Alternating Current Test

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Alternating Current Test - 45

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Weekly Quiz Competition

Question 1
1 / -0

In an LCR circuit the potential difference between the terminal of the inductance is $$60\ V$$, between the terminals of the capacitor is $$30\ V$$ and that between the terminals of the resistance is $$40\ V$$. The supply voltage will be equal to:

A
$$130\ V$$

B
$$10\ V$$

C
$$50\ V$$

D
$$70\ V$$

Solution

Supply voltage of an LCR circuit
$$V=\sqrt{V_R^2+(V_L-V_C)^2}$$

since inductor and capacitor potentials are out of phase with each other)

$$=\sqrt{40^2+(60-30)^2}\ V$$
$$=50\ V$$
Question 2
1 / -0

The peak value of an alternating e.m.f. E given by $$E={ E }_{ 0 } \cos\omega t$$ is 10 volt and frequency is 50 Hz. At time $$t = (1/600)$$ sec, the instantaneous value of e.m.f. is

A
$$10$$ volt

B
$$5\sqrt { 3 } $$ volt

C
$$5$$ volt

D
$$1$$ volt

Solution

$$E={ E }_{ 0 }cos\omega t=10cos50\times 2\pi \times t$$

$$=E(t=\dfrac { 1 }{ 600 } s)=10cos\dfrac { 100\pi }{ 600 } $$

$$=10cos\dfrac { \pi }{ 6 } =5\sqrt { 3 } $$ Volt
Question 3
1 / -0

A series LCR circuit is tuned to resonance. If the angular frequency of the applied AC voltage at resonance is $$\omega $$, the impedance of the circuit then is:

A
$$R+\omega L+(\frac { 1 }{ \omega C } )$$

B
R

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

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

Solution

At resonance $$\dfrac{1}{\omega C}=\omega L$$
$$\therefore impedance =R$$
Question 4
1 / -0

A series $$LCR$$ circuit is connected to a source of alternating emf $$50 \ V$$ and if the potential differences across inductor and capacitor are $$90\ V$$ and $$60\ V$$ respectively, the potential difference across resistor is:

A
$$400\ V$$

B
$$40\ V$$

C
$$80\ V$$

D
$$1600\ V$$

Solution

The potential difference across inductor and capacitor will be opposite to each other and that of the resistor will be perpendicular to them. The resultant of three must be $$50V$$.
As we know, $$ V=\sqrt{{(V_L-V_C)}^2 + {V_R}^2}$$
Hence,
$$\sqrt { { (90-60) }^{ 2 }+{ V }_{ R }^{ 2 } } =50$$

$$900+{ V }_{ R }^{ 2 }=2500$$

$${ V }_{ R }^{ 2 }=1600$$

$${ V }_{ R }=40\ V$$
Question 5
1 / -0

A $$100\Omega$$ resistor is connected to a $$220V,50Hz$$ AC supply. Find rms value of current in the circuit and the net power consumed for a complete cycle.

A
$$2.20A,484W$$

B
$$3.20A,564W$$

C
$$1.60A,278W$$

D
$$5.80A,646W$$

Solution

GIven: $$V_{rms}=220V$$
$$\therefore i_{rms}=\dfrac{V_{rms}}{R}=\dfrac{220V}{100\Omega}=2.20A$$
The power consumed for a complete cycle is:
$$P=(i_{rms})^2R=(2.20A)^2\times 100\Omega=484W$$
Question 6
1 / -0

A light bulb is rated at $$100W$$ for a $$220V$$ supply. Find the peak voltage of the source:

A
$$111V$$

B
$$211V$$

C
$$311V$$

D
$$411V$$

Solution

$$Answer:-$$ C
The rated voltage in bulb is rms voltage.
$$V_{rms}=\dfrac { { V }_{ 0 } }{ \sqrt { 2 } } \\ { V }_{ 0 }=\sqrt { 2 } \times 220=311.08V$$
Question 7
1 / -0

A sinusoidal voltage $$V=200sin314t$$ is applied to a $$10\Omega$$ resistor. Find rms current.

