Alternating Current Test

Result

Alternating Current Test - 41

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The adjoining figure shows an $$AC$$ circuit with resistance $$R$$, inductance $$L$$ and source voltage $$V_s$$. Then

A
the source voltage $$V_s=72.8V$$.

B
the phase angle between current and source voltage is $$\tan^{-1}(7/2)$$.

C
Both $$(a)$$ and $$(b)$$ are correct.

D
Both $$(a)$$ and $$(b)$$ are wrong.

Solution

$$\displaystyle V_S=\sqrt{V^2_R+V^2_L}$$
$$=\displaystyle \sqrt{(70)^2+(20)^2}=72.8V$$
$$\displaystyle \tan \phi=\frac{X_L}{R}=\frac{V_L}{V_R}=\frac{20}{70}=\frac{2}{7}$$
Question 2
1 / -0

In series $$L-C-R$$ circuit voltage drop across resistance is $$8V$$, across inductor is $$6V$$ and across capacitor is $$12V$$. Then

A
Voltage of the source will be leading in the circuit.

B
Voltage drop across each element will be less than the applied voltage.

C
Power factor of the circuit will be $$3/4$$.

D
None of the above.

Solution

$$\displaystyle V=\sqrt{V^2_R+(V_C-V_L)^2}=10V$$
$$V_C > V_L$$, hence current leads the voltage.
Power factor $$\displaystyle =\cos \phi=\frac{8}{10}=0.8$$
Question 3
1 / -0

An $$AC$$ voltage is applied across a series combination of $$L$$ and $$R$$. If the voltage drop across the resistor and inductor are $$20V$$ and $$15V$$ respectively, then applied peak voltage is :

A
$$25V$$

B
$$35V$$

C
$$25\sqrt 2V$$

D
$$5\sqrt 7V$$

Solution

$$\displaystyle V=\sqrt{V^2_R+V^2_L}$$
$$\displaystyle =\sqrt{(20)^2+(15)^2}=25V$$
But this is the rms value.
$$\therefore$$ Peak value $$=\displaystyle \sqrt 2 V_{rms}=25\sqrt 2V$$
Question 4
1 / -0

In the circuit shown in figure, the $$AC$$ source gives a voltage $$V=20\cos (2000t)$$. Neglecting source resistance, the voltmeter and ammeter readings will be

A
$$0 V, 2.0A$$

B
$$0V, 1.4A$$

C
$$5.6V, 1.4A$$

D
$$8V, 2.0A$$

Solution

Here, $$X_{L}=\omega L=2000\times 5\times 10^{-3}\Omega =10\Omega$$
$$X_{C}=\dfrac{1}{\omega C}=\dfrac{1}{2000\times 50\times 10^{-6}}\Omega=10\Omega$$
$$X_{L}=X_{C}$$. Hence this is a case of resonance.
$$\implies i=\dfrac{V}{R}=\dfrac{(20/\sqrt{2})}{10}=1.4A$$
Voltmeter reading=$$1.4\times 4 volt=5.6V$$
Question 5
1 / -0

When $$100\space V$$ dc is applied across a solenoid, a current of $$1.0\space A$$ flows in it. When $$100\space V$$ ac is applied across the same coil, the current drops to $$0.5\space A$$. If the frequency of the ac source is $$50\space Hz$$, the impedance and inductance of the solenoid are

A
$$200\Omega\space and\space 0.55\space H$$

B
$$100\Omega\space and\space 0.86\space H$$

C
$$200\Omega\space and\space 1.0\space H$$

D
$$200\Omega\space and\space 0.93\space H$$

Solution

impedance= $$\frac{V_{ac}}{I_{ac}}= \frac {100 V}{0.5 A} = 200 \Omega $$
R=$$ \frac{V_{dc}}{I_{dc}}=\frac{100}{1}=100 \Omega $$
$$L \omega $$ = $$ \sqrt{ {impedance}^{2} - R^2 } = \sqrt { 200^2-100^2 } =173.2 $$
$$L=\frac{173.2}{\omega}=\frac{173.2}{2\pi 50} =0.55 H $$
Question 6
1 / -0

An inductive coil has resistance of $$100\Omega$$. When an ac signal of frequency $$1000\space Hz$$ is fed to the coil, the applied voltage leads the current by $$45^{\small\circ}$$. What is the inductance of the coil?

A
$$2\space mH$$

B
$$3.3\space mH$$

C
$$16\space mH$$

D
$$\sqrt5\space mH$$

Solution

$$ tan(45) = 1 = \frac{L \omega}{R} $$
$$ L = R / \omega = R / ( 2 \pi 1000) = .016 H$$
Question 7
1 / -0

In a series $$LC$$ circuit, the applied voltage is $$V_0$$. If $$\omega$$ is very low, then the voltage drop across the inductor $$V_L$$ and capacitor $$V_C$$ are

A
$$\displaystyle V_L=\frac{V_0}{2}; V_C=\frac{V_0}{2}$$

B
$$\displaystyle V_L=0; V_C=V_0$$

C
$$\displaystyle V_L=V_0; V_C=0$$

D
$$\displaystyle V_L=-V_C=\frac{V_0}{2}$$

Solution

$$X_L=\omega L$$
If $$\omega$$ is very low, then $$X_L=0$$
$$\therefore V_L=0$$
or $$V=V_C=V_0$$
Question 8
1 / -0

