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Capacitance question help??

Question: In a simple photographic flashgun, a 0.20 F capacitor is charged by a 9.0 V battery. It is then discharged in a flash of duration 0.01 s. Calculate:
a) the charge and energy stored by the capacitor
b) the average power dissipated during the flash
c) the average current in the flash bulb
d) the approximate resistance of the bulb.

I can get a) which is 1.8C and 8.1J. I can also get c) which is 180A

But i need help with b) and c) the answer is 810W and 0.025 ohms but i got 1620W and 0.05 ohms. Any help please??

Reply 1

Stonebridge

b) The energy of 8.1J stored was dissipated in 0.01s according to the question. (average) Power = energy dissipated per second.
Just plug the numbers in.

d) R=V/I
Use the average current (part c) and the average pd (which is half the 9V value)

Reply 2

uberteknik

b) since:

Energy = Power x Time

Power = Energy/Time.

8.1/0.01 = 810W

c) Current = Charge/Time. 1.8/0.01 = 180A

d) Power = V*I

But V = I*R (ohms law) so

Power = (I*R)*I = I²R

Therefore R = Power / I²

R = 810 / 180²R = 810 / 32400

R = 0.025 Ohms.

Reply 3

Tobeadoc

Original post by uberteknik

When your working out part c would it be possible to use P=IV? You would have to assume that at the end of dissipation that there was no p.d. Thus the average voltage would have to be taken as 4.5V. Is this incorrect thinking??

Reply 4

uberteknik

Original post by Tobeadoc

Both are correct and give the same answer:

Energy stored in the capacitor is (CV²)/2

This must be converted to heat and light in 0.01 seconds

Current is rate of change of charge then I = dq/dt

But charge = CV and since C is constant

I = C*dv/dt

I = 0.2*9/0.01 = 180A i.e. this is the slope of the voltage and time over which all of the energy is dissipated.

This is therefore an average. In reality that the voltage and current will both fall at an exponential rate.

Power dissipated = Energy/Time

= CV²/2t

= 0.2*(9²)/2*0.1

P = 810W

Then if P = V*Iavg, this automatically implies:

Vavg = 810/180 = 4.5V

Hence using P = I^2_avg*R,

810 = (180²)R

R = 810/32400 = 0.025 Ohms.

Using V_avg = 4.5 V, you need to use P=V²/R

i.e. R = V²/Power

R = 4.5²/810 = 20.25/810

R = 0.025 Ohms.

(edited 11 years ago)