A
$$14.14A$$

B
$$28.28A$$

C
$$56.56A$$

D
$$100A$$

Solution

$$Answer:-$$ A
General sinusoidal voltage variation is given by:
$$V=V_0sin(\omega t)$$
Here, in this equation $$V_0$$ is peak voltage.
So, on comparing both equation we get:
$$V_0=200V ,\omega=314$$
Relation between rms voltage and peak voltage is:
$$V_{rms}=\dfrac { { V }_{ 0 } }{ \sqrt { 2 } } =\dfrac { 200 }{ \sqrt { 2 } } =141.4V$$
Relation between rms current and peak current is:
$$I_{rms}=\dfrac { { V }_{ rms } }{ R } =\dfrac { 141.4 }{ 10 } =14.14A$$
Question 8
1 / -0

A light bulb is rated at $$100W$$ for a $$220V$$ supply. Find the resistance of the bulb:

A
$$184\Omega$$

B
$$284\Omega$$

C
$$384\Omega$$

D
$$484\Omega$$

Solution

$$Answer:-$$ D
Power is given by:
$$P=\dfrac { { V }^{ 2 } }{ R } \\ 100=\dfrac { { (220 })^{ 2 } }{ R } \\ R=484\Omega $$
Question 9
1 / -0

A sinusoidal voltage $$V=200sin314t$$ is applied to a $$10\Omega$$ resistor. Find rms voltage.

A
$$141.4V$$

B
$$314.2V$$

C
$$519.6V$$

D
$$278.9V$$

Solution

$$Answer:-$$ A
General sinusoidal voltage variation is given by:
$$V=V_0sin(\omega t)$$
Here, in this equation $$V_0$$ is peak voltage.
So, on comparing both equation we get:
$$V_0=200V ,\omega=314$$
Relation between rms voltage and peak voltage is:
$$V_{rms}=\dfrac { { V }_{ 0 } }{ \sqrt { 2 } } =\dfrac { 200 }{ \sqrt { 2 } } =141.4V$$
Question 10
1 / -0

A sinusoidal voltage $$V=200sin314t$$ is applied to a $$10\Omega$$ resistor. Find the frequency of the supply.

A
$$40 Hz$$

B
$$50 Hz$$

C
$$60 Hz$$

D
$$80 Hz$$

Solution

$$Answer:-$$ B
General sinusoidal voltage variation is given by:
$$V=V_0sin(\omega t)$$
On comparing both equation we get:
$$\omega =314=2\pi f\\ f=\dfrac { 314 }{ 2\pi } =50Hz$$

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Alternating Current Test - 45

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Alternating Current Test - 45

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Question 1

$$130\ V$$

$$10\ V$$

$$50\ V$$

$$70\ V$$

Solution

Question 2

$$10$$ volt

$$5\sqrt { 3 } $$ volt

$$5$$ volt

$$1$$ volt

Solution

Question 3

$$R+\omega L+(\frac { 1 }{ \omega C } )$$

R

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

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

Solution

Question 4

$$400\ V$$

$$40\ V$$

$$80\ V$$

$$1600\ V$$

Solution

Question 5

$$2.20A,484W$$

$$3.20A,564W$$

$$1.60A,278W$$

$$5.80A,646W$$

Solution

Question 6

$$111V$$

$$211V$$

$$311V$$

$$411V$$

Solution

Question 7

$$14.14A$$

$$28.28A$$

$$56.56A$$

$$100A$$

Solution

Question 8

$$184\Omega$$

$$284\Omega$$

$$384\Omega$$

$$484\Omega$$

Solution

Question 9

$$141.4V$$

$$314.2V$$

$$519.6V$$

$$278.9V$$

Solution

Question 10

$$40 Hz$$

$$50 Hz$$

$$60 Hz$$

$$80 Hz$$

Solution

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