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Three capacitors C1 = 10.7 µF, C2 = 23.0 µF, and C3 = 29.3 µF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

User Engo
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Final answer:

To find the total energy stored in a series combination of capacitors, sum up the energies stored in each capacitor using the formula U = (1/2) * C * V^2. Plug in the values for capacitance and voltage for each capacitor, calculate the energy stored in each, and then add them together to find the total energy stored in the series combination.

Step-by-step explanation:

The total energy stored in a series combination of capacitors can be found by summing the individual energies stored in each capacitor. The energy stored in a capacitor is given by the formula:

U = (1/2) * C * V^2

where U is the energy, C is the capacitance, and V is the voltage across the capacitor. Using this formula, we can calculate the energy stored in each capacitor and then sum them up.

In this case, we have three capacitors connected in series with capacitances of 10.7 µF, 23.0 µF, and 29.3 µF, respectively. The maximum potential difference to which any capacitor can be charged is 125V. Plugging in the values into the formula, we get:

U1 = (1/2) * 10.7 µF * (125V)^2

U2 = (1/2) * 23.0 µF * (125V)^2

U3 = (1/2) * 29.3 µF * (125V)^2

The total energy stored in the series combination is the sum of U1, U2, and U3. Calculate each term separately and then sum them up to find the maximum energy stored in the series combination.

User Ye Win
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