You are building a small circuit that consists of a resistor in series with a ca

Question

You are building a small circuit that consists of a resistor in series with a capacitor as shown. You need the capacitor to store 60.7 mJ of energy when a 10.0 V battery is connected between terminals A and B for a long time. Then, you need the capacitor to discharge half of this stored in energy in exactly 1.91 seconds when the battery is removed and replaced by a 5.90 times 10^2 Ohm load. Determine the ideal values for the capacitor and resistor such that your circuit will achieve these design goals. C_1 = R_1 =

Explanation / Answer

The energy (E) stored in a capacitor is given by:

E = ½CV²
where C is the capacitance in farads and V is the voltage across the capacitor in volts.
60.7 * 10^-3 = ½ * C * 10.0²
C = 1.214x10^-3 F

Now we know C, we can use the equation again to find the capacitor's terminal voltage when the energy remaining in the capacitor has fallen to half its initial value:

E = ½CV²
½ * 60.7 * 10^-3 = ½ * 1.214x10^-3 * V²
V = 7.07 V

When the capacitor is discharging from some initial voltage (V), the voltage V(t) at some time (t) is given by:

V(t) = V e^(-t/CR)

Here, V(t) = 7.07 V; V = 10.0 V

V(t) / V =.7.07 / 10.0 = .707 = e^(-t/CR)
Take {natural, to base e} logs of both sides:

ln(0.707) = -t/CR

-.347= -t/CR
R = t / .347C

We are given t = 1.91 s, and have calculated C =1.214x10^-3F R= 4.534 * 10^3

But R is the equivalent resistance of R in series with RL

R = R - RL = (4.534 * 10^3 - 5.92 * 10^2) = 3.942 * 10^3

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