Electricity and Circuits Test

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Electricity and Circuits Test - 13

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Weekly Quiz Competition

Question 1
1 / -0

Look at the given figure. It consists of a cell, a bulb with the two terminals X and Y and wires with ends P and Q; and S and R. The direction of current will be

A
PQRS

B
SRQP

C
PRQS

D
SQRP

Solution

The direction of current will be PQRS because the current flows from positive terminal of the cell to the negative terminal.
Question 2
1 / -0

Why the filament of bulb glows ?

A
Excess of electricity passing through bulb filament, hence it glows

B
Current passes through the filament, it gets heated up and become red hot, hence glows

C
Current passing through filament cools it down, hence it glows

D
None of these

Solution

A filament has a lot of resistance to electricity. Therefore as a result of this, the filament heats up and become red hot and starts glowing, converting electrical energy to light energy. This is because of the Joule-effect, which means that resistances heat up when the electrical current runs through them.
Question 3
1 / -0

The bulb fuses because of _________ of ________ wire.

A
Melting, filament

B
Boiling, filament

C
Melting, circuit

D
Boiling, circuit

Solution

The bulb produces light by a filament wire that gets heated by electric current and fuse when that filament wire melts due to overheating. So the bulb fuses because of melting and filament wire.
Question 4
1 / -0

In following circuit the bulb will not glow if ends A and B are connected with ?

A
Steel clip

B
metal clip

C
plastic clip

D
copper wire

Solution

The bulb will not glow if ends A and B are connected with plastic clip because plastic is a poor conductor of electricity and it does not allow electricity to pass through it. Whereas, Steel clip, Metal clip as well as Copper wire conducts electricity through them
Question 5
1 / -0

Identify the closed circuit

A

B

C

D
Solution
Closed circuit is a complete electrical connection around which current flow. Current flow from positive to negative terminal of the battery. As option D is a closed circuit and bulb is glowing, it is the correct option.
In option A, the insulated handle of the screwdriver won't let the current flow.
In option B, the two cells are not connected properly.
In option C, the switch is open so current won't flow.
Question 6
1 / -0

Seema went to Usha's house to complete the project work given in school. Both were working on the project and suddenly the light bulb fuses. What could be the possible reason of fuse of bulb?

A
Electricity Failure

B
Breaking of filament of bulb due to overloading

C
Breaking of fuse wire

D
None of these

Solution

An electric bulb may fuse due to many reasons. One reason for a bulb to fuse is a break in its filament. A break in the filament of an electric bulb means a break in the path of the current between the terminals of the electric cell. When excess electricity or too much current passes through the circuit, it causes overheating and because of overheating the filament wire of bulb melts and then bulb gets fuse.
This is the reason why suddenly the light bulb fuses while Seema went to Usha's house to complete the project work given in the school.
Question 7
1 / -0

Which of the following circuits will the bulb glow?

A

B

C

D

Solution

In figure (A), both ends of the bulb are connected to the negative terminal of the cell. So, bulb will not glow while in figure (B) and (C), insulating objects like string and rubber band are used. So, the circuit is incomplete and bulb will not glow. Figure (D) represents the complete circuit with proper connection. Hence, the bulb glows in figure (D).
Question 8
1 / -0

Observe the given figure and answer the following questions. The bulb will glow when a ______ is placed in between the probes.

A
Crayon

B
Comb

C
Key

D
Piece of stone

Solution

Materials that allow electric current to pass through them are called conductors. Key is a metallic object. When it is placed in between probes then the current can easily flow in the circuit.
Question 9
1 / -0

To flow current in the circuit, one need a

A
Closed Circuit

B
Source of potential difference

C
Both A and B

D
Neither A nor B

Solution

To flow current through any circuit, Circuit need to be close and source of potential difference ie battery.
Hence option C.
Question 10
1 / -0

A $$100$$W bulb $$B_1$$ and two $$60$$W bulbs $$B_2$$ and $$B_3$$, are connected to a $$250$$V source, as shown in figure. Now $$W_1$$, $$W_2$$ and $$W_3$$ are the output powers of the bulbs $$B_1, B_2$$ and $$B_3$$, respectively. Then.

A
$$W_1 > W_2 = W_3$$

B
$$W_1 > W_2 > W_3$$

C
$$W_1 < W_2 = W_3$$

D
$$W_1 < W_2 < W_3$$

Solution

$$\therefore R_1 = \dfrac{V_1^2}{100}, R_2 = \dfrac{V_2^2}{60}, R_3=\dfrac{V^2}{60}$$,

$$W_1 = \dfrac{V_1^2}{R_1} = \dfrac{V^2R_1}{(R_1 + R_2)^2}$$,

$$W_2 = \dfrac{V_2^2}{R^2} = \dfrac{V^2R_2}{(R_1 + R_2)^2}$$

and $$W_3 = \dfrac{V^2}{R_3}$$

$$W_3 : W_2 : W_1 = \dfrac{(250)^2}{R_3} : \dfrac{(250)^2R_2}{(R_1 + R_2)^2} : \dfrac{(250)^2}{(R_1 + R_2)^2}R_1$$

or $$W_3 : W_2 : W_1$$.

$$= \dfrac{(250)^2}{V^2} \times 60:\dfrac{(250)^2}{\left[\dfrac{1}{100}+\dfrac{1}{60}\right]^2V^4} \times \dfrac{v^2}{60} : \dfrac{(250)^2V^2}{\left[\dfrac{1}{100} + \dfrac{1}{60}\right]^2V^4 \times 100}$$

or $$W_3 : W_2 : W_1$$

$$= 60 : \dfrac{100\times 100\times 60 \times 60}{160\times 160\times 60} : \dfrac{100\times 100\times 60 \times 60}{160\times 160\times 100}$$

$$= 64: 25: 15$$

Hence, $$W_1<W_2<W_3$$
Hence, the correct option is $$(D)$$