Electricity Test

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Electricity Test - 72

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Question 1
1 / -0

An electric lamp is marked $$60 \ W$$, $$230 \ V$$. The cost of $$1 \ kilowatt-hour$$ of power is $$Rs \ 1.25$$. The cost of using this lamp for $$8 \ hours$$ is

A
$$Rs. \ 1. 20$$

B
$$ Rs. \ 4. 00$$

C
$$Rs. \ 0. 25$$

D
$$Rs. \ 0. 60$$

Solution

Total energy consumed is given as $$E=P\times t$$
where $$P$$ is power and $$t$$ is time for which it is used.
Given, $$P=60\,W$$ and $$t=8\,h$$
$$E = 60 \times 8 \ Wh=\dfrac{60\times 8}{1000} = 0.48 \ kWh $$

Given that, cost of $$1 \ kWh$$ is $$Rs. \ 1.25$$

Hence, total cost is $$ \text{Cost}=0.48\times 1.25 = Rs. \ 0.6 $$
Question 2
1 / -0

The resistance of the filament of an electric bulb changes with temperature. If an electric bulb rated $$220 \ V$$ and $$100 \ W$$ is connected to $$(220\times 0.8) \ V$$ source, then the power would be :

A
$$100\times 0.8watt$$

B
$$100\times (0.8)^{2}watt$$

C
Between $$100\times 0.8watt$$ and 100 watt

D
Between $$100\times (0.8)^{2}watt$$ and $$100\times 0.8watt$$ watt

Solution

Since Power $$ P = \dfrac{V^2}{R}$$

For the first case, when $$V=220 \ V$$

$$P_{1}=\dfrac{(220)^2}{R_1} $$

For the second case, when $$V=220 \times 0.8 \ V$$

$$ P_{2}=\dfrac{(220\times 0.8)^{2}}{R_{2}}$$

Now, $$\dfrac{P_{2}}{P_{1}}=\dfrac{\left ( 220\times 0.8 \right )^{2}}{(220)^{2}}\times \dfrac{R_{1}}{R_{2}}$$

$$\Rightarrow \dfrac{P_{2}}{P_{1}}=(0.8)^{2}\times \dfrac{R_{1}}{R_{2}}$$

Since $$V_2<V_1$$, voltage has been decreased and is directly proportional to the resistance, from Ohm's law.
Hence, $$R_{2}< R_{1}$$

$$\Rightarrow \dfrac{R_1}{R_2} > 1$$
$$\Rightarrow \dfrac{P_2}{P_1} > (0.8)^2$$
$$\Rightarrow P_2 > 100 \times (0.8)^2 \ W$$

Also, Since Power $$P = Vi$$

Hence$$\dfrac{P_{2}}{P_{1}}=\dfrac{\left ( 220\times 0.8 \right )i_2}{220i_1},$$

Since $$V_2<V_1$$, voltage has been decreased and is directly proportional to the current, from Ohm's law.
Hence, $$i_{2}< i_{1}$$

$$\Rightarrow \dfrac{i_2}{i_1} <1$$

So $$\dfrac{P_{2}}{P_{1}}< 0.8\Rightarrow P_{2}< (100\times 0.8) \ W$$

Hence the actual power would be between $$100 \times (0.8)^{2} \ W$$ and $$100\times (0.8) \ W$$
Question 3
1 / -0

An electric bulb marked 40 W and 200 V, is used in a circuit of supply voltage 100 v, Now its power is

A
$$100 W$$

B
$$40 W$$

C
$$20 W$$

D
$$10 W$$

Solution

The power genrated is given by
$$P=\dfrac{V^{2}}{R}\Rightarrow \dfrac{P_{2}}{P_{1}}=\dfrac{V^{2}_{1}}{V^{2}_{1}}(\because R is constant )\Rightarrow \dfrac{P_{2}}{P_{1}}=\left ( \dfrac{100}{200} \right )^{2}=\dfrac{1}{4}\Rightarrow P_{2}=\frac{P_{1}}{4}=\dfrac{40}{4}=10W$$
Question 4
1 / -0

The current flowing through a lamp marked as $$50 \ W$$ and $$250 \ V$$ is

A
$$5 \ amp$$

B
$$2.5 \ amp$$

C
$$2 \ amp$$

D
$$0. 2 \ amp$$

Solution

The power dissipated is given by
$$P=Vi$$,
where $$V$$=voltage across lamp, and $$i$$=current flowing through lamp
Hence,

$$i = \dfrac{P}{V} = \dfrac{50}{250} = 0.2 \ amp$$
Question 5
1 / -0

The electric passing through a metallic wire produces heat because of

A
Collisions of conduction electrons with each other

B
Collisions of the atoms of the metal with each other

C
The energy released in the ionization of the atoms of the metal

D
Collisions of the conduction electrons with the atoms of the metallic wires

Solution

When an electric current passes through a metallic wire, electrons move inside the wire and they collide with each other during their motion which produces heat in the wire.
Question 6
1 / -0

