Alternating Current Test

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Alternating Current Test - 74

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Question 1
1 / -0

When $$100\space V$$ dc is applied across a coil, a current of $$1\space A$$ flows through it and when $$100\space V$$ ac of $$50\space Hz$$ is applied to the same coil, only $$0.5\space A$$ flows. The inductance of coil is

A
$$5.5\space H$$

B
$$3/\pi\space H$$

C
$$\sqrt3/\pi\space H$$

D
$$2.5\space H$$

Solution

$$ R = \frac{V_{DC}}{I_{DC} } = \frac{100}{1} = 100 \Omega $$
$$\frac{V_{AC}^2}{I_{AC}^2} = R^2 + X_L^2 \Rightarrow \frac{100^2}{0.5^2}= 100^2+ X_L^2 \Rightarrow X_L = \sqrt{30000}= 100\sqrt{3} \Omega$$
$$X_L= L 2 \pi f \Rightarrow 100\sqrt{3} = L 2 \pi 50 \Rightarrow L =\frac{\sqrt{3}}{\pi} H$$
Question 2
1 / -0

A charged capacitor discharges through a resistance R with time constant $$ \tau $$. The two are now placed in series across an AC source of angular frequency $$\displaystyle \omega = \frac {1} {\tau} $$ . The impedance of the circuit will be:

A
$$\displaystyle \frac {R} {\sqrt {2}} $$

B
R

C
$$ \sqrt {2} $$ R

D
2R

Solution

In RC circuit, impedance $$Z=\sqrt{R^2+\dfrac{1}{\omega^2C^2}}$$ and time constant $$\tau=CR$$
Since $$\omega=\dfrac{1}{\tau}$$ so $$Z=\sqrt{R^2+\dfrac{\tau^2}{C^2}}=\sqrt{R^2+\dfrac{C^2R^2}{C^2}}=\sqrt{2} R$$
Question 3
1 / -0

The resonant frequency of an L-C circuit is

A
$$\displaystyle \frac{1}{2\pi\sqrt{LC}}$$

B
$$\displaystyle \frac{1}{2\pi}\sqrt{\frac{L}{C}}$$

C
$$\displaystyle \frac{1}{2\,\pi}\sqrt{\frac{L}{C}}$$

D
$$\displaystyle \frac{1}{2\pi}\sqrt{\frac{C}{L}}$$

Solution
Question 4
1 / -0

A series LCR circuit is connected across a source of alternating emf of changing frequency and resonates at frequency $$\text{f}_o$$. Keeping capacitance constant, if the inductance (L) is increased by $$\sqrt{3}$$ times and resistance(R) is increased by 1.4 times, the resonant frequency now is:

A
$$3^{1/4}\text{f}_0$$

B
$$\sqrt{3}\text{f}_0$$

C
$$(\sqrt{3}-1)^{1/4}\text{f}_0$$

D
$$\left(\dfrac{1}{3}\right)^{1/4}\text{f}_0$$

Solution

Resonant frequency of a series LCR circuit is given by the formula

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

Hence, $${ f }_{ 0 }=\dfrac { 1 }{ 2\pi \sqrt { LC } } $$

In second case, $${ f }'=\dfrac { 1 }{ 2\pi \sqrt { L'C } } =\dfrac { 1 }{ 2\pi \sqrt { \sqrt { 3 } LC } } =\dfrac { 1 }{ { 3 }^{ \frac { 1 }{ 4 } } } { f }_{ 0 }$$
Question 5
1 / -0

In a LR circuit of $$3\;mH$$ inductane and $$4\;\Omega$$ resistance, emf $$E=4\;cos\;(1000\;t)$$ volt is applied. The amplitude of current is:

A
$$0.8\ {A}$$

B
$$\dfrac{4}{7}\ {A}$$

C
$$1.0\ {A}$$

D
$$\dfrac{4}{\sqrt{7}}\ {A}$$

Solution

Given : $$L = 3mH = 3 \times 10^{-3} H$$ $$R =4 \Omega$$
Emf $$E = 4 cos1000t$$
$$\therefore$$ $$E_o = 4$$ volt and $$\omega = 1000$$
Inductive reactance $$X_L = \omega L = 1000 \times 3 \times 10^{-3} = 3 \Omega$$
Impedance $$Z = \sqrt{R^2 + X_L^2} = \sqrt{4^2 + 3^2} = 5 \Omega$$
Current amplitude $$I= \dfrac{E_o}{Z} = \dfrac{4}{5} = 0.8 A$$
Question 6
1 / -0

