Electricity Test

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Electricity Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

In how many parts (equal) a wire of $$100\Omega$$ be cut so that a resistance of $$1\Omega$$ is obtained by connecting them in parallel?

A
$$10$$

B
$$5$$

C
$$100$$

D
$$50$$

Solution

The resistance of the conductor is proportional to length of the conductor.
That is, $$R=\rho \dfrac { l }{ A } $$.
The resistance of one wire of length l, R1 = 100 ohms.
If we cut it in a half: $$\dfrac { { R }_{ 1 } }{ l } =\dfrac { { R }_{ 2 } }{ \dfrac { l }{ 2 } } \quad \rightarrow \quad { R }_{ 2 }=\dfrac { { R }_{ 1 } }{ 2 } $$.
If we cut it in n parts the resistance will be $$R=\dfrac { { R }_{ 1 } }{ n } $$.
Because we shorten the length n times so it became $$\dfrac { l }{ n } $$.
When we connect these parts in parallel the output resistance must be 1 ohm.
So, $$1=\dfrac { 1 }{ R } +\dfrac { 1 }{ R } +...=n\dfrac { 1 }{ R } =n\dfrac { 1 }{ \dfrac { { R }_{ 1 } }{ n } } =\dfrac { { n }^{ 2 } }{ { R }_{ 1 } } =\dfrac { { n }^{ 2 } }{ 100 } \rightarrow n=10$$.
Hence, a resistance of 1 ohm is obtained by connecting 10 equal parts of the wire and connecting them in parallel.
Question 2
1 / -0

You are given two fuse wires $$A$$ and $$B$$ with current rating $$2.5 A$$ and $$6 A$$ respectively. Which of the two wires would you select for to use with a $$1100 W$$, $$220 V$$ room heater?

A
$$A$$

B
$$B$$

C
$$A$$ and $$B$$

D
None of these

Solution

Answer is B.
A fuse is a piece of wire of a material with a very low melting point. When a high current flows through the circuit due to overloading or a short circuit, the wires get heated and melts. As a result, the circuit is broken and the current stops flowing.
The fuse must always be connected to the mains and it must be of the correct value. For example, a 15-ampere fuse should trip when the current through it exceeds 15 amperes. A 20-ampere fuse should blow when the current through it exceeds 20 amps.
In this case, the electrical appliance is of 1100 W power and 220 V. So the current that passes through the appliance is calculated from $$P=VI$$. That is, $$I=P/V$$.

So, $$I = 1100\ W/220\ V = 5\ A$$.

Hence, wire B with a current rating of 6 A should be used for the fuse. so B is the answer
Question 3
1 / -0

Which one of the following will not conduct electricity?

A
solid NaCl

B
$$CuSO_4$$ solution

C
graphite

D
acidified water

Solution

A substance can only conduct electricity if it contains charged particles (electrons or ions) that are free to move around. In solid sodium chloride, there are ions but these ions are locked into the ionic lattice and are unable to move.
NaCl (common salt) is solid in state and solid ions or compounds don't conduct electricity. It needs to be either melted, molten or dissolved in a solution (i.e. water) first. Only then will the electrons be free to move to either the Cathode (-ve) or to the Anode (+ve). In a liquid state the electrons are free enough to move.
Question 4
1 / -0

The diagram shows two resistors connected in parallel across a voltage source. Which statement is true for this circuit?

A
The combined resistance is equal to the sum of the resistance

B
The current at every point in the circuit is the same

C
The current from the source is equal to the sum of the currents in X and in Y

D
The sum of the potential difference across X and across Y is equal to the potential difference across the voltage source

Solution

Both $$X$$ and $$Y$$ are in parallel so the total current delivered by source(battery) is divided into $$X$$ and $$Y$$. So total current from the source is equal to the sum of the currents in $$X$$ and in $$Y$$.
Question 5
1 / -0

The chemical reaction due to passage of electric current depends on:

A
electrodes.

B
magnitude of current.

C
density of liquid.

D
all of the above.

Solution

The chemical reaction due to passage of electric current depends on electrodes.
For example, electrolysis of aqueous copper sulphate solution using platinum electrodes will give oxygen gas at anode.
On the other hand, electrolysis of aqueous copper sulphate solution using copper electrodes will oxidize Cu to Cu(II) ions at anode.

Chemical reaction does not depend on magnitude of current and density of liquid .
Question 6
1 / -0

In a cell, by convention, the current is taken to be flowing from :

A
positive electrode to the negative electrode

B
negative electrode to the positive electrode

C
both A and B

D
none of the above

Solution

According to the convention, the direction of the current is opposite to the flow of negative charges (electrons). We take the direction of the flow of positive charges as the direction of the current. Negative charges flow from the negative electrode of a cell to the positive electrode, and hence we can say positive charges flow from the positive electrode to the negative electrode. Thus, by convention, the electric current flows from the positive electrode to the negative electrode.
Question 7
1 / -0

If $$R_1$$ and $$R_2$$ are the filament resistances of a 200 W bulb and a 100 W bulb respectively designed to operate on the same voltage, then

A
$$R_1 \,=\, 2R_2$$

B
$$R_2 \,=\, 2R_1$$

C
$$R_2 \,=\, 4R_1$$

D
$$R_1 \,=\, 4R_2$$

Solution

Given that,
Power of bulbs $$P_1=200\ W\\P_2=100\ W$$

The power $$P = VI$$, and $$I=V/R$$

On substituting we get, $$P=\dfrac { { V }^{ 2 } }{ R } $$.

