Electricity Test

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Electricity Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

A wire has resistance of $$\displaystyle 12\Omega $$. It is cut into two parts and both halve are connected in parallel. The new resistance is

A
$$\displaystyle 3\Omega $$

B
$$\displaystyle 1.5\Omega $$

C
$$\displaystyle 12\Omega $$

D
$$\displaystyle 6\Omega $$

Solution

As the wire is bent at its mid point, thus resistance of each segment $$R = \dfrac{12}{2} = 6\Omega$$
Both are in parallel
$$\therefore$$ Effective resistance between A and B $$R' = \dfrac{R_1\times R_2}{R_1+R_2} = \dfrac{6\times 6}{6+6} = 3\Omega$$
Question 2
1 / -0

A lamp rated $$20W$$ and an electric iron rated $$50W$$ are used for $$2$$ hour everyday. Calculate the total energy consumed in $$20 $$ days.

A
$$14kwh$$

B
$$2.8kwh$$

C
$$40kwh$$

D
None of these

Solution

Energy consumed by lamp in 2 hour $$E_l = 0.02\times 2 = 0.04$$ kWh per day
Energy consumed by iron in 2 hour $$E_i = 0.05\times 2 = 0.1$$ kWh per day
$$\therefore$$ Total energy consumed by both appliance in $$20$$ days,
$$E_T = (E_l+E_i)\times 20 = (0.04+0.1)\times 20 = 2.8$$ kWh
Question 3
1 / -0

Which of these units of energy is the largest?

A
$$1\ Calorie$$

B
$$1\ Joule$$

C
$$1\ Erg$$

D
$$1\ Kilowatt-hour$$

Solution

Energy is the ability to do work.
$$1\ {calorie} = 4.186\ J$$
$$1\ erg= 1\times 10^{-7}\ J$$
$$1\ kWh=3.6\times 10^{6}$$ $$J$$
$$1\ kilowatt-hour$$ is the largest unit of energy.
Question 4
1 / -0

If $$\displaystyle { R }_{ 1 }$$ and $$\displaystyle { R }_{ 2 }$$ are respectively the filament resistance of a $$200 \ W$$ bulb and $$100 \ W$$ bulb designed to operate on the same voltage then

A
$$\displaystyle { R }_{ 1 }$$ is two times $$\displaystyle { R }_{ 2 }$$

B
$$\displaystyle { R }_{ 2 }$$ is two times $$\displaystyle { R }_{ 1 }$$

C
$$\displaystyle { R }_{ 2 }$$ is four times $$\displaystyle { R }_{ 1 }$$

D
$$\displaystyle { R }_{ 1 }$$ is four times $$\displaystyle { R }_{ 2 }$$

Solution

Given :
$$P_1 = 200$$ W
$$P_2 = 100$$ W

Power, voltage and resistance are related as
$$P=\dfrac{V^2}{R} \Rightarrow R=\dfrac{V^2}{P}$$

For constant voltage,
$$\implies$$ $$R\propto \dfrac{1}{P}$$

$$\therefore$$ $$\dfrac{R_1}{R_2} = \dfrac{P_2}{P_1} = \dfrac{100}{200}$$ $$\implies R_2 = 2R_1$$
Question 5
1 / -0

Energy consumed in our house has a unit of

A
watts

B
horse power

C
kilowatt-hours

D
none
Question 6
1 / -0

Find the effective resistance when $$\displaystyle 1\Omega ,2\Omega $$ and $$\displaystyle 3\Omega $$ resistances are connected in series

A
$$\displaystyle 2\Omega $$

B
$$\displaystyle 1\Omega $$

C
$$\displaystyle 6\Omega $$

D
$$\displaystyle 5\Omega $$

Solution

Effective resistance of resistors connected in series $$R_{s} = R_1+R_2+R_3$$
$$\implies$$ $$R_s = 1+2+3 = 6\Omega$$
Question 7
1 / -0

For three resistors connected in series.

