Alternating Current Test

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Alternating Current Test - 72

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Question 1
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An ac voltage source $$V = V_0\sin(\omega t)$$ is connected across resistance $$R$$ and capacitance $$C$$ as shown in figure. It is given that $$R = \dfrac{1}{\omega C}$$. The peak current is $$I_0$$ . If the angular frequency of the voltage source is changed to $$\dfrac{\omega}{\sqrt{3}}$$ then the new peak current in the circuit is:

A
$$\dfrac{I_0}{2}$$

B
$$\dfrac{I_0}{\sqrt{2}}$$

C
$$\dfrac{I_0}{\sqrt{3}}$$

D
$$\dfrac{I_0}{3}$$

Solution

The peak value of the current is:

$$I_0 = \dfrac{V_0}{\sqrt{R^2+\dfrac{1}{{\omega}^2C^2}}} = \dfrac{V_0}{\sqrt{2}R}$$

When the angular frequency is changed to $$\dfrac{\omega}{\sqrt{3}}$$, the new peak value is:

$$I'_0 = \dfrac{V_0}{\sqrt{R^2+\dfrac{3}{{\omega}^2C^2}}}=\dfrac{V_0}{\sqrt{4R^2}}=\dfrac{V_0}{2R}$$

$$\therefore I'_0 = \dfrac{I_0}{\sqrt{2}}$$
Question 2
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A coil has resistance $$30\ \text{ohm}$$ and inductive reactance $$20\ \text{Ohm at}\ 50\ \text{Hz}$$ frequency. If an ac source, of $$200\ \text{volt},\ 100\ \text{Hz},$$ is connected across the coil, the current in the coil will be :

A
$$4.0\ \text{A}$$

B
$$8.0\ \text{A}$$

C
$$\dfrac {20}{\sqrt {13}}\ \text{A}$$

D
$$2.0\ \text{A}$$

Solution

For a given coil
R:Resistance$$=30\Omega $$
$${ X }_{ L }:$$ Inductive Reactance $$=\omega L\Omega =\left( 2\pi F \right) L\Omega $$
At $$50㎐,{ X }_{ L }=20\Omega $$
$$\therefore 2\pi \times 50\times L=20$$
$$\therefore L=\cfrac { 20 }{ 100\pi } $$
At $$100㎐$$
$${ X }_{ L }=2\pi \times 100\times L=2\pi \times 100\times \cfrac { 20 }{ 100\pi } =40\Omega $$
At $$100㎐$$
$$z=impedance=\sqrt { { R }^{ 2 }+{ { X }_{ L } }^{ 2 } } $$
$$\therefore Z=\sqrt { { 40 }^{ 2 }+{ 30 }^{ 2 } } =50\Omega $$
Current at $$200V,100㎐$$ is given as
$$I=\cfrac { V }{ z } =\cfrac { 200 }{ 50 } =4A$$
Therefore option A is correct
Question 3
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When $$100$$ volt $$DC$$ source is applied across a coil, a current of $$1A$$ flows through it. When $$100V$$ $$AC$$ source of $$50Hz$$ is applied to the same coil, only $$0.5A$$ current flows. Calculate the inductance of the coil.

A
$$\displaystyle (\pi/ \sqrt 3)H$$

B
$$\displaystyle (\sqrt 3/ \pi)H$$

C
$$\displaystyle (2/\pi)H$$

D
None of these

Solution

$$\displaystyle I_{dc}=\frac{V_{dc}}{R}$$

$$\displaystyle I=\frac{100}{R}\Rightarrow R=100\Omega$$

$$\displaystyle I_{ac}=\frac{V_{ac}}{\sqrt{R^2+X^2_L}}$$

$$\displaystyle 0.5=\frac{100}{\sqrt{(100)^2+X^2_L}}$$

or $$\displaystyle X_L=100\sqrt 3 \Omega =(2\pi fL)$$

$$\displaystyle \therefore L=\frac{100\sqrt 3}{2\pi f}$$

$$\displaystyle =\frac{100\sqrt 3}{2\pi (50)}$$

$$\displaystyle = \left(\frac{\sqrt 3}{\pi}\right)H$$
Question 4
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A coil, a capacitor and an $$AC$$ source of rms voltage $$24V$$ are connected in series. By varying the frequency of the source, a maximum rms current of $$6A$$ is observed. If coil is connected to a dc battery of emf $$12$$ volt and internal resistance $$4\Omega$$, then current through it in steady state is :

