Help on an exam question

[westos-electric]

Question
If a current of 200mA flows into a capacitor for 40ms how much charge will develop in the capacitor?

please help with the answer but i need to know how it's worked out

Thanks
Wayne

 

[pushrod]

Question
If a current of 200mA flows into a capacitor for 40ms how much charge will develop in the capacitor?

please help with the answer but i need to know how it's worked out

Thanks
Wayne

Q =I x t

 

[westos-electric]

Sorry i'm finding this harder to understand than I should, so is it 200x40?

Thanks
Wayne

 

[pushrod]

Sorry i'm finding this harder to understand than I should, so is it 200x40?

Thanks
Wayne

a milliamp is a 1000th of an amp and same for milliseconds so it is 0.200 x 0.040 = 0.008 coulombs



edit : to change milli anything into its whole unit just divide by 1000

hope that is clearer

 

[westos-electric]

Yeah that makes much more sense!!

If you have any other info/tips to help me through the exam let me know

Thanks

 

[pushrod]

Yeah that makes much more sense!!

If you have any other info/tips to help me through the exam let me know

Thanks

Don't Q u I t or Q = It



Eff the old Bill or F = BIL

try and make up silly ways to remember the formula

" Look there is a queue to buy old citroen CVs"

or Q = CV
(Q is charge, C is capacitance and V is voltage)

 

[westos-electric]

Hi

This is the last one.. if you can point me in the right direction?

Question
A small heating circuit draws 1.5kW of power and has an internal resistance of 0.38, what is the total voltage drawn.

Now the answer is 24v but i'm not 100% sure.

My workings are that P= v2/R so 1.5/0.38 = 5.92

then v= PxR but this is where i fall down.

 

[pushrod]

Hi

This is the last one.. if you can point me in the right direction?

Question
A small heating circuit draws 1.5kW of power and has an internal resistance of 0.38, what is the total voltage drawn.

Now the answer is 24v but i'm not 100% sure.

My workings are that P= v2/R so 1.5/0.38 = 5.92

then v= PxR but this is where i fall down.

On the right track but you must work in watts, volts, amps and ohms
so the 1.5 kW becomes 1500 watts

P =V²/R so V² = P x R so 1500 x 0.38 = 570 then find the √ of 570

should get something that rounds up to 24.

 

[JUD]

On the right track but you must work in watts, volts, amps and ohms
so the 1.5 kW becomes 1500 watts

P =V²/R so V² = P x R so 1500 x 0.38 = 570 then find the √ of 570

should get something that rounds up to 24.

In other words

V = √(P x R)


This may help

 

[westos-electric]

Awesome cheers this info has really helped and is starting to make sense again!

Cheers
Wayne

 

[westos-electric]

Hi sorry to ask but could you check this question & answer i worked out?

The voltage supplied to the primary winding of a step down transformer is 660v and the output voltage of 230v. If the primary is 12 A the value of the secondary will be?

660/230 = 2.86 then
12x2.86 = 34.32 the answer on the paper is 34.4V have I worked this out correct and just round up?

Thanks

 

[JUD]

You've got a slightly different answer due to you rounding off.

If you do it all in one calculation it works out at 34.4A

(660 ÷ 230) x 12 = 34.4A

 

[westos-electric]

Hi

I just wondered if you can assist on this question?

A circuit contains 3 capacitors in series of sizes 19, 24, and 32 micro farad. What is the total capacitance in the circuit?

I get 75 x10-6 == 7.5 the answer is 8 microfarad.

Are you able to say if this is correct?

Thanks
Wayne

 

[Guest123]

Calculating capacitors in series and parallell is exactly opposite to resistors in series and parallell.

The answer will be slightly smaller than your smallest capacitor.

So for caps in series its 1/ 1overC1 + 1overC2 + 1 overC3, and for parallell you just add them up. Exactly opposite to resistance.

 

[JUD]

The calculation for capacitance in series is:

1 ÷ (1 ÷ 19) + (1 ÷ 24) + (1 ÷ 32) = 8uF


**EDIT** Lenny beat me to it. Runner up prize again