Electricity Test

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Electricity Test - 29

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Weekly Quiz Competition

Question 1
1 / -0

What is the series equivalent of $1000 \ \Omega$ resistor and $$2700 \ \Omega$$ resistor in series?

A
$2700 \ \Omega$

B
$3000 \ \Omega$

C
$3700 \ \Omega$

D
$1000 \ \Omega$

Solution

Given : $R_1 = 1000\Omega$ $R_2 = 2700\Omega$

Equivalent resistance in series combination is given by
$R_{eq} = R_1+R_2 = 1000+2700 = 3700$ $\Omega$
Question 2
1 / -0

What is the total resistance when four resistors of $20, 40, 60$ , and $80$ respectively are connected in parallel (in ohm)?

A
$4.4$

B
$9.6$

C
$14.8$

D
$20$

E
$25.2$

Solution

Given : $R_1 =20\Omega$ $R_2 =40\Omega$ $R_3 =60\Omega$ $R_4 =80\Omega$
Equivalent resistance for parallel combination $\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} +\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}$
$\therefore$ $\dfrac{1}{R_{eq}} = \dfrac{1}{20} +\dfrac{1}{40}+\dfrac{1}{60}+\dfrac{1}{80}$ $\implies R_{eq} =9.6\Omega$
Question 3
1 / -0

A $10\ k\Omega$ resistor can be obtained by using

A
$3\ k\Omega,5\ k\Omega,2\ k\Omega$ resistors in series

B
$3\ k\Omega,5\ k\Omega,2\ k\Omega$ Resistors in parallel

C
$3\ k\Omega,5\ k\Omega$ in series and combination in parallel to $2\ k\Omega$ resistor in series

D
$5\ k\Omega,5\ k\Omega,2\ k\Omega,$ Resistors in series

Solution

When three resistors are connected in series, the equivalent resistance is $R_{eq}=R_1+R_2+R_3$ .
In the question, $R_1=3\ k\Omega$ , $R_2=5\ k\Omega$ , $R_3=2\ k\Omega$ .
$\therefore R_{eq}=3+5+2 = 10\ k\Omega$
Question 4
1 / -0

Current in circuit, which makes bulb glow or heats up the wire, is:

A
motion of charges

B
motion of atoms

C
motion of molecules

D
none of the above

Solution

An electric current is a stream of charged particles, such as electrons or ions, moving through an electrical conductor .It is often defined as rate of flow of charges.
Hence motion of charges through a conductor causes electric current.
Question 5
1 / -0

Which of the following best describe the current?

A
It is the rate of flow of charged particles through a conducting medium.

B
It's only present when there is no voltage.

C
It is the flow of a conductor through charged particles.

D
It is faster than the speed of light.

E
It is the random movement of electrons from atom to atom.

Solution

Current is defined as the rate of flow of charged particles (mainly electrons) in a particular direction through a conducting medium and is measured in Amperes.
$\therefore$ Current, $I = \dfrac{q}{t}$
Therefore, the correct option is A.
Question 6
1 / -0

The amount of charges that pass any section of the conductors in one second is called:

A
Current

B
Power dissipation

C
Electromotive force

D
Internal resistance

Solution

Current is defined as the amount of charges that passes any section of the conductors in one second.
$i = \dfrac{Q}{t}$
$Q$ ; the amount of charges passing through cross section area
Question 7
1 / -0

Identify the changes in a circuit on adding a light bulb to a series circuit. It will:

A
increase the total resistance

B
make the current through each light bulb different

C
decrease the total resistance

D
change the voltage of the battery

E
make the voltage lost in each light bulb equal

Solution

For series combination of two resistances $R_{eq} = R_1 + R_2$
Light bulb has its own resistance and thus the total resistance of the circuit increases when it is connected in series with the circuit.
Question 8
1 / -0

When does two elements are said to be in series?