A
$$2.4A$$

B
$$1.8A$$

C
$$1.5A$$

D
$$1.2A$$

Solution

$$\displaystyle I_{rms}=\frac{V}{R}$$ (at resonance)
$$\displaystyle 6=\frac{24}{R}$$
$$\displaystyle \therefore R=4\Omega$$
$$\displaystyle I_{dc}=\frac{V}{R+r}=\frac{12}{4+4}=1.5A$$
Question 5
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A $$50Hz$$ $$AC$$ source of $$20V$$ is connected across $$R$$ and $$C$$ as shown in figure. The voltage across $$R$$ is $$12V$$. The voltage across $$C$$ is

A
$$8V$$

B
$$16V$$

C
$$10V$$

D
Not possible to determine unless value of $$R$$ and $$C$$ are given.

Solution

$$\displaystyle V_C=\sqrt{V^2-V^2_R}=\sqrt{(20)^2-(12)^2}$$
$$=16V$$
Question 6
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Directions For Questions

A student in a lab took a coil and connected it to a $$12V$$ $$DC$$ source. He measures the steady state current in the circuit to be $$4A$$. He then replaced the $$12V$$ $$DC$$ source by a $$12V,$$ $$(\omega =50 rad/s)AC$$ source and observe that the reading in the $$AC$$ ammeter is $$2.4A$$. He then decides to connect a $$2500\mu F$$ capacitor in series with the coil and calculate the average power developed in the circuit. Further he also decides to study the variation in current (with the capacitor and the battery in series).
Based on the readings taken by the student, answer the following questions.

...view full instructions

The value of resistance of the coil calculated by the student is :

A
$$3\Omega$$

B
$$4\Omega$$

C
$$5\Omega$$

D
$$8\Omega$$

Solution

$$\displaystyle V_{DC}=I_{DC}R$$
$$\displaystyle \therefore R=\frac{V_{DC}}{I_{DC}}=\frac{12}{4}=3\Omega$$
$$\displaystyle I_{AC}=\frac{V_{AC}}{Z}=\frac{V_{AC}}{\sqrt {R^2+X^2_L}}$$
$$\displaystyle 2.4=\frac{12}{\sqrt {(3)^2+X^2_L}}$$
Solving this equaion we get,
$$\displaystyle X_L=4\Omega$$
$$\displaystyle X_C=\frac{1}{\omega C}=\frac{1}{50\times 2500\times 10^{-6}}$$
$$=8\Omega$$
$$\displaystyle Z=\sqrt{R^2+(X_C-X_L)^2}=5\Omega$$
$$\displaystyle \therefore I=\frac{V_{DC}}{Z}=\frac{12}{5}=2.4A=I_{rms}$$
$$\displaystyle P=I^2_{rms}R=(2.4)^2(3)$$
$$\displaystyle =17.28W$$
At

given frequency $$X_C > X_L$$. If $$\omega$$ is further decreased,

$$X_C$$ will increase $$\left( as X_C\propto \frac{1}{\omega}\right)$$

and $$X_L$$ will increase $$(as X_L \propto \omega)$$.
Therefore $$X_C-X_L$$ and hence $$Z$$ will increase. So, current will decrease.
Question 7
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Consider an $$L-C-R$$ circuit as shown in figure, with an $$AC$$ source of peak value $$V_0$$ and angular frequency $$\omega$$. Then the peak value of current through the $$AC$$ source is :

A
$$\displaystyle \frac{V_0}{\sqrt{\displaystyle \frac{1}{R^2}+\left(\omega L-\displaystyle\frac{1}{\omega C}\right)^2}}$$

B
$$\displaystyle V_0 \sqrt {1/ {R^2}+\left(\omega C-\displaystyle \frac{1}{\omega L}\right)^2 }$$

C
$$\displaystyle \frac{V_0}{\sqrt {R^2+\left(\omega L-\displaystyle \frac{1}{\omega C}\right)}^2}$$

D
None of these.

Solution

As R, L , C are in parallel, equivalent impedance is given by:
$$ \dfrac{1}{Z} = \dfrac{1}{Z_R } + \dfrac{1}{Z_L} + \dfrac{1}{Z_C} = \dfrac{1}{R} + \dfrac{1}{j L \omega} + j C \omega $$
$$ \left | \dfrac{1}{Z} \right| = \left | \dfrac{1}{R} + \dfrac{1}{j L \omega} + j C \omega \right | = \sqrt{ \dfrac{1}{R^2} + \left( \dfrac{1}{L \omega} - C \omega \right)^2 } $$

Peak value of current is given by:
$$ I_{peak} = \left | \dfrac{V_0}{Z} \right |= { V_0}{ \sqrt{ \dfrac{1}{R^2} + \left( \dfrac{1}{L \omega} - C \omega \right)^2 } }$$
Question 8
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When an alternating voltage of $$220V$$ is applied across a device $$P$$, a current of $$0.25A$$ flows through the circuit and it leads the applied voltage by an angle $$\pi/2$$ radian. When the same voltage source is connected across another device $$Q$$, the same current is observed in the circuit but in phase with the applied voltage. What is the current when the same source is connected across a series combination of $$P$$ and $$Q$$?

