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stellarik [79]
3 years ago
15

A circuit consists of a 6 ohm resistor, a 0.2 farad capacitor, and an AC voltage source supplying V(t) = 120 cos(20 t) volts. Wr

ite the differential equation for the charge on the capacitor. Write a formula for the solution, assuming q(0) = 0.
Physics
1 answer:
Zinaida [17]3 years ago
7 0

Answer:

q = 24 cos (20t) (1- e (-t / 1.2))

Explanation:

To work this circuit we use the mesh equation (Kirchoff) that states that the voltage in a closed circuit is zero

           Vg + Vr + Vc = 0

Where Vg is the generator voltage, Vr the resistance voltage and Vc the capacitor voltage

         Vg = 120 cos 20t  = V

         Vr = i R

         Vc = q / C

We see that in one term we have the current (i) and in another the load (q), but there is a relationship between the two

         i = dq / dt

Let's replace in the initial equation

       V + R dq / dt + q / C = 0

Reorder the terms

      Rdq / dt + q / C - V = 0

      dq / dt + q / rC - V / R = 0

      dq / dt = V / r -q / RC

       dq / (V / R -q / RC) = dt

we integrate

     ∫I dq / (V / R - 1 / RC q) = ∫ dt

We change variables

      u = (V / R - 1 / RC q)

      du = -1 / RC dq

     ∫ dq / (V / R - 1 / RC q) = -RC ∫ du / u

     -RC ln u = -RC ln (V / R - 1 / RC q)

We evaluate between the limits of integration, the lower t = 0 q (0) = 0

     -RC [ln (V / R - 1 / RC q) - ln (V / R)] = t

     [ln (V / R - 1 / RC q) - ln (V / R)] = -t / RC

     Ln (V / R - 1 / RC q) / (V / R)] = -t / RC

     Ln (1 - 1 / VC q)) = (-t / RC)

     Ln (VC- q) = ln (VC) (-t / RC)

     VC-q = VC e (-t / RC)

     q = VC - Vc e (-t / RC)

     q = VC (1- e (-t / RC)

We substitute the values ​​they give us

      q = (120 cos (20t) 0.2) (1- e (-t / 6 0.2))

      q = 24 cos (20t) (1- e (-t / 1.2))

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viktelen [127]

Answer:

W=2.76\times 10^{-6}\ J

Explanation:

Given that,

The distance between two spheres, r = 25 cm = 0.25 m

The capacitance, C = 26 pF = 26×10⁻¹² F

Charge, Q = 12 nC = 12 × 10⁻⁹ C

We need to find the work done in moving the charge. We know that, work done is given by :

U=\dfrac{Q^2}{2C}

Put all the values,

U=\dfrac{(12\times 10^{-9})^2}{2\times 26\times 10^{-12}}\\\\U=2.76\times 10^{-6}\ J

So, the work done is 2.76\times 10^{-6}\ J.

8 0
3 years ago
A 3.15-kg object is moving in a plane, with its x and y coordinates given by x = 6t2 − 4 and y = 5t3 + 6, where x and y are in m
ArbitrLikvidat [17]

Answer:

<h2>206.67N</h2>

Explanation:

The sum of force along both components x and y is expressed as;

\sum Fx = ma_x  \ and \ \sum Fy = ma_y

The magnitude of the net force which is also known as the resultant will be expressed as R =\sqrt{(\sum Fx)^2 + (\sum Fx )^2}

To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.

Given the position of the object along the x-component to be x = 6t² − 4;

a_x = \frac{d^2 x }{dt^2}

a_x = \frac{d}{dt}(\frac{dx}{dt} )\\ \\a_x = \frac{d}{dt}(6t^{2}-4  )\\\\a_x = \frac{d}{dt}(12t  )\\\\a_x = 12m/s^{2}

Similarly,

a_y = \frac{d}{dt}(\frac{dy}{dt} )\\ \\a_y = \frac{d}{dt}(5t^{3} +6 )\\\\a_y = \frac{d}{dt}(15t^{2}   )\\\\a_y = 30t\\a_y \ at \ t= 2.15s; a_y = 30(2.15)\\a_y = 64.5m/s^2

\sum F_x = 3.15 * 12 = 37.8N\\\sum F_y = 3.15 * 64.5 = 203.18N

R = \sqrt{37.8^2+203.18^2}\\ \\R = \sqrt{1428.84+41,282.11}\\ \\R = \sqrt{42.710.95}\\ \\R = 206.67N

Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N

7 0
3 years ago
Rubber rods charged by rubbing with cat fur repel each other. Glass rods charged by rubbing with silk repel each other. A rubber
puteri [66]

Answer:

C. A rubber rod and a glass rod charged this way have opposite charges on them.

Explanation:

When a rubber rod is rubbed against cat fur, it acquires a negative charge, it becomes negatively charged.

When you then try to bring two rubber rod's together, they repel because like charges repel.

Meanwhile, when you rub a glass rod against silk, it loses electrons to the silk material and becomes positively charged.

When you bring two positively charged glass rod's together, they repel, because like charges repel.

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5 0
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How to find the magnitude and direction of a resultant velocity?
mixas84 [53]
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2 years ago
The current in a lamp is 0.5 ampere when it is plugged into a standard wall outlet. What is the resistance of the lamp when it i
lidiya [134]

The resistance of the lamp plugged in to a standard wall outlet with a current of 0.5 amps is 240 Ω (ohms)

Explanation:

In the United States Of America the standard voltage is 120 v and their frequency is 60 Hz

Standard wall outlet voltage is 120 V

The current in the lamp is 0.5 ampere

Resistance (R) = V/ I

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Thus the resistance of the lamp plugged in to a standard wall outlet with a current of 0.5 amps is 240 Ω (ohms).

8 0
3 years ago
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