The energy stored in a capacitor with capacitance C, having an electric potential difference ΔV, is represented by W = 1/2 C ΔV2. One application of this is in the electronic photoflash of a strobe light, like the one in the figure below. In such a unit, a capacitor of 10.0 µF is charged to 340 V. Find the energy stored
W = (1/2) C V^2
W = .5 * 10 * 10^-6 * 340^2
= .578 Joules
Well, the formula for the energy stored in a capacitor is W = 1/2 C ΔV^2. So, let's plug in the values given in the problem - C = 10.0 µF and ΔV = 340 V.
W = 1/2 (10.0 µF)(340 V)^2
Now, let's do the calculation:
W = 1/2 (10.0 × 10^-6 F) (340^2 V^2)
W = 1/2 (10.0 × 10^-6)(115600 V^2)
W = 1/2 (1.156 × 10^-3) Joules
And there you have it! The energy stored in the capacitor is approximately 0.578 mJ (or about enough energy to make a lame joke).
To find the energy stored in the capacitor, we can use the formula:
W = 1/2 C ΔV^2
Given:
C = 10.0 µF = 10.0 x 10^-6 F
ΔV = 340 V
Substituting these values into the formula, we have:
W = 1/2 (10.0 x 10^-6 F) (340 V)^2
W = 1/2 (10.0 x 10^-6 F) (115600 V^2)
W = 0.5 x 10.0 x 10^-6 F x 115600 V^2
W = 5.78 J
Therefore, the energy stored in the capacitor is 5.78 Joules.
To find the energy stored in a capacitor, we will use the formula:
W = (1/2) C ΔV^2
where W represents the energy stored, C is the capacitance of the capacitor, and ΔV is the potential difference across the capacitor.
In this case, the given values are:
C = 10.0 µF (microfarads)
ΔV = 340 V
Now, let's substitute these values into the formula to calculate the energy stored:
W = (1/2) * (10.0 µF) * (340 V)^2
First, we need to convert the capacitance from microfarads (µF) to farads (F), as the standard unit for capacitance is farads.
1 µF = 10^-6 F
So, 10.0 µF = 10.0 × 10^-6 F = 1.0 × 10^-5 F
Now, let's substitute the new value of capacitance into the formula:
W = (1/2) * (1.0 × 10^-5 F) * (340 V)^2
Next, we'll calculate the value of (340 V)^2:
W = (1/2) * (1.0 × 10^-5 F) * (340^2 V^2)
W = (1/2) * (1.0 × 10^-5 F) * (115,600 V^2)
W = 0.5 * (1.0 × 10^-5 F) * (115,600 V^2)
Now, we can perform the multiplication:
W = 5.78 × 10^-1 * 10^-5 * 115,600 V^2
W = 5.78 × 10^-6 * 115,600 V^2
Finally, we'll square the voltage and multiply it by the constant:
W = 5.78 × 10^-6 * 13,324,000 V^2
W = 76.9 V^2
Therefore, the energy stored in the capacitor is approximately 76.9 Joules.