In a purely inductive circuit, the current is

A
In phase with the voltage

B
Out of phase with the voltage

C
Leads the voltage by $$\pi/ 2$$

D
Lags behind the voltage $$\pi/ 2$$
Question 7
1 / -0

If the power factor in a circuit is unit, then the impedance of the circuit is

A
Inductive

B
Capacitive

C
Partially inductive and partially capacitive

D
Resistive

Solution
Question 8
1 / -0

Inductive reactance of a coil is expressed in

A
Ampere

B
Ohm

C
Volt

D
Weber

Solution
Question 9
1 / -0

An AC generator producing $$10V$$ (rms) at $$200rad/s$$ is connected in series with a $$50\Omega $$ resistor, a $$400mH$$ inductor and a $$200\mu F$$ capacitor. The rms voltage across the inductor is

A
$$2.5V$$

B
$$3.4V$$

C
$$6.7V$$

D
$$10.8V$$

Solution

Given : $$E=10V, \ \ \ w=200, \ \ \ \ R=50\Omega , \ \ \ \ L=400mH, \ \ \ C=200\mu F\quad $$
So, capacitive reactance $$X_C = \dfrac{1}{wC} = \dfrac{1}{(200)(200\times 10^{-6})} = 25\Omega$$
Inductive reactance $$X_L = wL =(200)(400\times 10^{-3}) =80\Omega $$
Impedance $$\quad Z=\sqrt { { R }^{ 2 }+{ \left( { X }_{ L }-{ X }_{ C } \right) }^{ 2 } } =\sqrt { { 50 }^{ 2 }+{ (80-25) }^{ 2 } } $$
$$\quad Z=74.3\Omega $$
$$I=\cfrac { E }{ Z } =\cfrac { 10 }{ 74.3 } =0.13459A$$
$${ E }_{ L }=I{ X }_{ L }=0.1345\times 80=10.76V$$
Question 10
1 / -0

An inductor and a resistor are connected to an ac supply of $$50 V$$ and $$50 Hz$$. If the voltage across the resistor is $$40 V$$ the voltage across the inductor will be:

A
$$10\ V$$

B
$$20\ V$$

C
$$30\ V$$

D
$$60\ V$$

Solution

Given: Supply voltage, $$V_S=50\ V$$
Voltage across resistor, $$V_R=40\ V$$

For an LR circuit:
Applied voltage $$V_S=\sqrt { { V }_{ R }^{ 2 }+{ V }_{ L }^{ 2 } } $$
$$\Rightarrow { V }_{ L }=\sqrt { { 50 }^{ 2 }-{ 40 }^{ 2 } } =30\ V$$

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Re-Attempt Weekly Quiz Competition

Alternating Current Test - 74

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Alternating Current Test - 74

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Question 1

$$5.5\space H$$

$$3/\pi\space H$$

$$\sqrt3/\pi\space H$$

$$2.5\space H$$

Solution

Question 2

$$\displaystyle \frac {R} {\sqrt {2}} $$

R

$$ \sqrt {2} $$ R

2R

Solution

Question 3

$$\displaystyle \frac{1}{2\pi\sqrt{LC}}$$

$$\displaystyle \frac{1}{2\pi}\sqrt{\frac{L}{C}}$$

$$\displaystyle \frac{1}{2\,\pi}\sqrt{\frac{L}{C}}$$

$$\displaystyle \frac{1}{2\pi}\sqrt{\frac{C}{L}}$$

Solution

Question 4

$$3^{1/4}\text{f}_0$$

$$\sqrt{3}\text{f}_0$$

$$(\sqrt{3}-1)^{1/4}\text{f}_0$$

$$\left(\dfrac{1}{3}\right)^{1/4}\text{f}_0$$

Solution

Question 5

$$0.8\ {A}$$

$$\dfrac{4}{7}\ {A}$$

$$1.0\ {A}$$

$$\dfrac{4}{\sqrt{7}}\ {A}$$

Solution

Question 6

In phase with the voltage

Out of phase with the voltage

Leads the voltage by $$\pi/ 2$$

Lags behind the voltage $$\pi/ 2$$

Question 7

Inductive

Capacitive

Partially inductive and partially capacitive

Resistive

Solution

Question 8

Ampere

Ohm

Volt

Weber

Solution

Question 9

$$2.5V$$

$$3.4V$$

$$6.7V$$

$$10.8V$$

Solution

Question 10

$$10\ V$$

$$20\ V$$

$$30\ V$$

$$60\ V$$

Solution

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