Current Electricity Test

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Current Electricity Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

The resistance of germanium .................. with rise in temperature.

A
increases

B
decreases

C
remains the same

D
first increases then decreases

Solution

Germanium is a semiconductor. The resistivity of semiconductors, unlike that of resistors, decreases with increase in temperature.
Question 2
1 / -0

The specific resistance of a rod of copper as compared to that of thin wire of copper is

A
more

B
less

C
same

D
depends upon the length and area of wire

Solution

Specific resistance of a conductor depends on the nature of material but is independent of the dimension of the conductor. Thus specific resistance of rod of copper as compared to that of thin wire of copper is same.
Question 3
1 / -0

The length of a conductor is doubled and its radius is halved, its specific resistance is

A
halved

B
doubled

C
quadrupled

D
unchanged

Solution

Specific resistance ( or resistivity ) is an intrinsic property of a material i.e. it depends only on the nature of material but not on the dimensions of the conductor. Thus specific resistance remains the same even if the length of the conductor is doubled and its radius is halved.
Question 4
1 / -0

The maximum number of 40 W bulbs connected in parallel which can be lighted safely using a 240 V supply with a 5A fuse is.

A
30

B
45

C
8

D
44

Solution

Answer is A.

Fuses are safety devices that are to be built into our electrical system. If there were no fuses and we operated too many appliances on a single circuit, the cable carrying the power for that circuit would get extremely hot, short circuit, and possibly start a fire. To prevent electrical overloads, fuses are designed to trip or blow, stopping the flow of current to the overloaded cable.
The fuse must always be connected to the mains and it must be of 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 bulb is 40 W and the voltage is 240 V.
That is, P = VI, I = P/V = 40/240 = 0.16 A.
The current drawn from a single 40 W bulb is 0.16 A.
Therefore, with a fuse of 5 A, let the number of bulbs that can be used is given as follows.
I * x = 5 A, That is, x = 5/0.16 = 31.
So, 30 bulbs can burn with a 5 A fuse.
Question 5
1 / -0

In 220 V A.C., the value of 220 represents:

A
Normal voltage

B
Effective voltage

C
Constant voltage

D
Peak voltage

Solution

Answer is B.

The V in the 220 V represents the effective voltage the particular device, appliance or bulb will be able to operate.
It is also called the operating voltage.
Question 6
1 / -0

A radio-set of resistance 20 $$\Omega$$and a resistor of 4 $$\Omega$$ are connected in series with a 6V battery. What is the potential difference across the radio-set?

A
24V

B
120V

C
80V

D
5V

Solution

Answer is D.
In series.
$${ R }_{ total }={ R }_{ 1 }+{ R }_{ 2 }...{ R }_{ n }$$.
In this case, the total resistance is 20+4 = 24 ohms.
The total current in the circuit is calculated from the relation V = IR, that is, I=V/R.
I = 6 V/24 ohms = 0.25 A.
As it is a series connection, the current through each of the components is the same, the voltage across the circuit is the sum of the voltages across each component.
So, the voltage across the radio set is calculated as follows.
The current is 0.25 A and the resistance of the radio set is 20 ohms.
Therefore, V=IR = 0.25 * 20 = 5 V.
Hence, the potential difference across the radio-set is 5 V.
Question 7
1 / -0

Two resistors of $$30 \Omega$$ and $$20 \Omega$$ are joined together in series and then placed in parallel with a $$50 \Omega$$ resistor. What is the effective resistance of the combination?

A
$$25 \Omega$$

B
$$50 \Omega$$

C
$$5 \Omega$$

D
$$10 \Omega$$

Solution
Question 8
1 / -0

An electric cell does 5 joules of work in carrying 10-coulomb charge around the closed circuit. The electromotive force of the cell is :

A
2 volt

B
0.5 volt

C
4 volt

D
1 volt

Solution

Answer is B.

The electromotive force (e) or e.m.f. is the energy provided by a cell or battery per coulomb of charge passing through it, it is measured in volts (V). It is equal to the potential difference across the terminals of the cell when no current is flowing.
e = E/Q
e = electromotive force in volts, VE = energy in joules, JQ = charge in coulombs, C
In this case, work done = 5 J and amount of charge = 10 C.
Therefore, e = 5/10 = 1/2.
Hence, electromotive force of the cell is 1/2 volt.
Question 9
1 / -0

What is the effective resistance between points A and B?

A
$$\dfrac {R}{3}$$

B
$$\dfrac {4R}{3}$$

C
$$\dfrac {2R}{3}$$

D
$$\dfrac {R}{2}$$

Solution

The above figure shows that the three resistors are connected in parallel to each other.
$$\therefore$$ Equivalent resistance between A and B $$\dfrac{1}{R_{eq}} = \dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}$$
$$\implies$$ $$R_{eq} =\dfrac{R}{3}$$
Question 10
1 / -0

For a heater rated at $$4.4 \ kW \ , \ 220 \ V$$ . Calculate the current drawn by the heater.

A
$$40 \ A$$

B
$$20\ A$$

C
$$12.6 \ A$$

D
$$0 \ A$$

Solution

Given:
$$V =220$$ volts
$$P = 4.4 \ kW = 4400 \ W$$

Since, $$ P = VI$$
Current drawn by heater $$I =\dfrac{P}{V} = \dfrac{4400}{220} = 20 \ A$$

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Current Electricity Test - 28

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Current Electricity Test - 28

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Question 1

increases

decreases

remains the same

first increases then decreases

Solution

Question 2

more

less

same

depends upon the length and area of wire

Solution

Question 3

halved

doubled

quadrupled

unchanged

Solution

Question 4

30

45

8

44

Solution

Question 5

Normal voltage

Effective voltage

Constant voltage

Peak voltage

Solution

Question 6

24V

120V

80V

5V

Solution

Question 7

$$25 \Omega$$

$$50 \Omega$$

$$5 \Omega$$

$$10 \Omega$$

Solution

Question 8

2 volt

0.5 volt

4 volt

1 volt

Solution

Question 9

$$\dfrac {R}{3}$$

$$\dfrac {4R}{3}$$

$$\dfrac {2R}{3}$$

$$\dfrac {R}{2}$$

Solution

Question 10

$$40 \ A$$

$$20\ A$$

$$12.6 \ A$$

$$0 \ A$$

Solution

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