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Can Somone Help me with this.

My kettle needs to be able to give 672,000J672,000J of heat energy to water in 240s240s. Assuming that it is connected to the 240V240V mains, what current is needed? (Give your answer to 3 significant figures).
W=PtW = Pt

Where P is electrical power in this situation, so:

W=VItW = VIt

Solve for current.
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Original post by The_Big_E
W=PtW = Pt

Where P is electrical power in this situation, so:

W=VItW = VIt

Solve for current.


Sorry that not very helpful. please elaborate
Original post by 999sian
Sorry that not very helpful. please elaborate


Sure.

W=VItW = VIt

Where:
WW represents the work done by the kettle (the energy used to boil the water)
VV represents the 240V potential difference/voltage supplied to the kettle.
tt represents the time period for which the kettle is receiving electrical power.

You need to find the current, so you can rearrange the formula in the following way:

I=WVtI = \frac{W}{Vt}

Edit: If you're wondering where the W=PtW = Pt formula comes from, power is defined as the rate at which work is done:

P=WΔtP = \frac{W}{\Delta t}
(edited 7 years ago)
Reply 4
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2651019
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Original post by The_Big_E
Sure.

W=VItW = VIt

Where:
WW represents the work done by the kettle (the energy used to boil the water)
VV represents the potential difference/voltage 240V from mains electricity I presume.
tt represents the time period for which the kettle is receiving electrical power.

You need to find the current, so you can rearrange the formula in the following way:

I=WVtI = \frac{W}{Vt}

Edit: If you're wondering where the W=PtW = Pt formula comes from, power is defined as the rate at which work is done:

P=WΔtP = \frac{W}{\Delta t}


KK Thanks

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