2.) What is the voltage across the 2.0 kilo-ohm resistor when the capacitor is f

Question

2.) What is the voltage across the 2.0 kilo-ohm resistor when the capacitor is fully charged?

4.) What is the current 1 second after discharging begins?

Capacitors are used in several different ways in electronic circuits, and RC circuits are actually quite common in everyday life. For example,. they are used to control the speed of a car?s windshield wiper, the timing of the change of traffic lights, and they are even used in camera flashes and in many other electronic devices. In RO circuits, the interest is not In the final voltage and charge on the capacitor, but rather in how these variables change in time. Unlike a battery which could take minutes or hours to discharge, a capacitor can dump its entire charge in a tiny fraction of a second. The example below demonstrates this in terms of the flash unit on a camera. Note: flash units and lVs have warnings about opening them up because they contain large capacitors that can potentially harm or kill someone with the charge they contain Consider the flash unit of a camera which uses a 3.0V battery to charge a capacitor of 1500 micro-farads. The capacitor is then discharged through a flash-lamp which has a resistance 2 kilo-ohm. The figure shown is a circuit diagram of this flash. Please refer to it in answering the following questions: . 1.) Find the electric field intensity between the plates of the capacitor if the separation distance is 0.50 mm. (Hint: How is the magnitude of the electric field intensity related to the potential difference between two points separated by a known distance?) 2.) What is the voltage across the 2.0 kilo-ohm resistor when the capacitor is fully charged? 3.) What is the current at the instant discharging begins, at

Explanation / Answer

part A:

electric field Intneisty E = V/d

where V is Pot difff

and

d is distance

so

E = 3/0.5e-3

E = 6 *10^3 V/m

--------------------------------

apply current i = V/R

V = 3V

--------------------------

3. current i = v/R = 3/2000 = 1.5 mA

--------------------------------

apply I = Io e^-t/RC

RC = 2000*1500e-6

RC = 3 secs

I = 1.5 mA * e^(-1/3)

I = 1.074 mA

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