Alternating Current Test

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Alternating Current Test - 23

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Question 1
1 / -0

The capacitive reactance in an AC circuit is

A
effective resistance due to capacity

B
effective wattage

C
effective voltage

D
None of these

Solution

Capacitive reactance in an A.C circuit is: $${X}_{C}=\cfrac{1}{\omega

C}$$ ohm
where, $C$ is the capacitance of capacitor and $$\omega=2\pi

n $($ n$$ is the frequency of A.C source)
Question 2
1 / -0

If the inductance of a coil in 1 henry then its effective resistance in a $D.C$ circuit will be

A
$\displaystyle \infty$

B
zero

C
1 $\Omega$

D
2 $\Omega$

Solution

${ X }_{ L }=\omega L$
In DC, $\omega=0$
Therefore, effective inductive resistance is
${ X }_{ L }=0$
Question 3
1 / -0

The power loss in an $AC$ circuit is $E_{rms}$ $I_{rms}$ , when in the circuit there is only

A
$C$

B
$L$

C
$R$

D
$L,\ C$ and $R$

Solution

Inductors and capacitors bring a phase difference between the voltage and current in the circuit, hence changing the p.f. When only a resistance is present, $Poer\ factor= 1$ .
The power loss in an AC circuit $=E_{rms} I_{rms} Power\ factor$
Question 4
1 / -0

With increase in frequency of an A.C. supply, the inductive reactance

A
decreases

B
increases directly with frequency

C
increases as square of frequency

D
decreases inversely with frequency

Solution

The inductive reactance ${ X }_{ l }=\omega L$

Hence, ${ X }_{ l } \propto \omega$

As frequency increases $\rightarrow \omega$
Therefore, inductive reactance increases with frequency.
Question 5
1 / -0

Directions For Questions

The above figure shows a cross-over network in a loud speaker system. One branch consists of a capacitor $C$ and a resistor $R$ in series $($ the tweeter $).$ This branch is in parallel with a second branch $($ The woofer $)$ that consists of an. inductor $L$ and a resistor $R$ in series. The same source voltage with angular frequency $\omega ($ from an amplifier $)$ is applied across each parallel branch.

...view full instructions

The impedance of the tweeter and woofer branches are respectively

A
$\displaystyle \sqrt{R^2+X^2_C} , \sqrt{R^2+X^2_L}$

B
$\displaystyle \sqrt{R^2+X^2_L} , \sqrt{R^2+X^2_C}$

C
$\displaystyle \sqrt{R^2+(X_L - X_C)^2}$ each

D
$\displaystyle \sqrt{\left(\frac{1}{R^2}+\left(\frac{1}{X_C}-\frac{1}{X_L}\right)^2 \right)^{-1}}$ each

Solution

Impedance of tweeter $=\left | R + \frac{1}{jC\omega} \right | =\sqrt{R^2+X_C^2}$
Impedance of woofer $=\left | R + jL\omega \right | =\sqrt{R^2+X_L^2}$
Question 6
1 / -0

An inductor, a resistor and a capacitor are joined in series with an AC source. As the frequency of the source is slightly increased from a very low value, the reactance of the

A
inductor increases

B
resistor increases

C
capacitor increases

D
circuit increases

Solution

The reactance of inductor, ${X}_{L}=\omega L$
The reactance of capacitor, ${X}_{C}=\cfrac{1}{\omega C}$
where $\omega=2\pi n$ & $n$ is the frequency of A.C source.
Therefore, reactance of inductor increases.
Question 7
1 / -0

With increase in frequency of an AC supply, the impedence of an L-C-R series circuit

A
remains constant

B
increases

C
decreases

D
decreases at first, becomes minimum and then increases

Solution

We have the formula for Impedance

$\\ Z=\sqrt { { R }^{ 2 }+{ ({ { X }_{ l } }-{ X }_{ C }) }^{ 2 } } \\ { X }_{ C }=\dfrac { 1 }{ \omega C } \\ { X }_{ l }=\omega L$

And from the graph it can be easily seen that is $Decreases$ first and then $Increases$ .
Question 8
1 / -0

