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Archy [21]
3 years ago
15

You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int

ernal resistance. find the time constant of the circuit. what is the charge of the capacitor 1.95 time constants after the circuit is closed? what is the charge after a long time?
Physics
1 answer:
uysha [10]3 years ago
4 0
In a RC-circuit, with the capacitor initially uncharged,  when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
Q(t) = Q_0 (1-e^{-t/\tau})
where t is the time, Q_0 = CV is the maximum charge on the capacitor at voltage V, and \tau = RC is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using R=39.1 \Omega and C= 65.7 mF=65.7\cdot 10^{-3}F, the time constant of the circuit is:
\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s

2) To find the charge on the capacitor at time t=1.95 \tau, we must find before the maximum charge on the capacitor, which is
Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C
And then, the charge at time t=1.95 \tau is equal to
Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be Q_0 = 0.59 C. We can see this also from the previous formule, by using t=\infty:
Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C

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3 years ago
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Answer:

Q = 200800 Joules.

Explanation:

Given the following data;

Mass = 4kg

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Final temperature = 90.0°C

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To find the quantity of heat absorbed;

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Q = mcdt

Where;

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dt represents the change in temperature.

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Q = 200800 Joules.

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8 0
3 years ago
A runner completes the 300-meter dash in 38 seconds. What is the speed of the runner? Round your answer the answer to the neares
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6 0
3 years ago
1500 kg wrecking ball traveling at a speed of 3.5 m/s hits a wall that does not crumble but is pushed back 75 cm. If the wreckin
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The size of the force that pushes the wall is 12,250 N.

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F = 9187.5 / 0.75

F = 12,250 N

Therefore, the size of the force that pushes the wall is 12,250 N.

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