Here, the power $$P_1$$ used in case of $$R_1$$ is $$P_1=\dfrac { { V }^{ 2 } }{ R_1 } =200W$$..........(1)

The power $$P_2$$ used in case of $$R_2$$ is $$P_2=\dfrac { { V }^{ 2 } }{ R_2 } =100W$$.........(2)

On taking ration,
$$\dfrac { P_1 }{ P_2 } =\dfrac { R_2 }{ R_1 } =\dfrac { 200}{ 100 } $$.

Hence, $$R_2 = 2R_1$$.
Option B
Question 8
1 / -0

In how many parts (equal) a wire of 100$$\Omega$$ be cut so that a resistance of 1$$\Omega$$ is obtained by connecting them in parallel ?

A
10

B
5

C
100

D
50

Solution

The resistance of the conductor is proportional to length of the conductor.
That is, $$R=\rho \dfrac { l }{ A } $$.
The resistance of one wire of length l, R1 = 100 ohms.

If we cut it in a half: $$\dfrac { { R }_{ 1 } }{ l } =\dfrac { { R }_{ 2 } }{ \dfrac { l }{ 2 } } \quad \rightarrow \quad { R }_{ 2 }=\dfrac { { R }_{ 1 } }{ 2 } $$.

If we cut it in n parts the resistance will be $$R=\dfrac { { R }_{ 1 } }{ n } $$.
Because we shorten the length n times so it became $$\dfrac { l }{ n } $$.

When we connect these parts in parallel the output resistance must be 1 ohm.
So, $$\dfrac{1}{R_{eq}}=\dfrac11=\dfrac { 1 }{ R } +\dfrac { 1 }{ R } +...=n\dfrac { 1 }{ R } =n\dfrac { 1 }{ \dfrac { { R }_{ 1 } }{ n } } =\dfrac { { n }^{ 2 } }{ { R }_{ 1 } } =\dfrac { { n }^{ 2 } }{ 100 }$$

$$ \rightarrow n=10$$.
Hence, a resistance of 1 ohm is obtained by connecting 10 equal parts of the wire and connecting them in parallel.
Question 9
1 / -0

A car headlamp of 48 W works on the car battery of 12 V. The correct fuse for the circuit is

A
5A

B
2A

C
3A

D
15A

Solution

Answer is A.

Fuses are safety devices that are to be built into our electrical system.
In this case, the car headlamp is $$48\ W$$ and the voltage is $$12\ V$$.
We know that power $$P = V\times I$$
$$\therefore I =\dfrac{P}{V} = \dfrac{48}{12} = 4\ A$$
This is the maximum current that should pass through the device. If the current is more than this then the fuse must disconnect the circuit. Hence, the fuse of 5 A should be connected to the circuit.
Question 10
1 / -0

Three identical bulbs are connected in parallel with a battery. The current drawn from the battery is 6 A. If one of the bulbs gets fused, what will be the total current drawn from the battery ?

A
6A

B
2A

C
4A

D
5A

Solution

Answer is C.

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 this case, 3 identical bulbs are connected in parallel. Therefore, the voltage remains the same across each bulb but the current is equally shared by all the three bulbs, that is, each bulb needs 2 A of current out of total 6 A as given in the question.
When one bulb, is fused out, only bulbs burn and only 4 A current is used.
Hence, the total current drawn from the battery when one bulb fuses is 4 A.

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Re-Attempt Weekly Quiz Competition

Electricity Test - 44

Result

Electricity Test - 44

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SHARING IS CARING

Question 1

$$10$$

$$5$$

$$100$$

$$50$$

Solution

Question 2

$$A$$

$$B$$

$$A$$ and $$B$$

None of these

Solution

Question 3

solid NaCl

$$CuSO_4$$ solution

graphite

acidified water

Solution

Question 4

The combined resistance is equal to the sum of the resistance

The current at every point in the circuit is the same

The current from the source is equal to the sum of the currents in X and in Y

The sum of the potential difference across X and across Y is equal to the potential difference across the voltage source

Solution

Question 5

electrodes.

magnitude of current.

density of liquid.

all of the above.

Solution

Question 6

positive electrode to the negative electrode

negative electrode to the positive electrode

both A and B

none of the above

Solution

Question 7

$$R_1 \,=\, 2R_2$$

$$R_2 \,=\, 2R_1$$

$$R_2 \,=\, 4R_1$$

$$R_1 \,=\, 4R_2$$

Solution

Question 8

10

5

100

50

Solution

Question 9

5A

2A

3A

15A

Solution

Question 10

6A

2A

4A

5A

Solution

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