Electricity Test

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Electricity Test - 52

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Weekly Quiz Competition

Question 1
1 / -0

The equivalent resistance of parallel combination is:

A
Lower than the value of lowest resistance in the combination

B
Higher than the value of lowest resistance in the combination

C
Both A and B

D
None of these

Solution

Consider two resistors having resistances $$R_1$$ and $$R_2$$.
The equivalent resistance of $$R_1$$ and $$R_2$$ in parallel is,
$$\dfrac{1}{R_{eq}}= \dfrac{1}{R_1}+\dfrac{1}{R_2}$$

$$\implies R_{eq}=\dfrac{R_1R_2}{R_1+R_2}$$
For any values of $$R_1$$ and $$R_2$$, $$R_{eq}$$ will be less than $$R_1$$ and $$R_2$$.
For example,
Let $$R_1=2\ \Omega$$
$$R_2=6\ \Omega$$
Then,
Equivalent resistance of $$R_1$$ and $$R_2$$ in parallel,
$$ R_{eq}=\dfrac{R_1R_2}{R_1+R_2}$$

$$ R_{eq}=\dfrac{2\times 6}{2+8}$$

$$ R_{eq}=\dfrac{12}{8}$$

$$ R_{eq}=1.5\ \Omega$$
$$\implies R_{eq} \lt R_1\ \text{and}\ R_{eq} \lt R_2$$
Question 2
1 / -0

Smallest commercial unit of energy is

A
Kilowatt hour

B
Watt second

C
Watt hour

D
Watt minutes

Solution

Smallest commercial unit of energy is called Watt hour $$(Wh)$$. One watt hour is the energy consumed when $$1 \ watt$$ of power is used for $$1 \ hour.$$
Question 3
1 / -0

Convert $$28.8 kJ$$ into kilo watt hour.

A
$$8 \times {10}^{-3} kwh$$

B
$$8 kwh$$

C
$$8 \times {10}^{3} kwh$$

D
None of these

Solution

We know that $$1 kWh=3.6 \times 10^6$$ $$Joules$$
$$E=28.8 \times 10^3 $$ $$ Joules$$

$$E = \dfrac{28.8 \times {10}^{3}}{3.6 \times {10}^{6}}$$
$$= 8 \times {10}^{-3} kwh$$
Question 4
1 / -0

If n resistance (each R) are connected in parallel then the resultant will be

A
$$nR$$

B
$$R/n$$

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

D
$$\text{None}$$

Solution

Equivalent resistance of parallel combination

$$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+.........$$

Given, Resistanvce of all resistors $$=R$$
$$\therefore$$ $$\dfrac{1}{R_{eq}} = \dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}+.........n$$ terms

OR $$\dfrac{1}{R_{eq}} = \dfrac{n}{R}$$

$$\implies$$ $$R_{eq} = \dfrac{R}{n}$$
Question 5
1 / -0

A certain household has consumed $$250$$ units of energy during a month. How much energy is this in joules?

A
$$9 \times 10^{10}\ J$$

B
$$9 \times 10^{8}\ J$$

C
$$9 \times 10^{9}\ J$$

D
$$9 \times 10^{-8}\ J$$

Solution

Hint: Conversion of KWh to joule.

Step 1: Commercial unit of energy is KWh.

   $$1\; KWh = 10^3\; W = 10^3\; JS^{-1}$$

  $$1\;hr = 60\times60 = 3600\;sec$$

Therefore, $$1\;KWh = 1\; KW \times1\;hr$$
$$=10^3 \times 3600$$
$$=3.6 \times 10^6 \;J$$

Step 2: Calculation according to question

Energy consumption $$=250\;KWh$$
$$= 250\times3.6\times10^6\;J$$
$$=900\times10^6\;J$$
$$=9\times10^8\;J$$ or $$900\;MJ$$

Hence, Option (B) is correct.
Question 6
1 / -0

Calculate the monthly bill if a heater of 100 watt is used at the rate of rs. 1 per unit for 1 hour daily.

A
Rs. 200

B
Rs. 6000

C
Rs. 100

D
Rs. 3

Solution

Given : $$P=0.1$$ kW $$t = 1$$ hour per day
Energy consumed per day $$E = Pt = 0.1\times 1 = 0.1$$ kWh per day
Total energy consumed per month $$E_T = 30\times 0.1 = 3$$ kWh per month
Cost of electricity $$ =Re$$ $$ 1 $$ per unit i.e per kWh
$$\therefore$$ Monthly bill $$ = Re.$$ $$1\times 3 =$$Rs. $$3$$ per month
Question 7
1 / -0

Commercial unit of energy is .......................

A
Joule

B
Volt

C
Kilowatt hour

D
Watt second

Solution

Commercial unit of electrical energy is kilowatt hour $$\left(kWh\right)$$
$$1 kwh = 3.6 \times {10}^{6} J$$
Question 8
1 / -0

Whenever an electric current is passed through a conductor, it gets heated up. this means that a part of electrical energy given to the conductor gets converted to

A
potential energy

B
gravitational energy

C
heat energy

D
none of these

Solution

Whenever an electric current is passed through a conductor, it gets heated up due to the movement of electrons inside the conductor. This indicates that a part of electrical energy gets converted into heat energy.
Question 9
1 / -0

Imagine that you have a 100$$\Omega$$ resistor. You want to add a resistor in series with this 100$$\Omega$$ resistor in order to limit the current to 0.5A when $$110 $$ Volts is placed across two resistors in series. How much resistance should you use?

A
$$100\Omega$$

B
$$120 \Omega$$

C
$$250 \Omega$$

D
$$300 \Omega$$

Solution

Given : $$V = 110$$ volts $$I = 0.5$$ A $$R_1 = 100\Omega$$
Let a resistance $$R$$ is connected in series with $$R_1$$.
Using Ohm's law, $$V = I(R_1+R)$$
$$\therefore$$ $$110 = 0.5(100+R)$$ $$\implies R = 120\Omega$$
Question 10
1 / -0

The attached circuit diagram shows connection of 3 resistors. All are connected in parallel . Thus what is the equivalent resistance.

A
1 ohm

B
4 ohm

C
3ohm

D
2 ohm

Solution

Equivalent resistance of resistors connected in parallel $$\dfrac{1}{R_{eq}} =\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$$
$$\therefore$$ $$\dfrac{1}{R_{eq}} =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}$$ $$\implies R_{eq} = 1$$ ohm

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Electricity Test - 52

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Electricity Test - 52

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Question 1

Lower than the value of lowest resistance in the combination

Higher than the value of lowest resistance in the combination

Both A and B

None of these

Solution

Question 2

Kilowatt hour

Watt second

Watt hour

Watt minutes

Solution

Question 3

$$8 \times {10}^{-3} kwh$$

$$8 kwh$$

$$8 \times {10}^{3} kwh$$

None of these

Solution

Question 4

$$nR$$

$$R/n$$

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

$$\text{None}$$

Solution

Question 5

$$9 \times 10^{10}\ J$$

$$9 \times 10^{8}\ J$$

$$9 \times 10^{9}\ J$$

$$9 \times 10^{-8}\ J$$

Solution

Question 6

Rs. 200

Rs. 6000

Rs. 100

Rs. 3

Solution

Question 7

Joule

Volt

Kilowatt hour

Watt second

Solution

Question 8

potential energy

gravitational energy

heat energy

none of these

Solution

Question 9

$$100\Omega$$

$$120 \Omega$$

$$250 \Omega$$

$$300 \Omega$$

Solution

Question 10

1 ohm

4 ohm

3ohm

2 ohm

Solution

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