A
$$\displaystyle R_{eff}={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$$

B
$$\displaystyle V={ V }_{ 1 }={V }_{ 2 }={ V}_{ 3 }$$

C
$$\displaystyle \dfrac {1}{R_{eff}}=\dfrac {1}{{ R }_{ 1 }}+\dfrac {1}{ {R }_{ 2 }}+\dfrac {1}{ {R }_{ 3 }}$$

D
None of these

Solution

Same current flows through the resistors connected in series but potential difference is different across individual resistors and sum of individual potential difference is equal to the total potential difference.
$$\therefore$$ $$V = V_1+V_2+V_3$$
OR $$IR_{eff} = IR_1+IR_2+IR_3$$
$$\implies$$ $$R_{eff} = R_1+R_2+R_3$$
Question 8
1 / -0

A wire of resistance $$12\ ohm$$ per meter is bent to form a complete circle of radius $$10\ cm$$. The resistance b/w its two diametrically opposite points $$A$$ and $$B$$ as shown in the figure is

A
$$\displaystyle 6\pi \Omega $$

B
$$\displaystyle 3\Omega $$

C
$$\displaystyle 0.6\pi \Omega $$

D
$$\displaystyle 6\Omega $$

Solution

Length of semi-circle $$L = \pi r = \pi (0.1)$$
$$\therefore$$ Resistance of each semi-circle $$R = 12\times 0.1\pi = 1.2\pi \Omega$$
Equivalent resistance between point A and B $$R_{eq} = 1.2\pi\parallel 1.2\pi = \dfrac{1.2\pi}{2} = 0.6\pi \Omega$$
Question 9
1 / -0

Watt hour represents:

A
Electric energy

B
Current

C
Voltage

D
Power

Solution

The watt-hour (Wh) is a unit of electric energy equivalent to one watt (1 W) of power expended for one hour (1 h) of time. Watt-hour is commonly used in electrical applications.
$$1\ Wh = 1\times 60\times 60 = 3600\ J$$
Question 10
1 / -0

Unit used in selling electrical energy to consumer is:

A
$$Volt-Ampere$$

B
$$Kilowatt-hour$$

C
$$Volt/second$$

D
None of these

Solution

The commercial unit of electrical energy is $$Kilowatt-hour\ (kWh)$$.
$$1\ kWh$$ is equal to $$1\ unit$$ of energy.

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Re-Attempt Weekly Quiz Competition

Electricity Test - 50

Result

Electricity Test - 50

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SHARING IS CARING

Question 1

$$\displaystyle 3\Omega $$

$$\displaystyle 1.5\Omega $$

$$\displaystyle 12\Omega $$

$$\displaystyle 6\Omega $$

Solution

Question 2

$$14kwh$$

$$2.8kwh$$

$$40kwh$$

None of these

Solution

Question 3

$$1\ Calorie$$

$$1\ Joule$$

$$1\ Erg$$

$$1\ Kilowatt-hour$$

Solution

Question 4

$$\displaystyle { R }_{ 1 }$$ is two times $$\displaystyle { R }_{ 2 }$$

$$\displaystyle { R }_{ 2 }$$ is two times $$\displaystyle { R }_{ 1 }$$

$$\displaystyle { R }_{ 2 }$$ is four times $$\displaystyle { R }_{ 1 }$$

$$\displaystyle { R }_{ 1 }$$ is four times $$\displaystyle { R }_{ 2 }$$

Solution

Question 5

watts

horse power

kilowatt-hours

none

Question 6

$$\displaystyle 2\Omega $$

$$\displaystyle 1\Omega $$

$$\displaystyle 6\Omega $$

$$\displaystyle 5\Omega $$

Solution

Question 7

$$\displaystyle R_{eff}={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$$

$$\displaystyle V={ V }_{ 1 }={V }_{ 2 }={ V}_{ 3 }$$

$$\displaystyle \dfrac {1}{R_{eff}}=\dfrac {1}{{ R }_{ 1 }}+\dfrac {1}{ {R }_{ 2 }}+\dfrac {1}{ {R }_{ 3 }}$$

None of these

Solution

Question 8

$$\displaystyle 6\pi \Omega $$

$$\displaystyle 3\Omega $$

$$\displaystyle 0.6\pi \Omega $$

$$\displaystyle 6\Omega $$

Solution

Question 9

Electric energy

Current

Voltage

Power

Solution

Question 10

$$Volt-Ampere$$

$$Kilowatt-hour$$

$$Volt/second$$

None of these

Solution

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