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PSYCHO15rus [73]
3 years ago
9

A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2

.0-Ω resistor is connected between the two terminals of the capacitor. Find the charge on the capacitor at the following times. (When entering units, use micro for the metric system prefix µ.)
Physics
1 answer:
Grace [21]3 years ago
4 0

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S  

2) t = 1 mu S

3) t = 50 mu S

1) 22.4 \mu C

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

C=\frac{Q_0}{V_0}

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

C=4.4\mu F = 4.4\cdot 10^{-6}F

V_0 = 5.1 V

Solving for Q_0,

Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) 20.0\mu C

To find the charge on the capacitor at any other time t, we use the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

where

Q_0 = 22.4 \mu C

t is the time

R=2.0 \Omega is the resistance

C=4.4\mu F is the capacitance

Therefore, at time t=1 \mu s, we have:

Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C

3) 0.08 \mu C

As before, we use again the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

However, here the time to consider is

t=50 \mu C

Substituting into the formula,

Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C

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A race car traveling at 10 meters per second accelerates at 1.5 meters per second squared while moving a distance of 600 meters.
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A generator produces 38 mwmw of power and sends it to town at an rms voltage of 78 kvkv. part a what is the rms current in the t
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The rms current in the transmission lines is I = 487.18 A.

The root-imply-rectangular (rms) voltage of a sinusoidal supply of electromotive force  is used to represent the source. it is the rectangular root of the time average of the voltage squared.

Alternating-present day circuits. the root-imply-square (rms) voltage of a sinusoidal source of electromotive force is used to symbolize the supply. it's far the square root of the time average of the voltage squared.

Electric power is  by using present day or the waft of electric fee and voltage or the capacity of rate to deliver electricity. A given cost of power can be produced by using any combination of contemporary and voltage values

power = 38 M watt

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power = IV

I = power/V

I = (38 * 1000000)/78*1000

I = 487.18 A.

Learn more about rms current here:-brainly.com/question/20913680

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