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 26.5 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 3.17 seconds when the battery is removed and replaced by a 12200 ohm load. Determine the ideal values for the capacitor and resistor such that your circuit will achieve these design goals. While the rate at which a capacitor discharges in a sorios RC circuit depends on the values of both the resistor and capacitor, the energy stored on the capacitor only depends upon the capacitance and the applied potential difference, so start there. Keep in mind that the design constraints call for half of the energy to be discharged in the indicated time, rather than half of the charge How does this change your approach to the problem?

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. {26.5 V}

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

60.7 * 10^-3 = ½ * C * 26.5²
C = 1.728 * 10^-4 F or 172.9 uF

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 * 1.728 * 10^-4 * V²
V = 18.74 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) = 18.74 V; V = 26.5 V

V(t) / V =18.74 / 26.5 = 0.70716 = e^(-t/CR)
Take {natural, to base e} logs of both sides:

ln(0.70716) = -t/CR

-0.3465 = -t/CR
R = t / 0.3465C

We are given t = 3.17 s, and have calculated C = 1.728 * 10^-4 F
R = 52.943 * 10^3

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

R = R - RL = (52943.456 - 12200) = 40.743 * 10^3 or 40473.45

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