Alternating Current Test

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Alternating Current Test - 32

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Question 1
1 / -0

In the series L-C-R circuit figure, the voltmeter and ammeter readings are

A
$$V $$=$$ 100 volt, I $$=$$ 2 amp$$

B
$$V $$=$$ 100 volt, I $$=$$ 5 amp$$

C
$$V $$=$$1000 volt, I $$=$$ 2 amp$$

D
$$V $$=$$ 300 volt, I$$=$$1 amp$$

Solution

Here, $$V_L=V_C$$.

$$V_L$$ and $$V_C$$ are in oppositr phases. Hence they will cancel each other. Now the resultant potential difference is equal to the applied potential difference.

i.e. $$V=100V$$

Impedence, $$Z=R(X_L \, or \, X_C)$$

Therefore,

$$I_{rms}=\dfrac{V_{rms}}{Z}=\dfrac{V_{rms}}{R}=\dfrac{100}{50}=2\,A$$

So, the voltmeter reading is $$100\,V$$ and ammeter reading is $$2\,A$$.
Question 2
1 / -0

A resistor $$R$$ and capacitor $$ C$$ are connected in series across an AC source of rms voltage $$5 V$$. If the rms voltage across $$C$$ is $$3 V$$ then that across $$R$$ is :

A
$$1V$$

B
$$2V$$

C
$$3V$$

D
$$4V$$

Solution

Source voltage = $$\sqrt{(Capacitor\ voltage)^2+(Resistor \ voltage)^2}$$

$$[V_{s(rms)}]^2 = [V_{C(rms)}]^2+[V_{R(rms)}]^2$$

$$(5)^2 = (3)^2+[V_{R(rms)}]^2$$

$$25-9 = [V_{R(rms)}]^2$$

$$16 = [V_{R(rms)}]^2$$

$$ V_{R(rms)}=4V$$
Question 3
1 / -0

The potential difference between the ends of a resistance R is $$V_{R}$$, between the ends of capacitor is $$V_{C}$$ $$=$$ 2$$V_{R}$$ and between the ends of inductance is $$V_{L}$$ $$=$$ 3$$V_{R}$$ then the alternating potential of the source in terms of $$V_{R}$$ will be:

A
$$\sqrt{2}V_{R}$$

B
$$V_{R}$$

C
$$\dfrac{V_{R}}{\sqrt{2}}$$

D
$$5V_{R}$$

Solution

$$V_S = \sqrt{V_R^2+(V_L-V_C)^2}$$

$$= \sqrt{V_R^2+(3V_R-2V_R)^2}$$

$$= \sqrt{V_R^2+V_R^2}$$

$$= V_R \sqrt{2}$$.
Question 4
1 / -0

The natural frequency of an LC - circuit is $$1,25,000$$ cycles per second. Then the capacitor C is replaced by another capacitor with a dielectric medium of dielectric constant k. In this case, the frequency decreases by $$25 kHz$$. The value of k is:

A
$$3.0$$

B
$$2.1$$

C
$$1.56$$

D
$$1.7$$

Solution

Natural frequency $$= 125000 Hz$$

New frequency $$= 125000 - 25000$$

$$ = 100000 Hz$$

Frequency of L-C circuit

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

So, $$2\pi f_1 = \dfrac{1}{\sqrt{LC_1}}$$ ---(1)

$$2\pi f_2 = \dfrac{1}{\sqrt{LC_2}}$$ ---(2)

Dividing (1) by (2)

$$\dfrac{f_1}{f_2} = \sqrt{\dfrac{C_2}{C_1}}$$

$$\sqrt{\dfrac{C_2}{C_1}} = \dfrac{125000}{100000} = \dfrac{5}{4}$$

$$\dfrac{C_2}{C_1} = \dfrac{25}{16}$$

$$k = \dfrac{C_2}{C_1} = \dfrac{25}{16} = 1.56$$
Question 5
1 / -0

In the following circuit, the values of current flowing in the circuit at f=0 and f=$$\infty$$ will respectively be

A
8A and 0A

B
0A and 0A

C
8A and 8A

D
0A and 8A

Solution

In a LCR circuit current at $$t = 0$$ is zero because inductor does not allow flow of current,
and at $$t = \infty$$ also current is zero because capacitor does not allow flow of current.
Question 6
1 / -0

The reading of voltmeter and ammeter in the following figure will respectively be :

A
0 and 2A

B
2A and 0V

C
2V and 2A

D
0V and 0A

Solution

In the problem $$X_C = 4\Omega$$
and $$X_L = 4\Omega$$
So, V across $$X_C$$ and $$X_L$$ will be same and in opposite direction, So net voltage will be zero. Since voltmeter is connected parallel to Capacitor and inductor so, it will read 0 volts.
Current $$=$$ $$\dfrac{V}{impedance}$$
$$impedence= \sqrt { R^{2}+X_L^{2} } $$
Current$$= \dfrac{90} {\sqrt{45^{2}+4^{2}}}$$
$$= \dfrac{90}{45.17}$$ $$\simeq 2$$A.
Question 7
1 / -0

