Question 1 of 20 Incorrect Map Sapling Learning macmillan learning You are build

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Question 1 of 20 Incorrect Map Sapling Learning macmillan learning You are building a small circuit that consists ofa resistor in series with a capacitor as shown. You need the capacitor to store 97.7 mJ of energy when a 15.5 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 1.91 seconds when the battery is removed and replaced by a 1560 load bat L. C. Determine the ideal values for the capacitor and resistor such that your circuit will achieve these design goals Number Number R,

Explanation / Answer

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

E =1/2*C*V2

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. {15.5 V}

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

97.7 * 10-3 = 1/2 * C * (15.5)2

C = 0.8133*10-3 F

C = 813.3*10-6 F = 813.3 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 = 1/2*C*V2

1/2* 97.7 * 10-3 = 1/2 * 8.13*10-4 * V2

V = 10.96 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/RC)

Here, V(t) = 10.96 ; V = 15.5 V

V(t) / V =10.96 / 15.5 = 0.707 = e(-t/RC)

ln(0.707) = -t/CR

-0.346 = -t/CR

R = t / 0.346*C

R = 1.91/0.346*0.813*10-3

R = 6789.95 ohm

But R is the equivalent resistance of R1 in series with RL

R1 = R - RL

R1 =   6789.95 - 1560

R1 = 5229.95 ohm

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