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 21.0 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 4.43 seconds when the battery is removed and - replaced by a 7890 ohm load. Determine the ideal values for the capacitor and resistor such that your circuit will achieve these design goals. C_1= muF R_1= ohm

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.

That key phrase "..for a long time.." implies that a time period equivalent to many time constants has passed, so that the voltage across the terminals of the capacitor is essentially equal to the battery voltage. {21.0 V}

If we substitute the given values into the equation above, we get:

60.7 * 10-3 = ½ * C * 21²
C = 2.753 * 10-4 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 * 2.753 * 10^-4 * V²
V = 14.85 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) = 14.85 V; V = 21.0 V

V(t) / V =14.85 / 21.0 = 0.707 = e^(-t/CR)
Take {natural, to base e} logs of both sides:

ln(0.707) = -t/CR

-0.3466 = -t/CR
R = t / 0.3466C

We are given t = 4.43 s, and have calculated C = 2.753 * 10^-4 F
R = 4.642 * 10^4

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

R = R - RL = (4.642 * 10^4 - 7890) = 3.85 * 10^4

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