Capacitors are used in several different ways in electronic circuits, and RC cir

Question

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 cars 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 RC circuits, the interest is not in the final voltage and charge on the capacitor, but rather in how these variableschange 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 TVs have warnings about opening them up because they contain large capacitors that can potentially harm or kill someone with the charge they contain.

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 inparallel 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?

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 car½s 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 RC circuits, the interest is not in the final voltage and charge on the capacitor, but rather in how these variableschange 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 TVs 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 3.0 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: ? That is, at the instant point d is connected to point b. 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 inparallel 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? after points a and d are joined? 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? 12. What is the current at the instant discharging begins, at after points a and d are joined? 7. What is the magnitude of the voltage across the capacitor at the instant (i.e., when the knife blade switch first connects to a)? 6. What is the magnitude of the current flow to the capacitor at the instant (i.e., when points a and d are connected.)? 5. What is the magnitude of the voltage across the capacitor at the instant (i.e., when points a and d are joined)? 4. What is the magnitude of the current flow to the capacitor at the instant 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? 3. What is the magnitude of the current flow through the capacitor at the instant

Explanation / Answer

1) Qc = 0

2) VC = 0

3) Imax = Vmax/R

= 3/2*10^3

= 1.5*10^-3 A or 1.5 mA

4) Imax = Vmax/R

= 3/2*10^3

= 1.5*10^-3 A or 1.5 mA

5) VC = 0

6) I = 0

7) Vc = 0

8) Qmax = C*Vmax

= 1500*10^-6*3

= 4.5*10^-3 C

9) U = Q^2/(2*C)

= (4.5*10^-3)^2/(2*1500*10^-6)

= 6.75*10^-3 J

10) E = Vmax/d

= 3/0.5*10^-3

= 6000 N/c

11) VR = 0

12) Imax = 1.5*10^-3 A

13) T = R*C = 2000*1500*10^-6 = 3 s

I = Imax*e^(-t/T)

= 1.5*10^-3*e^(-1/3)

= 1.075*10^-3 A

14)
Cnet = 1500*3000/(1500+3000)

= 1000 micro F

15) Cnet = 1500 + 3000
= 4500 micro F

16) U_series = 0.5*Cnet*v^2

= 0.5*1000*10^-6*3^2

= 4.5*10^-3 J

U_paralle = 0.5*Cnet*V^2

= 0.5*4500*10^-6*3^2

= 2.025*10^-2 J

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