An $$AC$$ voltage source $$V=V_0\sin \omega t$$ is connected across resistance $$R$$ and capacitance $$C$$ as shown in the figure. It is given that $$R=1/\omega C$$ and the peak current is $$I_0$$. If the angular frequency of the voltage source is changed to $$\omega/\sqrt 3$$ then the new peak current in the circuit is

A
$$\displaystyle \frac{I_0}{2}$$

B
$$\displaystyle \frac{I_0}{\sqrt 2}$$

C
$$\displaystyle \frac{I_0}{\sqrt 3}$$

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

Solution

$$\displaystyle R=\frac{1}{\omega C}=X_C$$
$$\displaystyle \therefore Z=\sqrt{R^2+X^2_C}=\sqrt 2 R$$ (as $$X_C=R$$)
$$\displaystyle I_0=\frac{V_0}{Z}=\frac{V_0}{\sqrt 2 R}$$
When $$\omega$$ becomes $$\displaystyle \frac{1}{\sqrt 3}$$ times, $$X_C$$ will become $$\sqrt 3$$ times or $$\sqrt 3R$$.
$$\displaystyle Z'=\sqrt{(R^2)+(\sqrt 3R)^2}=2R$$
$$\displaystyle I'_0=\frac{V_0}{Z'}=\frac{V_0}{2R}=\frac{I_0}{\sqrt 2}$$
Question 9
1 / -0

A coil has an inductance of $$0.7\space H$$ and is joined in series with a resistance of $$220\Omega$$. When an alternating emf of $$220\space V$$ at $$50\space cps$$ is applied to it, then the wattless component of the current in the circuit is (take $$0.7\pi = 2.2)$$

A
$$5\space A$$

B
$$0.5\space A$$

C
$$0.7\space A$$

D
$$7\space A$$

Solution

wattless current = $$ { I }_{ max }sin({ tan }^{ -1 }\{ L\omega /R\} )=\frac { 220 }{ \sqrt { { R }^{ 2 }+{ L }^{ 2 }{ \omega }^{ 2 } } } sin({ tan }^{ -1 }\{ L\omega /R\} )=0.5A $$
Question 10
1 / -0

A resistor and an inductor are connected to an ac supply of $$120\space V$$ and $$50\space Hz$$. The current in the circuit is $$3\space A$$. If the power consumed in the circuit is $$108\space W$$, then the resistance in the circuit is

A
$$12\space\Omega$$

B
$$40\space\Omega$$

C
$$\sqrt{(52\times28)}\space\Omega$$

D
$$360\space\Omega$$

Solution

$$ I_{rms}= current in circuit = 3A $$
$$ P = 108 W = {I}_{rms}^{2} R =3^2 R$$
$$ R =108/9= 12 \Omega $$

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Alternating Current Test - 41

Result

Alternating Current Test - 41

Score

-

Rank

-

SHARING IS CARING

Question 1

the source voltage $$V_s=72.8V$$.

the phase angle between current and source voltage is $$\tan^{-1}(7/2)$$.

Both $$(a)$$ and $$(b)$$ are correct.

Both $$(a)$$ and $$(b)$$ are wrong.

Solution

Question 2

Voltage of the source will be leading in the circuit.

Voltage drop across each element will be less than the applied voltage.

Power factor of the circuit will be $$3/4$$.

None of the above.

Solution

Question 3

$$25V$$

$$35V$$

$$25\sqrt 2V$$

$$5\sqrt 7V$$

Solution

Question 4

$$0 V, 2.0A$$

$$0V, 1.4A$$

$$5.6V, 1.4A$$

$$8V, 2.0A$$

Solution

Question 5

$$200\Omega\space and\space 0.55\space H$$

$$100\Omega\space and\space 0.86\space H$$

$$200\Omega\space and\space 1.0\space H$$

$$200\Omega\space and\space 0.93\space H$$

Solution

Question 6

$$2\space mH$$

$$3.3\space mH$$

$$16\space mH$$

$$\sqrt5\space mH$$

Solution

Question 7

$$\displaystyle V_L=\frac{V_0}{2}; V_C=\frac{V_0}{2}$$

$$\displaystyle V_L=0; V_C=V_0$$

$$\displaystyle V_L=V_0; V_C=0$$

$$\displaystyle V_L=-V_C=\frac{V_0}{2}$$

Solution

Question 8

$$\displaystyle \frac{I_0}{2}$$

$$\displaystyle \frac{I_0}{\sqrt 2}$$

$$\displaystyle \frac{I_0}{\sqrt 3}$$

$$\displaystyle \dfrac{I_0}{3}$$

Solution

Question 9

$$5\space A$$

$$0.5\space A$$

$$0.7\space A$$

$$7\space A$$

Solution

Question 10

$$12\space\Omega$$

$$40\space\Omega$$

$$\sqrt{(52\times28)}\space\Omega$$

$$360\space\Omega$$

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.