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2 years ago

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A 10-µf capacitor is plugged into a 110 v-rms 60-hz voltage source. what is the rms value of the current through the capacitor?

what is the maximum value of the current through the capacitor?

1 answer:

ValentinkaMS [17]2 years ago

7 0

The rms value of current is0.414A, and the maximum value of current is 0.585A.

To find the answer, we have to study about the capacitive reactance.

<h3>How to find the rms and maximum value of current?</h3>

We have the expression for rms value of current as,

$I_{rms}=\frac{V_{rms}}{X_c}$

where; Xc is the capacitive reactance of the circuit.

We have the capacitive reactance as,

$X_c=\frac{1}{2\pi fC} =\frac{1}{2\pi *60*10*10^{-6}} =265.39 Ohm$

Thus, the value of rms current will be,

$I_{rms}=\frac{110}{265.39} =0.414A$

Thus, the maximum value of current will be,

$I_0=\sqrt{2}*I_{rms} =0.585A$

Thus, we can conclude that, the rms value of current is0.414A, and the maximum value of current is 0.585A.

Learn more about the capacitive reactance here:

brainly.com/question/27853857

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Answer:

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Explanation:

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The initial concentration of A can be calculated from the ideal gas equation:

$pV=nRT\\\\n/v=p/(RT)\\\\C_A=\dfrac{10atm}{0.08206(atm-dm^3/K-mol)/times 400K}\\\\C_A=0.304mol/liter$

Determine the conversion of substance A to substance C using an ICE table and the Kc constant:

A ⇄ 3C

I 0.304 0

C - x + 3x

E 0.304 - x 3x

$K_c=0.25=\dfrac{(3x)^3}{(0.304-x)}$

Solve for x:

You need to use a graphing calculator:

108x³ = 0.304 - x

108x³ + x - 0.304 = 0

x ≈ 0.1195 mol/liter

Then:

$C_A=0.304mol/liter-0.1195mol/liter=0.1845mol/liter\\\\C_C=3\times 0.304mol/liter=0.912mol/liter$

The equilbrium conversion is:

$\% = [0.304mol/liter-0.1845mol/liter]/(0.304mol/liter)\times 100$

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3 years ago

An electric motor spins at 1000 rpm and is slowing down at a rate of 10 t rad/s2 ; where t is measured in seconds. (a) If the mo

REY [17]

Answer:

a) The tangential component of acceleration at the edge of the motor at $t = 1.5\,s$ is -1.075 meters per square second.

b) The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

Explanation:

The angular aceleration of the electric motor ( $\alpha$ ), measured in radians per square second, as a function of time ( $t$ ), measured in seconds, is determined by the following formula:

$\alpha = -10\cdot t\,\left[\frac{rad}{s^{2}} \right]$ (1)

The function for the angular velocity of the electric motor ( $\omega$ ), measured in radians per second, is found by integration:

$\omega = \omega_{o} - 5\cdot t^{2}\,\left[\frac{rad}{s} \right]$ (2)

Where $\omega_{o}$ is the initial angular velocity, measured in radians per second.

a) The tangential component of aceleration ( $a_{t}$ ), measured in meters per square second, is defined by the following formula:

$a_{t} = R\cdot \alpha$ (3)

Where $R$ is the radius of the electric motor, measured in meters.

If we know that $R = 7.165\times 10^{-2}\,m$ , $\alpha = 10\cdot t$ and $t = 1.5\,s$ , then the tangential component of the acceleration at the edge of the motor is:

$a_{t} = (7.165\times 10^{-2}\,m)\cdot (-10)\cdot (1.5\,s)$

$a_{t} = -1.075\, \frac{m}{s^{2}}$

The tangential component of acceleration at the edge of the motor at $t = 1.5\,s$ is -1.075 meters per square second.

b) If we know that $\omega_{o} = 104.720\,\frac{rad}{s}$ and $\omega = 26.180\,\frac{rad}{s}$ , then the time needed is:

$26.180\,\frac{rad}{s} = 104.720\,\frac{rad}{s}-5\cdot t^{2}$

$5\cdot t^{2} = 104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s}$

$t^{2} = \frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s} }{5}$

$t = \sqrt{\frac{104.720\,\frac{rad}{s}-26.180\,\frac{rad}{s} }{5} }$

$t \approx 3.963\,s$

The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

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