Power Required to Charge 10.7 μF Capacitor to 0.513 J in 43s

In summary, the energy stored in a capacitor can be calculated by using the formula W= 1/2 CV^2, where C is the capacitance in Farads and V is the potential difference in volts. One practical application of this is in electronic photoflash or strobe lights. For example, a 10.7 μF capacitor charged to 311 V would have an energy stored of 0.513 J. To find the power required to charge the capacitor to this energy level in 43 seconds, one can use the formula P = W/t, which results in a power of 0.0119 W. However, this may not be entirely accurate as there may be some energy lost in the source power supply
  • #1
Punchlinegirl
224
0
The energy (in joules) stored in a capacitor with capacitance C (in Farads), having a potential difference V (in volts), is given by W= 1/2 CV^2. One application is in the electronice photoflash or strobe light. In such a unit, a capacitor of 10.7 [tex] \mu F [/tex] is charged to 311 V. Find the energy stored. Answer in units of J.
I got this part.. the answer was 0.513 J.
Next, it says, find the power required to charge the capacitor to this energy level in 43 s. Answer in units of W.
Since W is given in J/s, do I just do 0.513/43? which gives me .0119. We didn't do much of this in class, so I don't know how else to do it.
 
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  • #2
yep, what you did seems right to me.
 
  • #3
this wasn't right... is there another way to do it?
 
  • #4
Hmmm. Looks tricky. Maybe try a few examples with a real >=311V power supply and some source impedance. Try a source impedance that gives you one time constant at 43 seconds with that capacitance, and calculate the total power delivered, and the power lost in the source resistor. Then try a different source voltage and source resistance to see if you get the same answer or a different answer (I don't know which it will be offhand). The problem may be trying to point out that you can't charge a capacitor without burning some energy in the source power supply...
 

1. How is the power required to charge a capacitor calculated?

The power required to charge a capacitor is calculated by multiplying the capacitance (in Farads) by the voltage (in Volts) and dividing it by the time (in seconds).

2. What is the capacitance of a 10.7 μF capacitor?

The capacitance of a capacitor is a measure of its ability to store electric charge. A 10.7 μF capacitor has a capacitance of 10.7 microfarads, which is equivalent to 0.0000107 Farads.

3. How much energy does a 10.7 μF capacitor hold when charged to 0.513 J?

The energy stored in a capacitor is calculated by multiplying the capacitance by the square of the voltage. In this case, the energy stored in the 10.7 μF capacitor when charged to 0.513 J would be 2.747 μJ (microjoules).

4. How long does it take to charge a 10.7 μF capacitor to 0.513 J?

The time it takes to charge a capacitor depends on the power supply and the circuit in which it is connected. In this case, it took 43 seconds to charge the 10.7 μF capacitor to 0.513 J.

5. What is the significance of charging a capacitor to 0.513 J?

The amount of energy (in Joules) stored in a capacitor determines its ability to deliver a certain amount of power. Charging a 10.7 μF capacitor to 0.513 J means that it can hold this amount of energy and can potentially deliver it to a circuit in a short amount of time.

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