A charged capacitor,C=10e-6 F, with an initial stored energy of 3.5J, is discharged across a resistor, R. The charge on the capacitor drops to 40% of its maximum value in time, t=8.2s, from the instant the switch is closed (t=0).
Find the current through the resistor at a time t=10s.
How much thermal energy is dissipated across the resistor after two time constants?
Determine the electrostatic energy on the capacitor after two time constants.
C = Q/V
V = (1/C) integral i dt
dV/dt = (1/C) i
but
i = -V/R
so
dV/dt = (1/C) -V/R
or
dV/V = (-1/RC) dt (I bet you knew that)
so
ln V = (-1/RC) t
V = Vi e^(-t/RC) (which you also know)
and Q = CV = CVi e^(-t/RC)
at t = 8.2
Q(8.2) = Cv = .4 CVi = CVi e^(-8.2/RC)
so
.4 = e^(8.2/{R*10^-5} )
ln .4 = -8.2*10^5/R
solve for R
but how much energy is stored at V ?
E = (1/2) C V^2
that should get you started
To find the current through the resistor at t=10s, we can use Ohm's Law (V=IR) and the equation for the voltage across a discharging capacitor (V(t) = V(0) * exp(-t/(RC))).
1. First, we need to find the time constant (RC) of the circuit. The time constant is equal to the product of the resistance (R) and the capacitance (C). In this case, C = 10e-6 F. We need to find R, which we can calculate using the formula: R = -t / (C * ln(V(t)/V(0))) = -8.2s / (10e-6 F * ln(0.40)). Calculate the values inside the parentheses and substitute them into the formula to find R.
2. Once we have R, we can find the current through the resistor (I) at t=10s using Ohm's Law: I = V(t) / R. Substitute the values of V(t) and R into the formula to find I.
To find the amount of thermal energy dissipated across the resistor after two time constants, we need to calculate the energy dissipated in one time constant and then double that value.
3. The energy dissipated across a resistor is given by the formula: E = I^2 * R * t. Calculate the energy dissipated in one time constant using the values of I and R from step 2, and t = 2 time constants. Multiply this value by 2 to find the energy dissipated after two time constants.
To determine the electrostatic energy on the capacitor after two time constants, we can use the equation for the energy stored in a capacitor (E = 0.5 * C * V^2).
4. Calculate V(t) at the end of two time constants using the equation V(t) = V(0) * exp(-2). Substitute the value of V(0) and calculate V(t) at t = 2 time constants.
5. Finally, use the formula for the energy stored in a capacitor E = 0.5 * C * V^2. Substitute the value of C and V(t) from step 4 into the formula to find the electrostatic energy on the capacitor after two time constants.