Electricity Test

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Electricity Test - 43

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Question 1
1 / -0

If a bulb of 60 W is connected across a source of 220 V, the current drawn by it is:

A
0.3 A

B
0.27 A

C
3.67 A

D
0.5 A

Solution

The electric power in watts associated with a complete electric circuit or a circuit component represents the rate at which energy is converted from the electrical energy of the moving charges to some other form, e.g., heat, mechanical energy, or energy stored in electric fields or magnetic fields. The power is given by the product of applied voltage and the electric current. That is, P=VI. The power of the bulb is given as 60 W and the voltage is 220 V.
The current drawn in calculated as follows.
$${ I=\dfrac { P }{ V } =\dfrac { 60W }{ 220V } }=0.2727A$$
Hence, the current drawn by the bulb is 0.2727 A.
Question 2
1 / -0

Substances that do not allow free flow of charge through them are called __________.

A
Insulators

B
Conductors

C
Both A and B

D
None of these

Solution

Substances that do not allow free flow of charge (i.e. electron) through them are called insulator.
For example:- Plastic, wood
Question 3
1 / -0

The energy consumed by $$100 \ W$$ electric bulb in $$5 \ hours$$ is:

A
$$0.5 \ kWh $$

B
$$1.5 \ kWh $$

C
$$5 \ kWh $$

D
$$2.5 \ kWh $$

Solution

Given, power of the electric bulb,
$$P =100 \:W=\dfrac{100}{1000} \:kW=0.1 \:kW$$ ;

time for which bulb is used, $$t =5 \:h$$

As $$\displaystyle P =\frac{W}{t}$$ or $$W =Pt$$

or $$W =0.1 \:kW \times 5 \:h =0.5 \:kWh $$
Question 4
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Which of the following does not belong to the group formed by the others?

A
Aluminium

B
Salt solution

C
Ceramic articles

D
Silver

Solution

$$\rightarrow$$ Aluminium is a good conductor of electricity. It allows the flow of electricity.
$$\rightarrow$$ Salt solution is also a good conductor of electricity.
$$\rightarrow$$ Ceramic articles is an insulator.It does not allow electricity to pass.
$$\rightarrow$$ Silver is also good conductor of electricity
That's why ceramic articles are different as ceramic articles are insulator while others are conductors.
Question 5
1 / -0

Two wires of equal length, one of copper and the other of manganin (an alloy) have the same thickness. Which one can be used for electrical heating devices? Why?

A
Manganin because being an alloy it has more resistivity than copper

B
Copper because it has more conductivity.

C
Both because both are good conductors of heat.

D
None of them can be used as heating device.

Solution

Mangan can be used for electrical heating devices because it has very high resistance (about 25 times more than that of copper) and therefore, produces a lot of heat when current is passed through it.
Question 6
1 / -0

The given figure is a 'material tester'.
Which of the following objects will make the bulb glow when put on the gap shown?

A

B

C

D

Solution

The nail will make the bulb glow.
Because it is made up of metal, and we know that metals are good conductors of electricity. Other options are not good conductors of electricity.
So, only the nail will allow electricity to pass through it.
Question 7
1 / -0

The equivalent resistance for the '$$n$$' resistors with resistance $$R$$ connected in parallel is

A
$$nR$$

B
$$R$$

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

D
$$nR^2$$

Solution

Series circuit:
Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component. In a series circuit, every device must function for the circuit to be complete. One bulb burning out in a series circuit breaks the circuit. A circuit composed solely of components connected in series is known as a series circuit.The total resistance of resistors in series is equal to the sum of their individual resistances. That is, $${ R }_{ total }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$$.
Parallel circuit:
Components connected in parallel are connected so the same voltage is applied to each component. In a parallel circuit, the voltage across each of the components is the same, and the total current is the sum of the currents through each component. In parallel circuits, each light has its own circuit, so all but one light could be burned out, and the last one will still function. A circuit composed solely of components connected in parallel is known as a parallel circuit.To find the total resistance of all components, the reciprocals of the resistances of each component is added and the reciprocal of the sum is taken. Total resistance will always be less than the value of the smallest resistance That is, $$\dfrac { 1 }{ { R }_{ total } } =\dfrac { 1 }{ { R }_{ 1 } } +\dfrac { 1 }{ { R }_{ 2 } } +\dfrac { 1 }{ { R }_{ 3 } } \quad that\quad is,\quad R={ { R }_{ total } }^{ -1 }$$.
Question 8
1 / -0

If we touch a naked current carrying wire, we get a shock. This is because our body is a :

A
conductor of electricity

B
insulator of electricity

C
source of electricity

D
both B&C

Solution

We get a shock because electric current passes through our body because our body is a conductor of electricity.

Answer-(A)
Question 9
1 / -0

Three unequal resistors in parallel are equivalent to a resistant 1 ohm. If two of them are in the ratio 1 :2 and if no resistance value is fractional, the largest of the three resistance in ohm is :-

A
4

B
6

C
8

D
12

Solution

$$1={\dfrac {1}{R_1}}+{\dfrac {1}{R_2}}+{\dfrac {1}{R_3}}$$
$${\dfrac {R_1}{R_2}}={\dfrac {1}{2}}$$ (Given)
$$R_2=2R_1$$
$$1={\dfrac {2}{R_2}}+{\dfrac {1}{R_2}}+{\dfrac {1}{R_3}}$$
$$1={\dfrac {3}{R_2}}+{\dfrac {1}{R_3}}$$
$$1-{\dfrac {3}{R_2}}={\dfrac {1}{R_3}}$$
$${\dfrac {{R_2}-3}{R_2}}={\dfrac {1}{R_3}}$$
$${R_3}={\dfrac {R_2}{{R_2}-3}}$$
Now only $$R_2=6 \Omega$$ satisfies this above equation $$\Rightarrow R_3=2 \Omega$$ & $$R_1=3 \Omega$$
Question 10
1 / -0

Two 220 volt, 100 watt bulbs are connected first in series and then in parallel. Each time the combination is connected to a 220 volt a.c.supply line. The power drawn by the combination in each case respectively will be

A
50 watt, 100 watt

B
100 watt, 50 watt

C
200 watt, 150 watt

D
50 watt, 200 watt

Solution

Resistance of the 220V, 100W bulb is given by
$$R=\quad \frac { { V }^{ 2 } }{ P } =\frac { 220^{ 2 } }{ 100 } \\ \quad =484\Omega \quad $$
Thus, when connected in series, total resistance $$ R_{ s }=484+484\quad\Omega \quad\\=968\Omega \quad$$
Power drawn in case of series combination,
$${ P }_{ s }=\quad \frac { { V }^{ 2 } }{ R_{ s } } =\frac { 220^{ 2 } }{ 968 } \\ \quad =50W$$
When connected in parallel, total resistance $$ R_{ p }=\frac { 484 }{ 2 }\Omega \quad\\=242\Omega \quad$$
Power drawn in case of series combination,
$${ P }_{ p }=\quad \frac { { V }^{ 2 } }{ R_{ p } } =\frac { 220^{ 2 } }{ 242 } \\ \quad =200W$$