# Capacitors

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I'm stuck on part b(ii) in where it asks me to calculate the time taken for the capacitor to recharge from 11V to 12V. The work done from recharging from 11 to 12 volts is 230 joules. What I don't understand is the mark scheme divided 230 by 5%(which is the percentage of the output power by the solar panels that the capacitor has the power rating of).

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#2

Technically what the mark scheme has done there will give you the ratio of time spent charging the capacitor to the time spent collecting the solar power, multiplied by the solar energy collected.

In my opinion it's a very lazy solution to have in the mark scheme.

If you do the algebra you can see that it's true though:

The output power of the solar cells is equal to the energy they produce, per unit time. Normally that unit time is 1 second, but it makes it easier here to just leave it in terms of energy/time.

So P

We're told that the input power to the capacitor, is 5% of the solar cells output power.

Let's call the capacitor input power P

So P

so P

therefore t

Hope this helps

In my opinion it's a very lazy solution to have in the mark scheme.

If you do the algebra you can see that it's true though:

The output power of the solar cells is equal to the energy they produce, per unit time. Normally that unit time is 1 second, but it makes it easier here to just leave it in terms of energy/time.

So P

_{1 }= E_{1}/t_{1}We're told that the input power to the capacitor, is 5% of the solar cells output power.

Let's call the capacitor input power P

_{2}So P

_{1}= 20*P_{2 }The capacitor input power is then equal to the energy (work done) by charging the capacitor (E_{2}), divided by the time taken to charge it (t_{2}).so P

_{2}= E_{2}/t_{2 }combine the equations: E_{1}/t_{1}= 20*(E_{2}/t_{2})therefore t

_{2}/t_{1}= (230*20)/10,000Hope this helps

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(Original post by

Technically what the mark scheme has done there will give you the ratio of time spent charging the capacitor to the time spent collecting the solar power, multiplied by the solar energy collected.

In my opinion it's a very lazy solution to have in the mark scheme.

If you do the algebra you can see that it's true though:

The output power of the solar cells is equal to the energy they produce, per unit time. Normally that unit time is 1 second, but it makes it easier here to just leave it in terms of energy/time.

So P

We're told that the input power to the capacitor, is 5% of the solar cells output power.

Let's call the capacitor input power P

So P

so P

therefore t

Hope this helps

**james153**)Technically what the mark scheme has done there will give you the ratio of time spent charging the capacitor to the time spent collecting the solar power, multiplied by the solar energy collected.

In my opinion it's a very lazy solution to have in the mark scheme.

If you do the algebra you can see that it's true though:

The output power of the solar cells is equal to the energy they produce, per unit time. Normally that unit time is 1 second, but it makes it easier here to just leave it in terms of energy/time.

So P

_{1 }= E_{1}/t_{1}We're told that the input power to the capacitor, is 5% of the solar cells output power.

Let's call the capacitor input power P

_{2}So P

_{1}= 20*P_{2 }The capacitor input power is then equal to the energy (work done) by charging the capacitor (E_{2}), divided by the time taken to charge it (t_{2}).so P

_{2}= E_{2}/t_{2 }combine the equations: E_{1}/t_{1}= 20*(E_{2}/t_{2})therefore t

_{2}/t_{1}= (230*20)/10,000Hope this helps

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#4

(Original post by

I'm slightly confused on how you got the twenty everything else seemed good i prefer algebra way shows more working making it easier to understand.

**MSB47**)I'm slightly confused on how you got the twenty everything else seemed good i prefer algebra way shows more working making it easier to understand.

Thus the solar cells output power is 20 times greater than the input power to the capacitor. Hence, is the solar cells output power, etc etc.

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(Original post by

Just by reading his answer alone, it was mentioned that 'the input power to the capacitor, is 5% of the solar cells output power'.

Thus the solar cells output power is 20 times greater than the input power to the capacitor. Hence, is the solar cells output power, etc etc.

**Phichi**)Just by reading his answer alone, it was mentioned that 'the input power to the capacitor, is 5% of the solar cells output power'.

Thus the solar cells output power is 20 times greater than the input power to the capacitor. Hence, is the solar cells output power, etc etc.

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#6

(Original post by

Not familiar with the maths :/ whats the conversion method from 5% to 20?

**MSB47**)Not familiar with the maths :/ whats the conversion method from 5% to 20?

If you have 5% of A, then you have 0.05A.

If we say, B is 5% of A, then , or

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(Original post by

I think you're over complicating things.

If you have 5% of A, then you have 0.05A.

If we say, B is 5% of A, then , or

**Phichi**)I think you're over complicating things.

If you have 5% of A, then you have 0.05A.

If we say, B is 5% of A, then , or

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