Power wasted/delivered to a resistor

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Homework Help Overview

The problem involves two 8 ohm resistors connected in series with a 12.0V battery that has an internal resistance of 8 ohms. Participants are tasked with calculating the power delivered to each external resistor and the power wasted due to the internal resistance.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore various calculations for current and power, questioning the total resistance in the circuit and the impact of internal resistance. Some participants suggest sketching the circuit to clarify the flow of current and the total resistance encountered.

Discussion Status

There is an ongoing exploration of different approaches to calculating power and current. Some participants have provided guidance on the formulas for power, while others are attempting to reconcile their calculations with the circuit's configuration. Multiple interpretations of the problem are being discussed.

Contextual Notes

Participants express confusion over the correct application of formulas and the role of internal resistance in the overall calculations. There is a mention of the need to include units in results, and some participants are working through the implications of their calculations on the total power and distribution across the resistors.

chris858
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Homework Statement


two 8 ohm resistors and a battery of emf 12.0V and internal resistance of 8ohms are connected in series. Calculate the power delivered to each external resistor and then calculate the power wasted due to internal resistance.
I keep coming up with the answer 18 but apparently that is not right, and I am strugglying, so help would be much appreciated.

Homework Equations


W=IV
I=V/R

The Attempt at a Solution


VxV/R
12 X 12/8 =18...
current = 1.5A
 
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chris858 said:

Homework Statement


two 8 ohm resistors and a battery of emf 12.0V and internal resistance of 8ohms are connected in series. Calculate the power delivered to each external resistor and then calculate the power wasted due to internal resistance.
I keep coming up with the answer 18 but apparently that is not right, and I am strugglying, so help would be much appreciated.

Homework Equations


W=IV
I=V/R

The Attempt at a Solution


VxV/R
12 X 12/8 =18...
current = 1.5A
Sketch the circuit. How many resistors does the current flow through? What's the total resistance that the current flows through? So what then the current?
 
flows through two resistors, total resistance = 16 ohms. Current = 12/16 =0.75A
 
chris858 said:
flows through two resistors, total resistance = 16 ohms. Current = 12/16 =0.75A
What about the battery's internal resistance?
 
16+8=24 so 12/24=0.5A
 
chris858 said:
16+8=24 so 12/24=0.5A
Right. So now what will you do now that you know the current through each resistor?
 
pwer= voltage times voltage over current. p=VxV/R so 12x 12/0.5=2
 
chris858 said:
pwer= voltage times voltage over current. p=VxV/R so 12x 12/0.5=2
R is not current in the formula P = V2/R, it represents resistance.

There are other formulas for power. What are they?

Oh, and be sure to include units with all results.
 
power = work/time
P=IV
P=V²/R
P=RI²
 
  • #10
power is measured in watts
current in amps
resistance in ohms
voltage in volts
 
  • #11
Okay, so what is the power dissipated by one of the resistors?
 
  • #12
I get it now, so basically the equation i was using gave the total power watts, which is actually 6 not 2, and so there are 3 resistors, including the internal resistance, so therefore you have to divide 6 by 3 which gives 2.0Watts per resistor.
Thankyou very much!
 

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