A resistor with 780 is connected to the plates of a charged capacitor with capacitance 4.32 . Just before the connection is made, the charge on the capacitor is 7.80
1)What is the energy initially stored in the capacitor?
2)What is the electrical power dissipated in the resistor just after the connection is made?
3)What is the electrical power dissipated in the resistor at the instant when the energy stored in the capacitor has decreased to half the value calculated in part A?
Your numbers require units. (farads, ohms, Joules etc.) You should have learned that by now.
To answer these questions, we need to use formulas related to capacitors, resistors, charge, voltage, and energy. I will explain the process step by step for each question.
1) What is the energy initially stored in the capacitor?
The energy initially stored in the capacitor can be calculated using the formula: E = (1/2) * C * V^2, where E is the energy stored, C is the capacitance, and V is the voltage.
In this case, the capacitance is given as 4.32 μF (microfarads). The charge on the capacitor (just before the connection is made) is given as 7.80 μC (microcoulombs). To find the voltage, we can use the formula: V = Q/C, where Q is the charge and C is the capacitance.
To find the voltage: V = (7.80 μC) / (4.32 μF)
Now, we can substitute the values into the energy formula:
E = (1/2) * (4.32 μF) * [(7.80 μC / 4.32 μF)]^2
Calculate the expression inside the square brackets, square the result, and multiply it by (1/2) * (4.32 μF) to find the energy initially stored in the capacitor.
2) What is the electrical power dissipated in the resistor just after the connection is made?
The electrical power dissipated in the resistor can be calculated using the formula: P = V^2 / R, where P is the power, V is the voltage, and R is the resistance.
In this case, the resistance is given as 780 Ω (ohms). The voltage across the resistor can be found by subtracting the voltage across the capacitor (which is the same as before the connection) from the initial voltage on the capacitor.
To find the voltage across the resistor: V_resistor = V_initial - V_capacitor
Now, substitute the values into the power formula: P = (V_resistor)^2 / 780 Ω
Square the result of the voltage difference and divide it by the resistance to find the electrical power dissipated in the resistor just after the connection is made.
3) What is the electrical power dissipated in the resistor at the instant when the energy stored in the capacitor has decreased to half the value calculated in part 1?
To find the electrical power dissipated in the resistor at this instant, we need to first determine the energy stored in the capacitor at that point.
In part 1, we calculated the initial energy stored in the capacitor. To find the energy when it has decreased to half the initial value, we can simply multiply the initial energy by 0.5.
Now, substitute the calculated energy into the energy formula: E = (1/2) * C * V^2
Rearrange the formula to find V:
V = √ (2 * E / C)
Once you find the voltage, you can substitute it into the power formula from part 2: P = V^2 / R
Square the voltage and divide it by the resistor to find the electrical power dissipated in the resistor at the instant when the energy stored in the capacitor has decreased to half the value calculated in part 1.