Electricity Test

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Electricity Test - 58

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Question 1
1 / -0

You are given several identical resistance each of value $$R=10$$ and each capable of carrying a maximum current of one ampere. It is required to make a suitable combination of these resistance of $$5$$ which can carry a current of $$4$$ ampere. The minimum number of resistance's of the type $$R$$ that will be required for this job is

A
$$4$$

B
$$10$$

C
$$8$$

D
$$20$$

Solution

Resistance of each resistor $$R = 10$$
Equivalent resistance of each segment $$R' = R+R = 2R$$
$$\therefore$$ $$R' = 2\times 10 = 20$$
Such $$4$$ segments are connected in parallel to each other.
Thus equivalent resistance of the circuit $$R_p = \dfrac{R'}{4}$$
$$\implies$$ $$R_p = \dfrac{20}{4} = 5$$
Maximum current flowing through each resistor is one ampere i.e. $$i =1 A$$
Thus total current flowing through the circuit $$I = i+i+i+i$$
$$\therefore$$ $$I = 4i = 4 A$$
Thus minimum $$8$$ resistors are required to satisfy the above conditions.
Question 2
1 / -0

A $$30\ V,\ 90\ W$$ lamp is to be operated on a $$120\ V\ DC$$ line. For proper glow, a resistor of ..... should be connected in series with the lamp.

A
$$10$$

B
$$20$$

C
$$30$$

D
$$40$$

Solution

Resistance of the lamp $$R = \dfrac{V^2}{P}$$
$$\therefore$$ $$R = \dfrac{30^2}{90}$$
$$\implies$$ $$R = 10 \Omega$$
Current flowing through the lamp $$I = \dfrac{P}{V}$$
$$\therefore$$ $$I = \dfrac{90}{30} =3 A $$
DC line voltage $$V_d = 120$$ V
Resistance of the line must be $$R' = \dfrac{V_d}{I}$$
We get $$R' = \dfrac{120}{3} = 40 \Omega$$
Let $$R_{1}$$ be the external resistance to be connected in series with the lamp for proper glow.
$$\therefore$$ $$R + R_1 = R'$$
OR $$10 + R_1 = 40$$
$$\implies$$ $$R_1 = 30 \Omega$$
Question 3
1 / -0

$$1 Wh$$ (Watt hour) is equal to :

A
$$36\times { 10 }^{ 5 }J$$

B
$$36\times { 10 }^{ 4 }J$$

C
$$3600 J$$

D
$$3600 { Js }^{ -1 }$$

Solution

Watt is a unit of power such that $$1W =1\dfrac{J}{s}$$
Also we know $$1h = 3600$$ $$s$$
$$\therefore$$ $$1Wh = 1\dfrac{J}{s} \times 3600 $$ $$s = 3600$$ $$J$$
Question 4
1 / -0

Which of the following shows energy changes that takes place when an incandescent bulb glows using a battery?

A
Chemical>Electrical>heat>Light energy

B
Electrical>Chemical>Light>heat energy

C
Chemical>Electrical>Heat energy

D
Electrical>Chemical>Light energy

Solution

Energy stored in the battery in the form of chemical energy. Once the bulb is switched on the chemical energy changes to electrical, followed by heating of the filament then the bulb glows bright and gives light energy.
Question 5
1 / -0

A 30V-90W lamp is operated on a 120 V DC line. A resistor is connected in series with the lamp in order to glow it properly. The value of resistance is

A
$$10 \Omega$$

B
$$30 \Omega$$

C
$$20 \Omega$$

D
$$40 \Omega$$

Solution

Resistance of lamp
$$R_0 = \dfrac{V^2}{P} = \dfrac{(30)^2}{90} = 10 \Omega$$
Current in the lamp
$$I = \dfrac{V}{R_0} = \dfrac{30}{10} = 3A$$
As the lamp is operated on 120V DC, then resistance becomes
$$R' = \dfrac{V'}{i} = \dfrac{120}{3} = 40 \Omega$$
For proper glow, a resistance R is joined in series with the bul
$$R' = R + R_0$$
$$\Rightarrow R^{\alpha} = R' - R_0 = 40 - 10 = 30 \Omega$$
Question 6
1 / -0

Two resistors $$400\ \Omega $$ and $$800\ \Omega$$ are connected in series with a 6 V battery. the potential difference measured by voltmeter across $$400\ \Omega $$ resistor is:

A
$$2\ V$$

B
$$1.95\ V$$

C
$$3.8\ V$$

D
$$4\ V$$

Solution

Total resistance of the circuit (series combination) $$R_s = R_1+R_2$$
$$\therefore$$ $$R_s = 400+800 =1200\Omega$$