A
$$\displaystyle \frac{1}{4\sqrt 2}A$$ lagging in phase by $$\pi/4$$ with voltage.

B
$$\displaystyle \frac{1}{4\sqrt 2}A$$ leading in phase by $$\pi/4$$ with voltage.

C
$$\displaystyle \frac{1}{\sqrt 2}A$$ leading in phase by $$\pi/4$$ with voltage.

D
$$\displaystyle \frac{1}{4\sqrt 2}A$$ leading in phase by $$\pi/2$$ with voltage.

Solution

In first case, $$\displaystyle X_C=\frac{V}{I}=\frac{220}{0.25}=880\Omega$$
In the second case, $$\displaystyle R=\frac{V}{I}=\frac{220}{0.25}=880\Omega$$
In the combination of $$P$$ and $$Q$$
$$\displaystyle \tan\phi=\frac{X_C}{R}=1$$
$$\therefore \phi=45^o$$
Since the circuit is capacitive, current leads the voltage. Further,
$$\displaystyle Z=\sqrt{R^2+X^2_C}=880\sqrt 2\Omega$$
$$\displaystyle I=\frac{V}{Z}=\frac{220}{880\sqrt 2}=\frac{1}{4\sqrt 2}A$$
Question 9
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A signal generator supplies a sine wave of $$200V$$, $$5kHz$$ to the circuit as shown in the figure. Then choose the wrong statement:

A
The current in the resistive branch is $$0.2A$$.

B
The current in the capacitive branch is $$0.126A$$.

C
Total line current is $$=0.283A$$.

D
Current in both the branches is same.

Solution

$$\displaystyle I_R=\frac{V_{rms}}{R}=\frac{200}{100}=0.2A$$

$$\displaystyle X_C=\frac{1}{2\pi fC}=\frac{1}{(2\pi)(5\times 10^3)\left(\frac{1}{\pi}\times 10^{-6}\right)}$$

$$=100\Omega$$

$$\therefore \displaystyle I_C=\frac{V_{rms}}{X_C}=\frac{200}{100}=0.2A$$

$$I_C$$ is $$90^o$$ ahead of the applied voltage and $$I_R$$ is in phase with the applied voltage. Hence, there is a phase difference of $$90^o$$ between $$I_R$$ and $$I_C$$ too.

$$\therefore \displaystyle I=\sqrt{I^2_R+I^2_C}$$

$$=\displaystyle \sqrt{(0.2)^2+(0.2)^2}$$

$$=0.283A$$
Question 10
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An $$LCR$$ circuit contains resistance of $$100\Omega$$ and a supply of $$200\space V$$ at $$300\space rad/s$$ angular frequency. If only capacitance is taken out from the circuit and the rest of the circuit is joined, current lags behind the voltage by $$60^{\small\circ}$$. If on the other hand, only inductor is taken out, the current leads by $$60^{\small\circ{}}$$ with the applied voltage. The current flowing in the circuit is

A
$$1\space A$$

B
$$1.5\space A$$

C
$$2\space A$$

D
$$2.5\space A$$

Solution

if only capacitance is remove,
$$ Z=R+jL\omega \quad \Rightarrow \quad \left| Z \right| cos(60)=R=100$$
$$\Rightarrow \quad \left| Z \right| =200\quad \Omega$$
$$\Rightarrow \quad L\omega =\left| Z \right| sin(60)=173.205 $$
if only inductor is taken out,
$$ Z=R-j\frac { 1 }{ C\omega } \quad \Rightarrow \quad \left| Z \right| cos(-60)=R=100$$
$$\Rightarrow \quad \left| Z \right| =200\quad \Omega$$
$$\Rightarrow \quad \frac { 1 }{ C\omega } =\left| Z \right| sin(-60)=-173.205 $$
therefore actual impedance= $$ R+jL\omega - j \frac{1}{C\omega} = 100 \Omega $$
therefore current in circuit=$$\dfrac{200 V }{100 \Omega} = 2A $$