A
When same current physically flows through both elements

B
When different current physically flows through both elements

C
When no current physically flows through both elements

D
None of these

Solution

When two resistors are connected end to end are said to be connected in series. In a series connection, the same current physically flows through both elements.
Question 9
1 / -0

What is the equivalent resistance of three $1000 \ \Omega$ resistors in series?

A
$1000 \ \Omega$

B
$3000 \ \Omega$

C
$2000 \ \Omega$

D
$1100 \ \Omega$

Solution

Given : $3$ resistors of resistance $R = 1000 \ \Omega$

Equivalent resistance in series combinationis given by,
$R_{eq} = R_1+R_2+R_3$

$R_{eq} = R+R+R = 3R = 3000 \ \Omega$
Question 10
1 / -0

What is the power produced by an appliance marked "240 V, 2 A"?

A
8.3 mW

B
60 W

C
120 W

D
480 W

Solution

$V=240 , I=2A$
Power, $P=V\times I=240\times 2=480W$

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Re-Attempt Weekly Quiz Competition

Electricity Test - 29

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Electricity Test - 29

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Question 1

2700 Ω2700 \ \Omega2700 Ω

3000 Ω3000 \ \Omega3000 Ω

3700 Ω3700 \ \Omega3700 Ω

1000 Ω1000 \ \Omega1000 Ω

Solution

Question 2

4.44.44.4

9.69.69.6

14.814.814.8

202020

25.225.225.2

Solution

Question 3

3 kΩ,5 kΩ,2 kΩ3\ k\Omega,5\ k\Omega,2\ k\Omega3 kΩ,5 kΩ,2 kΩ resistors in series

3 kΩ,5 kΩ,2 kΩ3\ k\Omega,5\ k\Omega,2\ k\Omega3 kΩ,5 kΩ,2 kΩ Resistors in parallel

3 kΩ,5 kΩ3\ k\Omega,5\ k\Omega3 kΩ,5 kΩ in series and combination in parallel to 2 kΩ2\ k\Omega2 kΩ resistor in series

5 kΩ,5 kΩ,2 kΩ,5\ k\Omega,5\ k\Omega,2\ k\Omega,5 kΩ,5 kΩ,2 kΩ, Resistors in series

Solution

Question 4

motion of charges

motion of atoms

motion of molecules

none of the above

Solution

Question 5

It is the rate of flow of charged particles through a conducting medium.

It's only present when there is no voltage.

It is the flow of a conductor through charged particles.

It is faster than the speed of light.

It is the random movement of electrons from atom to atom.

Solution

Question 6

Current

Power dissipation

Electromotive force

Internal resistance

Solution

Question 7

increase the total resistance

make the current through each light bulb different

decrease the total resistance

change the voltage of the battery

make the voltage lost in each light bulb equal

Solution

Question 8

When same current physically flows through both elements

When different current physically flows through both elements

When no current physically flows through both elements

None of these

Solution

Question 9

1000 Ω1000 \ \Omega1000 Ω

3000 Ω3000 \ \Omega3000 Ω

2000 Ω2000 \ \Omega2000 Ω

1100 Ω1100 \ \Omega1100 Ω

Solution

Question 10

8.3 mW

60 W

120 W

480 W

Solution

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$2700 \ \Omega$

$3000 \ \Omega$

$3700 \ \Omega$

$1000 \ \Omega$

$4.4$

$9.6$

$14.8$

$20$

$25.2$

$3\ k\Omega,5\ k\Omega,2\ k\Omega$ resistors in series

$3\ k\Omega,5\ k\Omega,2\ k\Omega$ Resistors in parallel

$3\ k\Omega,5\ k\Omega$ in series and combination in parallel to $2\ k\Omega$ resistor in series

$5\ k\Omega,5\ k\Omega,2\ k\Omega,$ Resistors in series

$1000 \ \Omega$

$3000 \ \Omega$

$2000 \ \Omega$

$1100 \ \Omega$