Electricity Test

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Electricity Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Current in a circuit flows:

A
in direction from high potential to low potential

B
in direction from low potential to high potential

C
in direction of flow of electrons

D
in any direction.

Solution

Electric current flows from positive electrode of cell to negative electrode of cell, i.e., from high potential to low potential.
Question 2
1 / -0

To increase the effective resistance in a circuit the resistors are connected in ______.

A
series

B
parallel

C
both ways

D
none of these

Solution
Question 3
1 / -0

The current which is assumed to be flowing in a circuit from positive terminal to negative, is called:

A
direct current

B
pulsating current

C
conventional current

D
alternating current

Solution

The current flowing from positive terminal to negative terminal is due to flow of positive charge, which is the conventional current.
Question 4
1 / -0

A dynamo develops $0.5 A$ at $6 V$ . The energy which is generated in one second is

A
$0.083 J$

B
$3 J$

C
$12 J$

D
none of these

Solution

Given:
Voltage(V) = $6V$
Current(I) = $0.5A$
Time(t) = $1sec$
Energy generated = $VIt$
$\Rightarrow$ Energy $= 6\times 0.5 \times 1=3 J$
Question 5
1 / -0

The device used for producing current is called a _____.

A
Voltmeter

B
Ammeter

C
Galvanometer

D
Generator

Solution

In electricity generation A generator is a device which uses mechanical energy to produce electrical energy which flows in form of electric current in circuits. hence Generator can produce current.
Question 6
1 / -0

The insulator of electricity is :

A
copper

B
silk

C
human body

D
acidulated water

Solution

Silk does not allow electric current to pass through it.
Question 7
1 / -0

An electric lamp is marked $60\ W,\ 220\ V$ . The cost of kilo watt hour of electricity is $Rs.\ 1.25$ . The cost of using thing lamp on $220\ V$ for $8\ hrs$ is:

A
$Rs.\ 0.25$

B
$Rs.\ 0.60$

C
$Rs.\ 1.2$

D
$Rs.\ 4.00$

Solution

Given:
Power, $P=60\ W$
Time, $t=8\ hrs$

Energy consumed per day, $E=Pt=60\times 8=480\ Whr=\dfrac{480}{1000}=0.48\ kW hr$
Hence total cost, $C=0.48\times 1.25=Rs. 0.60$
Question 8
1 / -0

1 kilowatt hour $=$ ___________ .

A
$10.6 \times 10^6$ Joule

B
$3.6 \times 10^6$ Joule

C
$30.6 \times 10^6$ Joule

D
$3.6 \times 10^5$ Joule

Solution

$1\ watt = 3600\ joule$
$1\ kW = 1000\ watt$
$1\ h=60\times 60 = 3600\ s$
$1\ kWh= 1\ kW\times 1\ hour=1000\ W\times 3600\ s = 1000 \times 3600 joule = 3.6 \times 10^{6}\ J$
Question 9
1 / -0

The unit of current is :

A
ampere

B
volt

C
ohm

D
coulomb

Solution

Current is measured in Coulomb/sec or Ampere.
Question 10
1 / -0

The smallest resistance that can be obtained by the combination of $n$ resistors, each of resistance $R$ is:

A
$n^2R$

B
$nR$

C
$\dfrac{R}{n^2}$

D
$\dfrac{R}{n}$

Solution

For smallest resistance, the resistors are to be connected in parallel. The effective resistance $=$ R/n.

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Re-Attempt Weekly Quiz Competition

Electricity Test - 21

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Electricity Test - 21

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SHARING IS CARING

Question 1

in direction from high potential to low potential

in direction from low potential to high potential

in direction of flow of electrons

in any direction.

Solution

Question 2

series

parallel

both ways

none of these

Solution

Question 3

direct current

pulsating current

conventional current

alternating current

Solution

Question 4

0.083J0.083 J0.083J

3J3 J3J

12J12 J12J

none of these

Solution

Question 5

Voltmeter

Ammeter

Galvanometer

Generator

Solution

Question 6

copper

silk

human body

acidulated water

Solution

Question 7

Rs. 0.25Rs.\ 0.25Rs. 0.25

Rs. 0.60Rs.\ 0.60Rs. 0.60

Rs. 1.2Rs.\ 1.2Rs. 1.2

Rs. 4.00Rs.\ 4.00Rs. 4.00

Solution

Question 8

10.6×10610.6 \times 10^610.6×106Joule

3.6×1063.6 \times 10^63.6×106Joule

30.6×10630.6 \times 10^630.6×106Joule

3.6×1053.6 \times 10^53.6×105Joule

Solution

Question 9

ampere

volt

ohm

coulomb

Solution

Question 10

n2Rn^2Rn2R

nRnRnR

Rn2\dfrac{R}{n^2}n2R​

Rn\dfrac{R}{n}nR​

Solution

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$0.083 J$

$3 J$

$12 J$

$Rs.\ 0.25$

$Rs.\ 0.60$

$Rs.\ 1.2$

$Rs.\ 4.00$

$10.6 \times 10^6$ Joule

$3.6 \times 10^6$ Joule

$30.6 \times 10^6$ Joule

$3.6 \times 10^5$ Joule

$n^2R$

$nR$

$\dfrac{R}{n^2}$

$\dfrac{R}{n}$