Alternating Current Test

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Alternating Current Test - 49

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Question 1
1 / -0

If the rms current through a $$6.8 k \Omega$$ resistor is $$8 mA$$, the rms voltage drop across the resistor is

A
$$5.44 V$$

B
$$54.4 V$$

C
$$7.07 V$$

D
$$8 V$$

Solution

Rms current $$I_{rms} = 8 mA$$
Resistance of resistor $$R = 6.8k\Omega$$
Thus rms voltage $$V_{rms} = I_{rms} R$$
$$\therefore$$ $$V_{rms} = (8\times 10^{-3}) \times (6.8\times 10^3)$$
$$\implies$$ $$V_{rms} = 54.4 V$$
Question 2
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An L-C-R circuit contains $$R=50\Omega $$, $$L=1 mH$$ and $$C=0.1\mu F$$. The impedence of the circuit will be minimum for a frequency of

A
$$\dfrac { { 10 }^{ 5 } }{ 2\pi } Hz$$

B
$$\dfrac { { 10 }^{ 6 } }{ 2\pi } Hz$$

C
$$2\pi \times { 10 }^{ 5 }Hz$$

D
$$2\pi \times { 10 }^{ 6 }Hz$$

Solution

Impedance of L-C-R circuit will be minimum for a resonant frequency so,
$${ v }_{ 0 }=\dfrac { 1 }{ 2\pi \sqrt { LC } } $$
$$=\dfrac { 1 }{ 2\pi \sqrt { 1\times { 10 }^{ -3 }\times 0.1\times { 10 }^{ -6 } } } =\dfrac { { 10 }^{ 5 } }{ 2\pi } Hz$$
Question 3
1 / -0

The natural frequency of the circuit shown in adjoining figure is

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

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

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

D
Zero

Solution

In the given circuit, two capacitor and the two inductor are in series.
$$\therefore =L_s=L+L=2L$$
and $$\dfrac{1}{C_s}=\dfrac{1}{C}+\dfrac{1}{C}=\dfrac{2}{C}\Rightarrow C_s=\dfrac{C}{2}$$
$$\therefore$$ Natural frequency of the circuit
$$v=\dfrac{1}{2\pi \sqrt{L_sC_s}}=\dfrac{1}{2\pi \sqrt{2L}\times \sqrt{C/2}}$$
$$=\dfrac{1}{2\pi \sqrt{LC}}$$
Question 4
1 / -0

The value of alternating emf $$E$$ in the given circuit will be

A
$$220 V$$

B
$$140 V$$

C
$$100 V$$

D
$$20 V$$

Solution

Value of alternating emf $$E = \sqrt{V_R^2 + (V_c-V_L)^2}$$
$$\therefore$$ $$E= \sqrt{(80)^2 + (100 - 40)^2}$$
Or $$E = \sqrt{80^2 + 60^2}$$
$$\implies$$ $$E = \sqrt{10000} = 100V$$
Question 5
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If L-R circuit connected to a battery of constant emf $$E$$ switch $$S$$ is closed at time $$t=0$$. If $$e$$ denotes the induced emf across inductor and $$I$$ the current in the circuit at any time $$t$$. Then which of the following graphs shows the variation of $$e$$ with $$I$$?

A

B

C

D

Solution

$$V_L$$ or $$e=E-V_R=E-iR$$
Thus, $$e-i$$ graph is a straight line with negative slope and positive intercept.
Question 6
1 / -0

In L-C-R circuit, $$f=\dfrac { 50 }{ \pi } Hz$$, $$V=50 V$$, $$R=300\Omega$$. If $$L=1H$$ and $$C=20\mu C$$, then the voltage across capacitor is :

