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Cerrena [4.2K]
3 years ago
13

Now solve the differential equation V(t)=−CRdV(t)dt for the initial conditions given in the problem introduction to find the vol

tage as a function of time for any time t. Express your answer in terms of q0, C, R, and t.
Physics
1 answer:
mina [271]3 years ago
8 0

Answer:

V(t) = (q0/C) * e^(−t/RC )

Explanation:

If there were a battery in the circuit with EMF  E , the equation for  V(t)  would be  V(t)=E−(RC)(dV(t)/dt) . This differential equation is no longer homogeneous in  V(t)  (homogeneous means that if you multiply any solution by a constant it is still a solution). However, it can be solved simply by the substitution  Vb(t)=V(t)−E . The effect of this substitution is to eliminate the  E  term and yield an equation for  Vb(t)  that is identical to the equation you solved for  V(t) . If a battery is added, the initial condition is usually that the capacitor has zero charge at time  t=0 . The solution under these conditions will look like  V(t)=E(1−e−t/(RC)) . This solution implies that the voltage across the capacitor is zero at time  t=0  (since the capacitor was uncharged then) and rises asymptotically to  E  (with the result that current essentially stops flowing through the circuit).

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Which statement about energy transformations is not true?
frutty [35]

Answer:

C

Explanation:

Let's consider the conservation of energy

Energy can not be destroyed or created energy can be transferred one form energy to a another form of energy.

(A) Then all form of energy can be transfer into any form of energy . There for all forms of energy can be transformed to thermal energy

(B) This is the law of conservation of energy

(C)Potential chemical energy can be transferred into any form of energy. It's not only to kinetic energy

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3 years ago
The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential diffe
bazaltina [42]

To solve this problem we will apply the concepts related to the calculation of power, from the two electrical forms:

P = VI

P = \frac{V^2}{R}

Here,

V = Voltage

I = Current

R= Resistance

P = Power

PART A)

R = \frac{V^2}{P}

Replacing,

R = \frac{(120V)^2}{150W}

R = 96\Omega

Resistance of bulb is 96\Omega

PART B)

I = \frac{P}{V}

I = \frac{150W}{120V}

I = 1.25A

The bulb will draw 1.25A current

7 0
3 years ago
A go cart starts from REST and accelerates uniformly over a time of 7 seconds for a distance of 100m. Determine the acceleration
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Answer:

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Explanation:

Download pdf
6 0
3 years ago
A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential
ruslelena [56]

Answer:

(a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is 1.022\times10^{-16}\ J

(c). The work done on the proton is -8\times10^{-18}\ J.

Explanation:

Given that,

Speed v= 3.50\times10^{5}\ m/s

Initial potential V=100 V

Final potential = 150 V

(a). We need to calculate the change in the protons electric potential

Potential energy of the proton is

U=qV=eV

Using conservation of energy

K_{i}+U_{i}=K_{f}+U_{f}

\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}

]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})

\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V

\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}

Put the value into the formula

\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}

\Delta V=639.2=0.639\ kV

(b). We need to calculate the change in the potential energy of the proton

Using formula of potential energy

\Delta U=q\Delta V

Put the value into the formula

\Delta U=1.6\times10^{-19}\times639.2

\Delta U=1.022\times10^{-16}\ J

(c). We need to calculate the work done on the proton

Using formula of work done

\Delta U=-W

W=q(V_{2}-V_{1})

W=-1.6\times10^{-19}(150-100)

W=-8\times10^{-18}\ J

Hence, (a). The change in the protons electric potential is 0.639 kV.

(b). The change in the potential energy of the proton is 1.022\times10^{-16}\ J

(c). The work done on the proton is -8\times10^{-18}\ J.

5 0
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Verdich [7]

Answer:

A

Explanation:

5 0
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