You recently purchased a used camera with an electric flash. After taking a roll

Question

You recently purchased a used camera with an electric flash. After taking a roll of pictures you are
disappointed that the flash isn’t bright enough. You look in the camera and notice that the flash works
by allowing a battery to slowly charge a capacitor, and then quickly releasing the capacitor’s stored
electrical energy through a light bulb when a photo is taken. You think that the problem with your
camera may be that not enough energy is stored in the capacitor to properly light the flash bulb. You
have another capacitor with different capacitance, but aren’t sure if you should connect it in series or in
parallel with the original capacitor in order to store the most energy. You make an educated guess, and
decide to test your prediction with circuits consisting of one or two initially uncharged capacitors, a
battery, and a light bulb. You plan to measure the amount of time the bulb stays lit for one capacitor
and for each of the possible arrangements of two capacitors, reasoning that if capacitors in a circuit can
store more energy, they will take longer to fully charge.

How to find the equation for energy stored related to time?

Explanation / Answer

here,

for maximum brightness , the capacitor are connected in parallel

two initially uncharged capacitor , a light bulb and a battery is connected

these type of circuits are known as RC circuit

time constant , T = resistancce of bulb * ( C1 + C2)

the potential difference , V = V0 * ( 1 - e^ -t/T)

where , Vo is the peak voltage

the energy stored , E = 0.5 * C * V^2

E = 0.5 * (C1 + C2 ) * (V0 * ( 1 - e^ -t/(R * ( C1 + C2))))^2

the equation for the energy stored is E = 0.5 * (C1 + C2 ) * (V0 * ( 1 - e^ -t/(R * ( C1 + C2))))^2

Related Questions

Navigate

• Browse (All)
• Browse Y
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.