# Lost with simple electricity q

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#1
I just don't know how to solve this. I can't get the diagram on here but this is all of the info in the q.

The diagram shows a circuit containing a battery and three identical resistors X, Y and Z.

---battery--
|y------- z----|
| x |

The total power supplied by the battery is 18 W. What is the power dissipated as heat in resistor X?
0
1 month ago
#2
(Original post by nsaahelp)
I just don't know how to solve this. I can't get the diagram on here but this is all of the info in the q.

The diagram shows a circuit containing a battery and three identical resistors X, Y and Z.

---battery--
|y------- z----|
| x |

The total power supplied by the battery is 18 W. What is the power dissipated as heat in resistor X?
Use the formula Power = V2 / R
If the resistance of X is R, say, then the resistance of the top branch is 2R and the lower branch is R
Total power is 18W and V is the same for top and bottom branches.
You can do it using the ratio of the powers in the two branches now.
0
#3
(Original post by Stonebridge)
Use the formula Power = V2 / R
If the resistance of X is R, say, then the resistance of the top branch is 2R and the lower branch is R
Total power is 18W and V is the same for top and bottom branches.
You can do it using the ratio of the powers in the two branches now.
I think I might have drawn the circuit wrong before. Ive attached the actual diagram now. Wouldn't the resistance through the top branch ( passing through X and Z ) be 2R and the resistance through the lower branch ( through Y and Z) be 2R as well? How would the lower branch be just R?
0
4 weeks ago
#4
(Original post by nsaahelp)
I think I might have drawn the circuit wrong before. Ive attached the actual diagram now. Wouldn't the resistance through the top branch ( passing through X and Z ) be 2R and the resistance through the lower branch ( through Y and Z) be 2R as well? How would the lower branch be just R?
Well yes, the diagram is completely different now.
The resistance of the parallel section (X and Y) is R/2 and the other resistor is R. (As they are identical and all equal to R)
There are a couple of ways of approaching this. One is:
Use the power formulas P=I2R
On the 2 separate sections (the X&Y and the Z)
This will give you the ratio of the power in those 2 sections.
When you have that, you know the total power is 18W so you can work out the power in the individual sections.
For example, if the ratio was 2:1 the power would be 12W:6W
When you have the power in the XY section, you know that each resistor is half that power as they are identical and the current divides equally.
1
#5
(Original post by Stonebridge)
Well yes, the diagram is completely different now.
The resistance of the parallel section (X and Y) is R/2 and the other resistor is R. (As they are identical and all equal to R)
There are a couple of ways of approaching this. One is:
Use the power formulas P=I2R
On the 2 separate sections (the X&Y and the Z)
This will give you the ratio of the power in those 2 sections.
When you have that, you know the total power is 18W so you can work out the power in the individual sections.
For example, if the ratio was 2:1 the power would be 12W:6W
When you have the power in the XY section, you know that each resistor is half that power as they are identical and the current divides equally.
Got it now! Thanks for the help
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