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vodka [1.7K]
3 years ago
6

An AC source drives a sinusoidal current of amplitude I0and frequency ? into an inductor having inductance L and a capacitor hav

ing capacitance C that are connected in series. The current as a function of time is given by I(t)=I0sin(?t)
Recall that the voltages VL(t) and VC(t) across the inductor and capacitor are not in phase with the respective currents IL(t) and IC(t). In particular, which of the following statements is true for a sinusoidal current driver?



A. VL(t) and VC(t) both lag their respective currents.
B. VL(t) and VC(t) both lead their respective currents.
C. VL(t) lags IL(t) and VC(t) leads IC(t).
D. VL(t) leads IL(t) and VC(t) lags IC(t).
Physics
1 answer:
lukranit [14]3 years ago
5 0

Answer:

option (D)

Explanation:

In an LC circuit which are connected in series, the current lags by 90 by inductor and current leads by 90 by the capacitor.

So, VL(t) leads IL(t) and Vc(t) lags Ic(t).

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Answer:

Accuracy

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From this, we can agree that precision neglects the most important factor, closeness or say, exactness. Precision isn't bothered by it. And while that can be excused in a few instances, it certainly can not be permitted when it comes to life, or organs of the body

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According to cell theory which of the following is not true?
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Answer:

first one

Explanation:

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2 years ago
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Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha
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Answer:

F=18.58\times 10^{-11}\ N

\theta=30.276^{\circ}

Explanation:

Given:

mass of first particle, m_1=6.7\ kg

mass of second particle, m_2=5.1\ kg

mass of third particle, m_3=3.7\ kg

coordinate position of first particle in meters, (x_1,y_1)\equiv(4.2,0)

coordinate position of second particle in meters, (x_2,y_2)\equiv(0,2.8)

coordinate position of third particle in meters, (x_3,y_3)\equiv(0,0)

<u>Now, gravitational force on particle 3 due to particle 1:</u>

F_{31}=G\frac{m_1.m_3}{r_{31}^2}

F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}

F_{31}=9.37\times 10^{-11}\ N

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

F_{32}=G\frac{m_2.m_3}{r_{21}^2}

F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}

F_{32}=16.05\times 10^{-11}\ N

towards positive X axis.

<u>Now the net force</u>

F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}

F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}

F=18.58\times 10^{-11}\ N

<em>For angle in counterclockwise direction from the +x-axis</em>

tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}

\theta=30.276^{\circ}

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