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Anestetic [448]

2 years ago

10

A 1.5 V battery is connected to a 1,000 μF capacitor in series with a 150 Ω resistor. a. What is the maximum current that flows

through the resistor during charging? b. What is the maximum charge on the capacitor? c. How long does the capacitor take to reach a potential of 1.0V?

1 answer:

Elza [17]2 years ago

3 0

Answer:

$0.01\ \text{A}$

$0.0015\ \text{C}$

$0.0608\ \text{s}$

Explanation:

$V_0$ = Voltage = 1.5 V

$C$ = Capacitance = $1000\ \mu\text{F}$

$R$ = Resistance = $150\ \Omega$

Current is given by

$I=\dfrac{V_0}{R}\\\Rightarrow I=\dfrac{1.5}{150}\\\Rightarrow I=0.01\ \text{A}$

Current flowing in the resistor is $0.01\ \text{A}$ .

Charge is given by

$Q=CV\\\Rightarrow Q=1000\times 10^{-6}\times 1.5\\\Rightarrow Q=0.0015\ \text{C}$

The charge on the capacitor is $0.0015\ \text{C}$ .

Voltage is given by

$V=V_0e^{-\dfrac{t}{RC}}\\\Rightarrow t=-RC\ln\dfrac{V}{V_0}\\\Rightarrow t=-150\times 1000\times 10^{-6}\times\ln\dfrac{1}{1.5}\\\Rightarrow t=0.0608\ \text{s}$

Time taken to reach 1 V is $0.0608\ \text{s}$ .

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Answer:

the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$

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Explanation:

Let use m₁ to represent the mass of the block and m₂ to represent the mass of the cylinder

The radius of the cylinder be = R

The distance between the center of the pulley to center of the block to be = x

Also, the angles of inclinations of the cylinder and the block with respect to the ground to be $\phi$ and $\beta$ respectively.

The velocity of the block to be = v

The equivalent mass of the system = $m_e$

In the terms of the equivalent mass, the kinetic energy of the system can be written as:

$K.E = \frac{1}{2} m_ev^2$ --------------- equation (1)

The angular velocity of the cylinder = $\omega$ : &

The inertia of the cylinder about its center to be = I

The angular velocity of the cylinder can be written as:

$v = \omega R$

$\omega =\frac{v}{R}$

The kinetic energy of the system in terms of individual mass can be written as:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I\omega^2$

By replacing $\omega$ with $\frac{v}{R}$ ; we have:

$K.E = \frac{1}{2}m_1v^2+\frac{1}{2} m_2v^2+\frac{1}{2}I(\frac{v}{R})^2$

$K.E = \frac{1}{2}(m_1+ m_2+ \frac{I}{R} )v^2$ ------------------ equation (2)

Equating both equation (1) and (2); we have:

$m_e = m_1+m_2+\frac{I}{R^2}$

Therefore, the equivalent mass : $m_e = m_1+m_2+\frac{I}{R^2}$ which is read as;

The equivalent mass is equal to the mass of the block plus the mass of the cylinder plus the inertia by the square of the radius.

The expression for the force acting on equivalent mass due to the block is as follows:

$f_{block }=m_1gsin \beta$

Also; The expression for the force acting on equivalent mass due to the cylinder is as follows:

$f_{cylinder} = m_2gsin \phi$

Equating the above both equations; we have the equation of motion of the equivalent system to be

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$(m_1+m_2+\frac{I}{R^2} )(\bar x) = (m_2gsin \phi) -(m_1gsin \beta)$

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