual peice?
  
  I know that the total current is gonna be V = IR..which equals 2.46..
  
  But how can I solve for individual currents? Since you don't have V..what can you use to get the I?
  
  Last edited: Jul 25, 2006
  5. Jul 26, 2006
  
  ### Hootenanny
  
  Staff Emeritus
  Think Kirchoff's laws, you can use these the calculate the voltage or current at each point in the circuit.
  
  6. Sep 8, 2006
  
  ### physics_girl
  
  I have a similar question with multiple batteries and resistors in one big circuit. I have been given two labelled junctions but within one smaller loop, there are two batteries and one resistor...
  I don't know where to begin, any advice???
explanation: The document is a series of posts from a homework help forum, where a user named MD2000 asks for help with a circuit problem. The user is trying to determine the power dissipated in a 6.0-ohm resistor in a circuit with given resistances and voltage. The post includes a question and subsequent responses from other users providing guidance and advice on how to approach the problem.
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  1. Briefly describe the document and mention if it contains a question or a question and answer.
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distilabel_metadata.role: system
distilabel_metadata.content: # Homework Help: Circuit Related Q
  
  1. Jul 24, 2006
  
  ### MD2000
  
  Determine the power dissipated in the 6.0 resistor in the circuit shown in the drawing. (R1 = 4.0 , R2 = 6.0 and V1 = 15 V.)
  
  Am I supposed to be using P = V^2/R..is the power flowing through each resistor gonna be diff?
  
  Anyone wanna help me out where to start?
  
  2. Jul 24, 2006
  
  You can solve this many different ways. First note that the current through the 2ohm and the 1ohm resistor is the same, and that the current through R2 and the 1ohm resistor is the same.
  
  I would first solve for the currents, and then you know that:
  
  $$P=IV$$
  
  $$V=IR$$
  
  Thus,
  
  $$P=I(IR)=I^2R$$
  
  Do you know KVL, KCL?
  
  3. Jul 25, 2006
  
  ### sdekivit
  
  4. Jul 25, 2006
  
  ### MD2000
  
  How exactly can I get the currents for each individual peice?
  
  I know that the total current is gonna be V = IR..which equals 2.46..
  
  But how can I solve for individual currents? Since you don't have V..what can you use to get the I?
  
  Last edited: Jul 25, 2006
  5. Jul 26, 2006
  
  ### Hootenanny
  
  Staff Emeritus
  Think Kirchoff's laws, you can use these the calculate the voltage or current at each point in the circuit.
  
  6. Sep 8, 2006
  
  ### physics_girl
  
  I have a similar question with multiple batteries and resistors in one big circuit. I have been given two labelled junctions but within one smaller loop, there are two batteries and one resistor...
  I don't know where to begin, any advice???
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document is a series of posts from a homework help forum, where a user named MD2000 asks for help with a circuit problem. The user is trying to determine the power dissipated in a 6.0-ohm resistor in a circuit with given resistances and voltage. The post includes a question and subsequent responses from other users providing guidance and advice on how to approach the problem.
  
  Label: question-and-answer