# Lost with simple electricity q

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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?

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?

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(Original post by

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?

**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?

^{2}/ 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.

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(Original post by

Use the formula Power = V

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.

**Stonebridge**)Use the formula Power = V

^{2}/ RIf 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.

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#4

(Original post by

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?

**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?

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=I

^{2}R

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.

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(Original post by

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=I

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.

**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=I

^{2}ROn 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.

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