Multifine
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In ‘normal driving conditions’, an electric car has a range of 150km. This uses all of the 200MJ energy stored in its batteries.
With the batteries initially fully charged, the car is driven 100 km in ‘normal driving conditions’. The batteries are then recharged from a household electrical supply delivering a constant current of 13.0 A at a potential difference (p.d.) of 230 V.
What is the minimum time required to recharge the batteries?

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The answer is 12.4 hours but I get 18.6 hours instead.
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Kevin De Bruyne
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(Original post by Multifine)
In ‘normal driving conditions’, an electric car has a range of 150km. This uses all of the 200MJ energy stored in its batteries.
With the batteries initially fully charged, the car is driven 100 km in ‘normal driving conditions’. The batteries are then recharged from a household electrical supply delivering a constant current of 13.0 A at a potential difference (p.d.) of 230 V.
What is the minimum time required to recharge the batteries?

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The answer is 12.4 hours but I get 18.6 hours instead.
Are you taking into account the bit where it drives 100km?

If in doubt, try dividing 12.4 by 18.6 and see what you notice about the number you get
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Kazi Aunaf
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150km---200 Mj
100km---400/3 Mj
P=IV
P=2990 W
P=W/t2990=(400/3 * 10^6)/t
T=44593.08 secs / 3600 = 12.4 hours
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Rhyme283664828
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Check how much energy is used by the engine after driving 100 km...To do that, simply do this calc- ((200×10^6)/150)×100This is the amount of energy usedNow,use formula of energy,which is VIt ...energy supplied=VItNow you'll supply that much energy to battery which you used driving 100 km...Hence Vit=((200×10^6)/150)×100Or, 230×13×t=((200×10^6)/150)×100After you've found t,just divide by 3600 and you'll find the answer in hrs..which in this case is BNote-You could've used F×d method to find out the constant Driving force and then multiply it with 100km to get energy lost,but since D.F isn't mentioned constant,use unitary method instead
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