Consider the flash unit of a camera which uses a 3.0 V battery to charge a capac

Question

Consider the flash unit of a camera which uses a 3.0 V 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)What is the initial charge on the positive plate of the capacitor just BEFORE point d is connected to point a?

2)What is the potential difference across the capacitor just BEFORE point d is connected to point a?

8)Find the charge on the positive plate when the capacitor is fully charged.

9)How much energy is stored in the fully charged capacitor?

10)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?)

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

13)What is the current 1 second after discharging begins?

14)Find the equivalent capacitance when another capacitor of 3000 micro-farad is connected in series with the 1500 micro-farad capacitor.

15)Now, the 3000 micro-farad capacitor is removed from the system. Find the equivalent capacitance when the 3000 micro-farad capacitor is connected in parallel with the 1500 micro-farad capacitor.

16)Calculate how much energy is stored in each system of capacitors (i.e. series and parallel) in the previous two questions. What does this indicate to you as to a possible reason for connecting capacitors in series or parallel?

Explanation / Answer

1) initial charge 0

2) intial potential difference =0 v

initial current=i=v/R=3/2000=0.0015 A

Vc=q/c

q=q0(1-exp(-t/Rc)

1/2 cV^2

electric field=V/d

i(t)=V0/R*exp(-t/T)    (T=Rc)

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