The self inductance of a coil is 1/2 henry. At what frequency will its inductive reactance be 3140 $\Omega$

A
$100 Hz$

B
$10 Hz$

C
$1000 Hz$

D
$10000 Hz$

Solution

We know that,

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

$\omega =\dfrac { { X }_{ L } }{ L }$

$\omega = 6280$

$f=\dfrac { \omega }{ 2\pi }$

$f=1000$ Hz.
Question 9
1 / -0

In a series combination of $R,L$ and $C$ to an A.C source at resonance. If $R=20\ ohm$ , then impedence $Z$ of the combination is

A
$20$ ohm

B
Zero

C
$1$ ohm

D
$400$ ohm

Solution

At resonance, impedence $Z=R$
So, $Z= R = 20\ ohm$
Question 10
1 / -0

In an $AC$ circuit, a resistance of $R$ ohm is connected in series with an inductance $L$ . If phase angle between voltage and current be ${45}^{o}$ , the value of inductive reactance will be

A
$R/4$

B
$R/2$

C
$R$

D
$cannot\ be\ found\ with\ given\ the\ data$

Solution

$\tan { \phi } =\cfrac { \omega L }{ R } =\cfrac { { X }_{ L } }{ R }$

Given: $\phi ={ 45 }^{ o }\quad \Rightarrow { X }_{ L }=R$

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Alternating Current Test - 23

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Alternating Current Test - 23

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Question 1

effective resistance due to capacity

effective wattage

effective voltage

None of these

Solution

Question 2

∞\displaystyle \infty∞

zero

1 Ω\OmegaΩ

2 Ω\OmegaΩ

Solution

Question 3

CCC

LLL

RRR

L, CL,\ CL, C and RRR

Solution

Question 4

decreases

increases directly with frequency

increases as square of frequency

decreases inversely with frequency

Solution

Question 5

R2+XC2,R2+XL2\displaystyle \sqrt{R^2+X^2_C} , \sqrt{R^2+X^2_L}R2+XC2​​,R2+XL2​​

R2+XL2,R2+XC2\displaystyle \sqrt{R^2+X^2_L} , \sqrt{R^2+X^2_C}R2+XL2​​,R2+XC2​​

R2+(XL−XC)2\displaystyle \sqrt{R^2+(X_L - X_C)^2}R2+(XL​−XC​)2​ each

(1R2+(1XC−1XL)2)−1\displaystyle \sqrt{\left(\frac{1}{R^2}+\left(\frac{1}{X_C}-\frac{1}{X_L}\right)^2 \right)^{-1}}(R21​+(XC​1​−XL​1​)2)−1​ each

Solution

Question 6

inductor increases

resistor increases

capacitor increases

circuit increases

Solution

Question 7

remains constant

increases

decreases

decreases at first, becomes minimum and then increases

Solution

Question 8

100Hz100 Hz100Hz

10Hz10 Hz10Hz

1000Hz1000 Hz1000Hz

10000Hz10000 Hz10000Hz

Solution

Question 9

202020 ohm

Zero

111 ohm

400400400 ohm

Solution

Question 10

R/4R/4R/4

R/2R/2R/2

RRR

cannot be found with given the datacannot\ be\ found\ with\ given\ the\ datacannot be found with given the data

Solution

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$\displaystyle \infty$

1 $\Omega$

2 $\Omega$

$C$

$L$

$R$

$L,\ C$ and $R$

$\displaystyle \sqrt{R^2+X^2_C} , \sqrt{R^2+X^2_L}$

$\displaystyle \sqrt{R^2+X^2_L} , \sqrt{R^2+X^2_C}$

$\displaystyle \sqrt{R^2+(X_L - X_C)^2}$ each

$\displaystyle \sqrt{\left(\frac{1}{R^2}+\left(\frac{1}{X_C}-\frac{1}{X_L}\right)^2 \right)^{-1}}$ each

$100 Hz$

$10 Hz$

$1000 Hz$

$10000 Hz$

$20$ ohm

$1$ ohm

$400$ ohm

$R/4$

$R/2$

$R$

$cannot\ be\ found\ with\ given\ the\ data$