Alternating Current Test

Result

Alternating Current Test - 51

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

A 44mH inductor is connected to a 220V, 50Hz AC supply. Then, the inductive reactance of the inductor is :

A
$$3.82 \, \Omega $$

B
$$10.8 \, Hz$$

C
$$13.82 \, \Omega $$

D
$$21.5 \, Hz$$

Solution

Inductive reactance $$X_L$$ is given by
$$X_L = \omega L = 2\pi fL$$
Given, f=50Hz, L=44mH
$$= 44\times 10^{-3}H$$
$$X_L = 2\pi \times 44 \times 10^{-3} \times 50= 13.82 \Omega$$
Question 2
1 / -0

In the given circuit $$R$$ in pure resistance and $$X$$ is unknown circuit element. An AC voltage source is applied across $$A$$ and $$C$$. If $${ V }_{ AB }={ V }_{ AC }$$, then $$X$$ is

A
Pure resistance

B
Pure inductance

C
Combination of inductance and capacitance at resonance

D
None of the above

Solution

Since the voltage across the resistive element is same as the voltage applied, the voltage drop across $$BC$$ is zero. This is possible only when the ohmic value of the element connected across $$BC$$ is zero. So, $$X$$ should be combination of inductance and capacitance at resonance.
Question 3
1 / -0

The frequency of the output signal becomes ________ times by doubling the value of the capacitance in the LC oscillator circuit.

A
$$\sqrt { 2 } $$

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

C
$$\dfrac { 1 }{ 2 } $$

D
$$2$$

Solution

In the L-C circuits
For the first condition
$${ f }_{ 1 }=\dfrac { 1 }{ 2\pi } \sqrt { \dfrac { 1 }{ LC } } $$ .....(i)
For the second condition, when $$C=$$ doubling
$${ f }_{ 2 }=\dfrac { 1 }{ 2\pi } \sqrt { \dfrac { 1 }{ 2LC } } $$ ....(ii)
From the equation (i), we get
$${ f }_{ 2 }=\dfrac { 1 }{ 2\pi } \sqrt { \dfrac { 1 }{ LC } } \times \dfrac { 1 }{ \sqrt { 2 } } \Rightarrow { f }_{ 2 }={ f }_{ 1 }\times \dfrac { 1 }{ \sqrt { 2 } } $$
So, the frequency of the output signal becomes $${ 1 }/{ \sqrt { 2 } }$$ times by doubling the value of the capacitance in the L-C oscillator circuit.
Question 4
1 / -0

In a series resonant circuit, the AC voltage across resistance R, inductor L and capacitor C are 5 V, 10 V and 10 V respectively. The AC voltage applied to the circuit will be

A
10 V

B
25 V

C
5 V

D
20 V

Solution

Given, $$V_R=5\, V, V_L=10$$ and $$V_C=10 \, V$$
In the MR circuit, the AC voltage applied to the circuit will be
$$V=\sqrt{V^2_R+(V_L-V_C)^2}$$
$$=\sqrt{(5)^2+(10-10)^2}$$
$$=5\, V$$
Question 5
1 / -0

The diagram given show the variation of voltage and current in an AC circuit. The circuit contains

A
Only a resistor

B
Only a pure inductor

C
Only a capcacitor

D
A capacitor and and inductor

Solution

The given circuit shows that the current lags the applied voltage. This is possible if the circuit has inductive element. So, the circuit contain a pure inductor.
Question 6
1 / -0

The power loss in an AC circuit can be minimized by.

A
Decreasing resistance and increasing inductance

B
Decreasing inductance and increasimg resistance

C
Increasing both inductance and resistance

D
Decreasing both inductance and resistance

Solution

In an AC circuit, power loss can be minimized by decreasing in resistance and by increasing in inductance.
Question 7
1 / -0

What is the range of the characteristic impedance of a coaxial cable?

A
Between $$150\Omega$$ to $$600\Omega$$

B
Between $$50\Omega$$ to $$70\Omega$$

C
Between $$0\Omega$$ to $$50\Omega$$

D
Between $$100\Omega$$ to $$150\Omega$$

Solution

Characteristic impedance of a coaxial cable is between $$50\Omega$$ to $$70\Omega$$.
Question 8
1 / -0

A $$220$$V main supply is connected to a resistance of $$100$$k$$\Omega$$. The effective current is?

A
$$2.2$$mA

B
$$2.2\sqrt{2}$$mA

C
$$\displaystyle\frac{2.2}{\sqrt{2}}$$mA

D
None of these

Solution

Effective current is the rms value. Here, $$220$$V is the labelled value of AC which is also the rms value. Hence,
$$I_{rms}=\displaystyle\frac{E_{rms}}{R}$$
$$I_{rms}=\displaystyle\frac{220}{100\times 10^3}$$
$$I_{rms}=2.2mA$$.
Question 9
1 / -0

Which one of the following curves represents variation of current i with frequency f in series LCR circuit?

A

B

C

D

Solution

In series LCR circuit, the graph(c) represents the curve between current (i) and frequency(f).
Question 10
1 / -0

Consider two series resonant circuits with components $$L_1C_1$$ and $$L_2C_2$$ with same resonant frequency , $$\omega$$.When connected in series, the resonant frequency of the combination is

A
$$2\omega$$

B
$$\dfrac{\omega}{2}$$

C
$$3\omega$$

D
$$\omega$$

Solution

Initially, $$\omega = \dfrac{1}{\sqrt{L_1C_1}}$$ and

$$\omega = \dfrac{1}{\sqrt{L_2C_2}}$$

Finally, $$L_{eq} = L_1 + L_2$$

$$C_{eq} = \dfrac{C_1C_2}{C_1 + C_2}$$

$$\omega' = \dfrac{1}{\sqrt{L_{eq}C_{eq}}}$$

from above equation, we will get $$\omega' = \omega$$