A
$$50 V$$

B
$$20 V$$

C
Zero

D
$$30 V$$

Solution

For an L-C-R circuit, the impedance $$\left( Z \right) $$ is given by
$$\because Z=\sqrt { { R }^{ 2 }+{ \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 } } $$
where, $${ X }_{ L }=\omega L=2\pi { f }_{ 1 }$$
and $${ X }_{ C }=\dfrac { 1 }{ \omega C } =\dfrac { 1 }{ 2\pi fC } $$
Given, $$f=\dfrac { 50 }{ \pi } Hz$$, $$R=300\Omega $$ and $$L=1H$$
and $$C=20\mu C=20\times { 10 }^{ -6 }C$$
$$Z=\sqrt { { \left( 300 \right) }^{ 2 }+\left( 2\pi \times \dfrac { 50 }{ \pi } \times 1-\dfrac { 1 }{ 2\pi \times \dfrac { 50 }{ \pi } \times 20\times { 10 }^{ -6 } } \right) } $$
$$Z=\sqrt { 90000+{ \left( 100-500 \right) }^{ 2 } } $$
$$\Rightarrow Z=\sqrt { 90000+160000 } =\sqrt { 250000 } $$
$$\Rightarrow Z=500\Omega $$
Hence, the current in the circuit is given by
$$i=\dfrac { V }{ Z } =\dfrac { 50 }{ 500 } =0.1A$$
Voltage across capacitor is
$${ V }_{ C }=i{ X }_{ C }=\dfrac { i }{ 2\pi fC } =\dfrac { 0.1 }{ 2\pi \times \dfrac { 50 }{ \pi } \times 20\times { 10 }^{ -4 } } $$
$$=\dfrac { 0.1\times { 10 }^{ 6 } }{ 100\times 20 } \Rightarrow { V }_{ C }=50V$$
Question 7
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A coil of inductive reactance $${ 1 }/{ \sqrt { 3 } }\Omega$$ and resistance $$ 1\Omega $$ is connected to $$200 V$$, $$50 Hz$$ A.C. supply. The time lag between maximum voltage and current is

A
$$\dfrac { 1 }{ 600 } s$$

B
$$\dfrac { 1 }{ 200 } s$$

C
$$\dfrac { 1 }{ 300 } s$$

D
$$\dfrac { 1 }{ 500 } s$$

Solution

$$\omega=2\pi f=314 rad/s$$
$$tan\phi=\dfrac{1}{1/\sqrt{3}}\rightarrow \phi=\dfrac{\pi}{6}$$
$$\omega t=\dfrac{\pi}{6}$$
$$t=\dfrac{1}{600}s$$
Question 8
1 / -0

In the adjoining circuit, if the reading of voltmeter $$V_1$$ and $$V_2$$ are 300 volts each, then the reading voltmeter $$V_3$$ and ammeter A are respectively

A
220 V, 2.2 A

B
100 V, 2.0 A

C
220 V, 2.0 A

D
100 V, 2.2 A

Solution

Given, $$V_1=V_2=300V;V_3=?,i=?$$
As, We know,
$$V=\sqrt{V^2_3+(V_1-V_2)^2}$$
We now $$(V_1-V_2)^2=(300-300)^2=0$$
$$\therefore 220=\sqrt{V_3^2}=V_3\Rightarrow V_3=220V$$
$$\therefore I=\dfrac{V_3}{R}=\dfrac{220}{100}=2.2A$$
So the reading of voltmeter $$V_3=220V $$ and Ammeter $$A=2.2A$$.
Question 9
1 / -0

An inductive coil has a resistance of 100 When an AC signal of frequency 1000 Hz is applied to the coil, the voltage leads the current by $$45^o$$. The inductance of the coil is

A
$$\dfrac{1}{10 \pi}$$

B
$$\dfrac{1}{20 \pi}$$

C
$$\dfrac{1}{40 \pi}$$

D
$$\dfrac{1}{60 \pi}$$

Solution

Given : $$f = 1000$$ Hz
Resistance of inductive coil $$X_L = 100\Omega$$
Thus inductance of coil $$L = \dfrac{X_L}{2\pi f}$$
$$\therefore$$ $$L = \dfrac{100}{2\pi \times 1000} = \dfrac{1}{20\pi }$$ Henry
Question 10
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The instantaneous values of current and voltage in an AC circuit are $$i=100\sin 314 t$$ amp and $$e=200\sin (314t+\dfrac{\pi}{3})V$$ respectively. If the resistance is $$1\Omega$$, then the reactance of the circuit will be:

A
$$\sqrt{3}\Omega$$

B
$$100\sqrt{3}\Omega$$

C
$$-200\sqrt{3}\Omega$$

D
$$-200/ \sqrt{3}\Omega$$

Solution

We have, $$V_0=i_0Z$$
$$\Rightarrow 200=100Z$$
$$Z=2\Omega$$
So, the impedance of the circuit
$$Z^2=R^2+X^2_L$$
$$\Rightarrow (2)^2=(1)^2+X^2_L$$
$$X_L=\sqrt{3}\Omega$$