In the following diagram, the value of emf of A.C. source will be :

A
$$40 V$$

B
$$40\sqrt{2}V$$

C
$$\dfrac{40}{\sqrt{2}}V$$

D
$$160 V$$

Solution

$$ E_{rms} = \sqrt{ V_R^2 + ( V_L - V_c )^2 } $$

$$= \sqrt{ 40^2 + (40-80)^2}$$

$$ = \sqrt{ 40^2 + 40^2} $$

$$= 40 \sqrt{2} V $$
Question 8
1 / -0

An inductance of 0.2 H and a resistance of 100$$\Omega $$ are connected in series to an A.C. 180 V, 50 Hz supply. The apparent current flowing in the circuit will be

A
0.525 A

B
5.25 A

C
1.525 A

D
15.25 A

Solution

$$Z= \sqrt{X_L^2+R^2}$$

$$= \sqrt{(2\pi fL)^2+R^2}$$

$$= \sqrt{(2\pi 50\times 0.2)^2+(100)^2}$$

$$= \sqrt{13947.84}$$

$$= 118.1 \Omega$$

$$i = \dfrac{V}{Z}$$

$$= \dfrac{180}{118.1}$$

$$= 1.525A$$
Question 9
1 / -0

An $$LCR$$ circuit has $$L=10\ mH, R=3\Omega$$, and $$C =1\mu$$ $$F$$ connected in series to a source of $$15 \cos\omega t$$ volt. The current amplitude at a frequency that is $$10\%$$ lower than the resonant frequency is:

A
$$0.5 A$$

B
$$0.7 A$$

C
$$0.9 A$$

D
$$1.1 A$$

Solution

Resonant frequency, $$w_o=\dfrac{1}{\sqrt{LC}}=\dfrac{1}{\sqrt{(10\times 10^{-3})(10^{-6})}}=10^4 rad/s$$

New frequency, $$\omega=(0.9)\omega_o=9\times 10^3 rad/s$$

We have new $${ X }_{ L }=\omega L=9\times { 10 }^{ 3 }\times 10\times { 10 }^{ -3 }=90\Omega $$ and $$X_C= \dfrac {1}{\omega C}=111.11$$ $$ohm$$.

Thus we calculate new $$Z$$ as

$$\sqrt{3^2+[90-111.11]^2}$$

$$=21.32 \Omega$$

Current amplitude $$=\dfrac{V_o}{Z}=\dfrac{15}{21.32}=0.7 A$$
Question 10
1 / -0

The values of $$X_{L}, X_{C}$$ and R in series with an A.C. circuit are 8$$\Omega $$ , 6$$\Omega $$ and 10$$\Omega $$ respectively. The
total impedance of the circuit will be ________$$\Omega $$

A
$$10.2$$

B
$$12.2$$

C
$$10$$

D
$$24.4$$

Solution

$$Z = \sqrt{R^2+(X_L-X_C)^2}$$

$$= \sqrt{(10)^2+(8-6)^2}$$

$$= \sqrt{10^2+2^2}$$

$$= \sqrt{100+4}$$

$$= 10.2 \Omega$$

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Alternating Current Test - 32

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Alternating Current Test - 32

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Question 1

$$V $$=$$ 100 volt, I $$=$$ 2 amp$$

$$V $$=$$ 100 volt, I $$=$$ 5 amp$$

$$V $$=$$1000 volt, I $$=$$ 2 amp$$

$$V $$=$$ 300 volt, I$$=$$1 amp$$

Solution

Question 2

$$1V$$

$$2V$$

$$3V$$

$$4V$$

Solution

Question 3

$$\sqrt{2}V_{R}$$

$$V_{R}$$

$$\dfrac{V_{R}}{\sqrt{2}}$$

$$5V_{R}$$

Solution

Question 4

$$3.0$$

$$2.1$$

$$1.56$$

$$1.7$$

Solution

Question 5

8A and 0A

0A and 0A

8A and 8A

0A and 8A

Solution

Question 6

0 and 2A

2A and 0V

2V and 2A

0V and 0A

Solution

Question 7

$$40 V$$

$$40\sqrt{2}V$$

$$\dfrac{40}{\sqrt{2}}V$$

$$160 V$$

Solution

Question 8

0.525 A

5.25 A

1.525 A

15.25 A

Solution

Question 9

$$0.5 A$$

$$0.7 A$$

$$0.9 A$$

$$1.1 A$$

Solution

Question 10

$$10.2$$

$$12.2$$

$$10$$

$$24.4$$

Solution

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