This is a trick question that comes up often in basic capacitor problems. Assume some source impedance for the battery, and calculate the energy that is burned in that resistance as the battery charges the cap. Then try a different source resistance and re-calculate...do you see a pattern?dan greig said:I think I have found the answer to the first question using,
E = 1/2 CV^2 or E = 1/2 QV
and I think the reason the two answers wouldn't be the same is because the rate of charge decreases as the charge increases therefore the capacitor can never fully charge.
I don't see how I can get a different answer in the second question?
rammstein said:well, the second answer is different coz there might be energy dissipation in the form of heat in the capacitors and the wires connecting them...Thats all I can think of.Otherwise the energy given out by the battery should be equal to the energy gained by the capacitor-energy conservation. What do u say?
The energy stored in a capacitor can be calculated using the formula: E = 1/2 * CV^2, where E is the energy in joules, C is the capacitance in farads, and V is the voltage in volts. In this case, the energy stored in the capacitor would be 1/2 * 10μF * (24V)^2 = 2.88μJ.
A 10μF capacitor has a capacitance of 10 microfarads (μF). This means that it can store a certain amount of electrical charge at a given voltage. A higher capacitance means that the capacitor can store more charge, while a lower capacitance means it can store less charge.
Yes, a 10μF capacitor can be charged to any voltage, as long as it is within the capacitor's voltage rating. However, the energy stored in the capacitor will vary depending on the voltage it is charged to.
To convert microfarads (μF) to farads (F), you can use the formula: 1μF = 10^-6F. This means that to convert from microfarads to farads, you would divide the value in microfarads by 1,000,000.
If the voltage in a capacitor is doubled, the energy stored in the capacitor will increase by a factor of four. This is because the formula for calculating energy in a capacitor (E = 1/2 * CV^2) includes the voltage squared. Therefore, a change in voltage has a significant impact on the amount of energy stored in a capacitor.