A fully charged 98 uF capacitor is discharged through a 125 ohms resistor. How long will it take for the capacitor to lose 75% of its initial energy
The time constant is RC = 0.0123 seconds
To lose 75% of its energy, the voltage on the capacitor must fall in half.
V/Vo = e^(-t/0.0123) = 0.500
Solve for t.
To find out how long it will take for the capacitor to lose 75% of its initial energy, we need to calculate the time constant (τ) for the RC circuit using the given values.
The time constant (τ) is equal to the product of the resistance (R) and the capacitance (C) in the circuit. We can calculate it using the formula:
τ = RC
Given:
C = 98 μF
R = 125 Ω
Converting the capacitance to Farads:
C = 98 μF = 98 × 10^(-6) F (since 1 μF = 10^(-6) F)
Now, we can calculate the time constant:
τ = (98 × 10^(-6)) F × 125 Ω
τ ≈ 12.25 × 10^(-3) s
Next, we need to find the time it takes for the capacitor to lose 75% of its initial energy. Since energy stored in a capacitor is proportional to the square of the voltage across it, we can say that the voltage across the capacitor will decrease to 75% of its initial value.
Using the formula for exponential decay in an RC circuit:
V(t) = V0 * e^(-t/τ)
where:
V(t) = voltage across the capacitor at time t
V0 = voltage across the capacitor at t = 0
t = time
τ = time constant
We can rearrange the formula to solve for time:
t = -τ * ln(V(t)/V0)
Given that V(t)/V0 = 0.75 (75% of initial energy) and τ ≈ 12.25 × 10^(-3) s, we can calculate the time:
t = - (12.25 × 10^(-3) s) * ln(0.75)
t ≈ - (12.25 × 10^(-3) s) * (-0.2877)
t ≈ 3.3 × 10^(-3) s
Therefore, it will take approximately 3.3 milliseconds for the capacitor to lose 75% of its initial energy.
To find the time it takes for the capacitor to lose 75% of its initial energy, we need to use the formula for the energy stored in a capacitor:
E = (1/2) * C * V^2
where E is the energy, C is the capacitance, and V is the voltage.
Given that the capacitance (C) is 98 uF and the voltage (V) is unknown, we can rearrange the formula to solve for V:
V = sqrt(2 * E / C)
To find the voltage when the capacitor has lost 75% of its initial energy, we can multiply the initial voltage by 0.25:
V_final = V_initial * 0.25
Next, we can substitute the values into the formula to find the initial voltage:
V_initial = sqrt(2 * E_initial / C)
Now, we can solve for V_initial:
V_initial = sqrt(2 * E_initial / C)
= sqrt(2 * (1/2) * C * V_initial^2 / C)
= sqrt(V_initial^2)
= V_initial
Since V_initial = V_final * 0.25, we can substitute it into the equation:
V_final * 0.25 = V_initial
0.25 = V_initial / V_final
V_initial = 0.25 * V_final
Now, we can find the time it takes for the capacitor to lose 75% of its initial energy by calculating the time constant (t) using the formula:
t = R * C
Given that the resistance (R) is 125 ohms and the capacitance (C) is 98 uF, we can substitute the values into the formula:
t = R * C
= 125 * (98 * 10^-6)
= 12.25 * 10^-3
= 12.25 ms
Therefore, it will take approximately 12.25 milliseconds for the capacitor to lose 75% of its initial energy.