Current, $$I=\dfrac{E}{R_s}=\dfrac{6}{1200}$$

$$I=0.005\ A$$

Potential difference across $$R_1$$, $$V' = IR_1$$
$$\therefore$$ $$V' = 0.005 \times 400$$
$$V' =2\ V$$
Question 7
1 / -0

In the following circuit, the 1 $$\Omega$$ resistor dissipates power P. If the resistor is replaced by 9 $$\Omega$$, the power dissipated in it is

A
$$P$$

B
$$3P$$

C
$$9P$$

D
$$P/3$$

Solution

Equivalent resistance in the circuit $$= (3+1)=4\Omega$$

By Ohm's law,
$$10 \ V=(4 \ \Omega)(i \ Ampere)$$
$$i=2.5 \ A$$

Power through a resistance is given by $$P=I^2R$$
Power through $$1 \Omega,$$
$$P_{1 \Omega}=i^2\times 1=6.25 \ W$$
Given, $$P_{1\Omega}=P$$

When $$1\Omega$$ is replaced by $$9\Omega$$, equivalent resistance $$=(3+9)=12\Omega$$
By Ohm's law,
$$10 \ V=(12 \ \Omega)(i' \ Ampere)$$
Current $$i'=\dfrac{10}{12} \ A$$

Power dissipated through replaced resistor, $$P_{9\Omega}=(\dfrac{10}{12})^2\times 9=6.25=P_{1\Omega}=P$$

Option A is correct.
Question 8
1 / -0

Which of the following are the properties of fuse wire?

A
Made of alloy of tin

B
Has a low melting point

C
Connected in series with main supply

D
All of the above
Solution
All the given options hold true for fuse wire. A fuse wire should have these characteristics:
low melting point,
high conductivity and
least deterioration due to oxidation
It is connected in series with main supply.
Question 9
1 / -0

The ratio of resistance of two copper wire of the same length and of same cross-sectional area, when connected in series to that when connected in parallel, is

A
$$1 : 1$$

B
$$1 : 2$$

C
$$2 : 1$$

D
$$4 : 1$$

E
$$1 : 4$$

Solution

Let $$R$$ be the resistance of copper wire. In the first condition,
$$R_{series} = R + R = 2R .... (i)$$
In the second condition,
$$\dfrac {1}{R_{parallel}} = \dfrac {1}{R} + \dfrac {1}{R} = \dfrac {1}{R_{parallel}} = \dfrac {1 + 1}{R}$$
$$R_{parallel} =\dfrac {R}{2} ..... (ii)$$

On dividing Eq. (i) by Eq. (ii), we get

$$\dfrac {R_{series}}{R_{parallel}} = \dfrac {2R}{\dfrac {R}{2}} = \dfrac {4}{1}$$.
Question 10
1 / -0

An electric circuit contains an operating heating element and a lit lamp. Which statement best explains why the lamp remains lit when the heating element is removed from the circuit?

A
The lamp and heating element were connected in parallel.

B
The lamp has less resistance than the heating element.

C
The lamp has more resistance than the heating element.

D
The lamp and heating element were connected in series.

Solution

When the components of an electrical circuit are connected in parallel, same potential difference appears across each of the element and hence, removal of an element does not affect the working of other elements. Hence, A is the correct option.

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Electricity Test - 58

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Question 1

$$4$$

$$10$$

$$8$$

$$20$$

Solution

Question 2

$$10$$

$$20$$

$$30$$

$$40$$

Solution

Question 3

$$36\times { 10 }^{ 5 }J$$

$$36\times { 10 }^{ 4 }J$$

$$3600 J$$

$$3600 { Js }^{ -1 }$$

Solution

Question 4

Chemical>Electrical>heat>Light energy

Electrical>Chemical>Light>heat energy

Chemical>Electrical>Heat energy

Electrical>Chemical>Light energy

Solution

Question 5

$$10 \Omega$$

$$30 \Omega$$

$$20 \Omega$$

$$40 \Omega$$

Solution

Question 6

$$2\ V$$

$$1.95\ V$$

$$3.8\ V$$

$$4\ V$$

Solution

Question 7

$$P$$

$$3P$$

$$9P$$

$$P/3$$

Solution

Question 8

Made of alloy of tin

Has a low melting point

Connected in series with main supply

All of the above

Solution

Question 9

$$1 : 1$$

$$1 : 2$$

$$2 : 1$$

$$4 : 1$$

$$1 : 4$$

Solution

Question 10

The lamp and heating element were connected in parallel.

The lamp has less resistance than the heating element.

The lamp has more resistance than the heating element.

The lamp and heating element were connected in series.

Solution

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