Answer :
Charging the bank of capacitors to 60000 V would require a certain amount of energy, which can be calculated using the formula: Energy = (1/2) * C * V^2, where C is the total capacitance and V is the voltage. Substituting the given values, the energy required is 1.86 × 10^6 J. To find the cost, we convert this energy to kilowatt-hours and multiply by the cost per kilowatt-hour, resulting in a total cost of $55.80.
To determine the energy required to charge the bank of capacitors, we use the formula for the energy stored in a capacitor, which is given by: Energy = (1/2) * C * V^2, where C is the total capacitance and V is the voltage. Substituting the given values of C = 3100 * 5.00 µF and V = 60000 V, we calculate the energy to be 1.86 × 10^6 J.
To find the cost of this energy, we need to convert it from joules to kilowatt-hours (kW·h) using the conversion factor 1 kW·h = 3.6 × 10^6 J. Dividing the energy in joules by this conversion factor gives us the energy in kilowatt-hours. Multiplying this by the cost per kilowatt-hour, which is $3.0/kW·h, we find the total cost to be $55.80.
Therefore, charging the bank of capacitors to 60000 V would cost $55.80. This cost represents the energy consumed in the charging process and is calculated based on the capacitance of the capacitors, the voltage applied, and the cost per unit of energy.
