# Calculating capacitance required for burst current when V, t and i are known

#### electronice123

Joined Oct 10, 2008
325
This is in reference to the question below:

The circuit has a burst current of 20A @ 12V for 5 seconds with a 1V drop, calculate the required capacitance?

I get the 2V * 20A * 5s = 1200 Joules, but the formula below that one makes no sense to me. I've tried to calculate it over and over and I get .724Farads, not 104Farads like the answer on Stackex. What am I doing wrong?

My attempt at a solution:
C/2 (12^2-11^2)V^2 =1200

(2) C/2 (12^2 -11^2) V^2 = 1200(2)

C(12^2-11^2)V^2=2400

C(144-121)V^2=2400

C(23)V^2=2400

C23(144) =2400

C(3312) =2400

C(3312)/ 3312= 2400/3312 = .724 F

I've been racking my brain so much I've only confused myself further, any help will be greatly appreciated.

#### crutschow

Joined Mar 14, 2008
31,580
Calculating the energy unnecessarily complicates the calculations (and obviously gave you the incorrect answer).

20A for 5 seconds is 100 coulombs of charge (q) .
The charge on a capacitor is C = Q / V, so the required capacitance for a 1V drop is C = 100q / 1V = 100F.

• electronice123

#### WBahn

Joined Mar 31, 2012
28,192
This is in reference to the question below:

The circuit has a burst current of 20A @ 12V for 5 seconds with a 1V drop, calculate the required capacitance?

I get the 2V * 20A * 5s = 1200 Joules, but the formula below that one makes no sense to me. I've tried to calculate it over and over and I get .724Farads, not 104Farads like the answer on Stackex. What am I doing wrong?
Where does this 2 V come from?

As I look at it, it appears to be a typo and that you meant 12 V.

While you can certainly get to the answer using energy arguments, it is FAR from the simplest and most straight forward.

But, since you went down that road, you want to learn how to do it properly.

Your 1200 J appears to be your attempt to get the amount of energy consumed by the load during this five seconds. However, it's just an approximation since the voltage is dropping during that 5 s. What you don't know is whether the current drops in proportion to the voltage drop or not. So you have to make some assumptions (and should state them in your solution). You've made a couple of assumptions here, but they are both reasonably defensible. Do you recognize what they are?

My attempt at a solution:
C/2 (12^2-11^2)V^2 =1200

(2) C/2 (12^2 -11^2) V^2 = 1200(2)

C(12^2-11^2)V^2=2400

C(144-121)V^2=2400

C(23)V^2=2400

C23(144) =2400

C(3312) =2400

C(3312)/ 3312= 2400/3312 = .724 F
Your very sloppy use of units is what has caused you problems. You put in some units, but not all of them. Then you later treated a unit like a variable, which totally destroyed any validity you might have had, and then because you didn't track units, but instead just tacked on whatever you unit you wanted the answer to have at the end, you deprived yourself of the ability to catch your error.

So let's go through, tracking the units properly, and see what you get.

CV²/2 = E

C/2 ( (12 V)² - (11 V)² ) =1200 J
C/2 ( (144 V² - 121 V² ) =1200 J
C/2 ( 23 V² ) =1200 J

C = (2·1200 J) / (23 V²)
C = 2400/23 J/V²
C = 2400/23 J/V² (1 J = 1 A · 1 V · 1 s = 1 coulomb·V) I'm writing out coulomb to avoid confusion with the variable C
C = 104.3 coulomb/V
C = 104.3 F

I've been racking my brain so much I've only confused myself further, any help will be greatly appreciated.
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