A heater coil is cut into two equal parts and only one part is now use in the heater. The heat generated will now be

A
One fourth

B
Halved

C
Doubled

D
Four times

Solution

The heat generation is given by
$$Heat=Power\times time$$
$$Power=\dfrac{V^2}{R}$$
$$H=\dfrac{V^2 \times t}{R}$$
After cutting equally, length becomes half.
The resistance is directly proportional to $$l$$
It means, resistance becomes half.
$$H=\dfrac{V^2 \times t}{R/2}$$
$$H=2\ \dfrac{V^2 \times t}{R}$$

Therefore the heat generated will be doubled.
Question 7
1 / -0

The conductors of resistance $$1 \Omega, 2 \Omega$$ and $$3\Omega$$ are in series in a circuit. What is the equivalent resistance in the circuit ?

A
Less than $$1\Omega$$

B
Less than $$3\Omega$$

C
More than $$1 \Omega$$

D
More than $$3 \Omega$$

Solution

Given:
$$R_1=1 \Omega$$
$$R_2=2 \Omega$$
$$R_3=3 \Omega$$
All are in series
So $$R_{eq}=R_1+R_2+R_3=1+2+3=6 \Omega$$
Question 8
1 / -0

If a $$5 \,V$$ battery is provoding $$2 \,A$$ current in a conductor then what is the resistance of conductor ?

A
$$3$$ Ohm

B
$$2.5$$ Ohm

C
$$10$$ Ohm

D
$$2$$ Ohm

Solution

Given that:
Battery supplies a voltage of $$V=5 \ V$$
Current provided by the battery $$I=2 \ A$$
Resistance= ?
By Ohm's law,
$$V=IR$$
$$\Rightarrow R=\dfrac{V}{I}=\dfrac{5}{2}=2.5 \Omega$$
Question 9
1 / -0

Kilowatt hour is unit of

A
Energy

B
Power

C
Impulse

D
Force

Solution

The kilowatt-hour (symbolized kWh) is a unit of energy equivalent to one kilowatt (1 kW) of power expended for one hour. The kilowatt-hour is commercially used as a billing unit for energy delivered to consumers by electric utilities.
Question 10
1 / -0

What happens to the current in each resistor when resistors of different values in parallel combination are connected to a source of electricity?

A
Values of current and potential difference are different

B
Values of current and potential difference are same

C
Values of current are different but value of potential difference is same

D
Values of current are the same but values of potential difference are different

Solution

In a parallel combination, all the resistors are connected across the same voltage source. Thus, all of them will have the same potential difference.
But since the values of resistors are different, they will draw different current from the source.

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Electricity Test - 72

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Electricity Test - 72

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Question 1

$$Rs. \ 1. 20$$

$$ Rs. \ 4. 00$$

$$Rs. \ 0. 25$$

$$Rs. \ 0. 60$$

Solution

Question 2

$$100\times 0.8watt$$

$$100\times (0.8)^{2}watt$$

Between $$100\times 0.8watt$$ and 100 watt

Between $$100\times (0.8)^{2}watt$$ and $$100\times 0.8watt$$ watt

Solution

Question 3

$$100 W$$

$$40 W$$

$$20 W$$

$$10 W$$

Solution

Question 4

$$5 \ amp$$

$$2.5 \ amp$$

$$2 \ amp$$

$$0. 2 \ amp$$

Solution

Question 5

Collisions of conduction electrons with each other

Collisions of the atoms of the metal with each other

The energy released in the ionization of the atoms of the metal

Collisions of the conduction electrons with the atoms of the metallic wires

Solution

Question 6

One fourth

Halved

Doubled

Four times

Solution

After cutting equally, length becomes half.

It means, resistance becomes half.

Question 7

Less than $$1\Omega$$

Less than $$3\Omega$$

More than $$1 \Omega$$

More than $$3 \Omega$$

Solution

Question 8

$$3$$ Ohm

$$2.5$$ Ohm

$$10$$ Ohm

$$2$$ Ohm

Solution

Question 9

Energy

Power

Impulse

Force

Solution

Question 10

Values of current and potential difference are different

Values of current and potential difference are same

Values of current are different but value of potential difference is same

Values of current are the same but values of potential difference are different

Solution

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