Alternating Current Test

Directions For Questions

An alternating source

V\ \text{volt},\ f\ \text{Hz}

is connected across an lnductance

L(

in series wlth resistance

R)

in parallel with a capacitance

C

as shown above.
The admittance Y of the circuit is then defined through I= YV

where

|\displaystyle \mathrm{Y}|=\frac{[\mathrm{R}^{2}\mathrm{X}_{\mathrm{c}}^{2}+(\mathrm{R}^{2}+\mathrm{X}_{L}^{2}-X_{L}X_{\mathrm{c}})^{2}]}{\mathrm{X}_{\mathrm{c}}(\mathrm{R}^{2}+\mathrm{X}_{L}^{2})}\frac{1}{2}

The current

I

leads the source voltage

V

by 6,

such that

\displaystyle \tan\partial=\frac{(\mathrm{R}^{2}+X_{L}-X_{L}\mathrm{X}_{\mathrm{c}})}{\mathrm{R}\mathrm{X}_{\mathrm{c}}}

with

\mathrm{X}_{\mathrm{L}}=\omega \mathrm{L}

and

\displaystyle X_{C}=\dfrac{1}{\omega \mathrm{C}}

...view full instructions

Alternating Cur...

Question 1

6e−20t from B to A6e^{-20t}\ \text{from B to A}6e−20t from B to A

3e−40t from A to B3e^{-40t}\ \text{from A to B}3e−40t from A to B

3e−40t from B to A3e^{-40t}\ \text{from B to A}3e−40t from B to A

6e−20t from A to B6e^{-20t}\ \text{from A to B}6e−20t from A to B

Question 2

12π1LC+R2L2\dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}+\dfrac{R^{2}}{L^{2}}}2π1​LC1​+L2R2​​

12π1LC\dfrac{1}{2\pi }\dfrac{1}{\sqrt{LC}}2π1​LC​1​

12π1LC−R2L2\dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}}2π1​LC1​−L2R2​​

12πRL\dfrac{1}{2\pi }\dfrac{R}{L}2π1​LR​

Question 3

120 V

100 V

140 V

160 V

Question 4

2 A

22A2 \sqrt{2} A22​A

0.5 A

0.4A\sqrt{0.4} A0.4​A

Question 5

Question 6

6.28×10−3rad/s6.28\times{10}^{-3}rad/s6.28×10−3rad/s; 10−3s{10}^{-3}s10−3s

6.28×103rad/s6.28\times{10}^{3}rad/s6.28×103rad/s; 10−3s{10}^{-3}s10−3s

7.28×103rad/s7.28\times{10}^{3}rad/s7.28×103rad/s; 10−3s{10}^{-3}s10−3s

7.28×10−3rad/s7.28\times{10}^{-3}rad/s7.28×10−3rad/s; 10−3s{10}^{-3}s10−3s

Question 7

glows more brightly

gets dimmer

glows with the same brightness

gets momentarily dimmer and then glows more brightly

Question 8

8sin⁡(50πt+37π180)8 \sin (50\pi t+37\dfrac{\pi }{180})8sin(50πt+37180π​)

8sin⁡(50πt−37π180)8 \sin (50\pi t-37\dfrac{\pi }{180})8sin(50πt−37180π​)

10sin⁡(50πt)10 \sin (50\pi t )10sin(50πt)

10cos⁡(50πt)10 \cos (50\pi t )10cos(50πt)

Question 9

Question 10

13LC\dfrac {1}{3\sqrt {LC}}3LC​1​

16πLC\dfrac {1}{6\pi \sqrt {LC}}6πLC​1​

1LC\dfrac {1}{\sqrt {LC}}LC​1​

13πLC\dfrac {1}{3\pi \sqrt {LC}}3πLC​1​

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Question Analysis

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$6e^{-20t}\ \text{from B to A}$

$3e^{-40t}\ \text{from A to B}$

$3e^{-40t}\ \text{from B to A}$

$6e^{-20t}\ \text{from A to B}$

$\dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}+\dfrac{R^{2}}{L^{2}}}$

$\dfrac{1}{2\pi }\dfrac{1}{\sqrt{LC}}$

$\dfrac{1}{2\pi }\sqrt{\dfrac{1}{LC}-\dfrac{R^{2}}{L^{2}}}$

$\dfrac{1}{2\pi }\dfrac{R}{L}$

$2 \sqrt{2} A$

$\sqrt{0.4} A$

$6.28\times{10}^{-3}rad/s$ ; ${10}^{-3}s$

$6.28\times{10}^{3}rad/s$ ; ${10}^{-3}s$

$7.28\times{10}^{3}rad/s$ ; ${10}^{-3}s$

$7.28\times{10}^{-3}rad/s$ ; ${10}^{-3}s$

$8 \sin (50\pi t+37\dfrac{\pi }{180})$

$8 \sin (50\pi t-37\dfrac{\pi }{180})$

$10 \sin (50\pi t )$

$10 \cos (50\pi t )$

$\dfrac {1}{3\sqrt {LC}}$

$\dfrac {1}{6\pi \sqrt {LC}}$

$\dfrac {1}{\sqrt {LC}}$

$\dfrac {1}{3\pi \sqrt {LC}}$