Question

Capacitors are used in many applications where one needs to supply a short burst of relatively large current. A 164-μF capacitor in an electronic flash lamp supplies a burst of current that dissipates 10.1 J of energy (as light and heat) in the lamp.

To what potential difference must the capacitor initially be charged?

What is its initial charge?

Approximately what is the resistance of the lamp if the current reaches 5.00% of its original value in 1.37 ms?

Answer 1

Solution: Given that C 164uF E-10.1 J (a)We know that E-cv2 2E C V164x10-5 V 350.956 V (b) The initial charge is Q-CV Q-164x10-x350.956 Q-0.0575 C c) The voltage and current equation will be similar, -137 10 τ-0.457x 10-3 RC 0.457x103 R 2786Ω τ RC time constant in RC circuit)