Electricity Test

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Electricity Test - 45

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Question 1
1 / -0

An electric iron is connected to a $$200 V$$ mains supply and draws a current of $$4.0 A$$ What is the power rating of the iron ?

A
$$800W$$

B
$$50W$$

C
$$106W$$

D
$$112W$$

Solution

Operating voltage $$V=200 V$$

Current drawn$$I=4 A$$

Power rating $$P=V\times I=200\times 4=800 W$$
Question 2
1 / -0

A potential difference of 20V is needed to make a current of 0.05A flow through a resistor. What potential difference is needed to make a current of 300 mA flow through the same resistor?

A
60V

B
120V

C
40V

D
150V

Solution

Answer is B.

According to Ohm's law, $$V = IR$$. That is, $$R = \dfrac VI$$.
In this case, the voltage is $$20\ V$$ and the current is $$0.05\ A$$.
So, the resistance $$R = 20/0.05 = 400$$ ohms.
Now, to make a current of 300 mA flow through the same resistor of 400 ohms, the potential difference to be applied is calculated as follows.
Here, $$I = 300 mA = 0.3\ A$$ and $$R = 400$$ ohms.
Therefore, $$V = 0.3\times 400 = 120\ V$$.
Hence, to make a current of 300 mA flow through the same resistor of 400 ohms a potential difference of 120 V is applied.
Question 3
1 / -0

An $$8 \Omega$$ resistance wire is doubled up by folding. What is the new resistance of the wire?

A
$$0.5 \Omega$$

B
$$1.5 \Omega$$

C
$$2\Omega$$

D
$$4\Omega$$

Solution

Resistance of the conductor depends on the following factors:
1. Length of the conductor
2. Area of the cross-section of the conductor
3. Material used as a conductor.

Resistance is directly proportional to the length of the wire, and inversely proportional to the cross sectional area of the wire.
$$R=\rho \dfrac { L }{ A } $$ where R is resistance, $$\rho$$ is the material's resistance in ohms, $$ l$$ is the length, and $$A$$ is the cross sectional area in $${m}^2$$.
Question 4
1 / -0

The value of equivalent resistance between the points $$A\;and\;B$$ in the given circuit, will be

A
$$6\;R$$

B
$$\displaystyle\frac{4\;R}{11}$$

C
$$\displaystyle\frac{11\;R}{4}$$

D
$$\displaystyle\frac{R}{6}$$

Solution

$$\text{Refer image-1}$$
In the given circuit resistor cd, de and ef are in series.
$$\therefore R_{eq}=cd+de+ef=R+R+R=3R$$

Now, the circuit will be like $$\text{Refer image-2}$$
$$\text{Refer image-2}$$
In this circuit cf and de are parallel
$$\therefore R_{eq}=\dfrac{R_{cf}\times R_{de}}{R_{cf}+R_{de}}$$

$$=\dfrac{R\times 3R}{R+3R}$$

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

Now, the circuit will be like $$\text{Refer image-3}$$
$$\text{Refer image-3}$$
In this circuit, $$Ac,cf,fB$$ are in series

$$\therefore R_{AB}=R+\dfrac{3R}{4}+R=2R+\dfrac{3R}{4}=\dfrac{11R}{4}$$

$$\therefore$$ Correct option is C.
Question 5
1 / -0

The minimum resistance that can be obtained by connecting 5 resistance of $$\displaystyle \frac{1}{4} \Omega$$ each is

A
$$\displaystyle \frac{4}{5} \Omega$$

B
$$\displaystyle \frac{4} \Omega$$

C
$$20 \Omega$$

D
$$0.05 \Omega$$

Solution

Minimum resistance will be obtained from the combination if they are connected in, parallel
$$\therefore$$ When 5 resistance of $$\displaystyle \frac{1}{4} \Omega$$ each are connected in parallel then
$$\displaystyle \frac{1}{R_p} =\frac{5}{\frac{1}{4}} =20$$
$$\Rightarrow R_p = \displaystyle \frac{1}{20} = 0.05 \Omega$$
Question 6
1 / -0

Two electric bulbs whose resistance are in the ratio of $$1 : 2$$ are connected in parallel to a constant voltage source. The ratio of the power dissipated in them will be:

A
$$1 : 4$$

B
$$1 : 2$$

C
$$1 : 1$$

D
$$2 : 1$$

Solution

Given : $$\dfrac{R_1}{R_2} =\dfrac{1}{2}$$
And, the voltage difference across the resistors is the same in parallel connection.
And, $$ P= \frac{V^2}{R} $$
Ratio of power dissipated $$\dfrac{P_1}{P_2} = \dfrac{V^2/R_1}{V^2/R_2} = \dfrac{R_2}{R_1} = 2$$
Question 7
1 / -0

A wire of resistance 12 ohm is bent in the form of a circular ring. The effective resistance between the two points on any diameter of the circle is:

A
24 ohm

B
12 ohm

C
6 ohm

D
3 ohm

Solution

Resistance of wire is 12 ohm. Therefore, resistance of each semicircle = 6 ohm
Two resistances are in parallel.
Their equivalent resistance = $$\dfrac{6 \times 6}{6+6}=3$$ ohm
Question 8
1 / -0

The temperature of a metal wire rises when an electric current is passed through it because:

A
collision of conduction electrons with the atoms of metal gives them energy which appears as heat

B
.when electrons fall from higher energy level to lower energy level, heat energy is released

C
collisions of metal atoms with each other releases heat energy.

D
collisions of conduction electrons with each other releases heat energy.

Solution

Answer is A.

A metallic conductor has a large number of free electrons in it. When a potential difference is applied across the ends of a metallic wire, the free electrons begin to drift from the low potential to the high potential region. These electrons collide with the positive ions (the atoms which have lost their electrons). In these collisions, energy of the electrons is transferred to the positive ions and they begin to vibrate more violently. As a result, heat is produced. Greater the number of electrons flowing per second, greater will be the rate of collisions and hence more heat is produced.
Hence, the temperature of a metal wire rises when an electric current is passed through it because collision of conduction electrons with the atoms of metal gives them energy which appears as heat.
Question 9
1 / -0

What is the minimum resistance that one can obtain by connecting all the five resistances each of $$0.5\Omega$$?

A
$$\frac{1}{10} \Omega$$

B
$$\frac{1}{5} \Omega$$

C
$$ \frac{1}{50} \Omega$$

D
$$\frac{1}{25} \Omega$$

Solution

Given $$R=0.5=\dfrac{1}{2}\Omega$$

For obtaining minimum resistance, all five resistances should be connected in parallel.
And the equivalent resistance =$$\dfrac{R}{5}$$

$$=\dfrac{0.5}{5}=\dfrac{1}{10}\Omega$$

Answer-(A)
Question 10
1 / -0

An electric bulb is rated 220 volt and 100 watt, Power consumed by it when operated on 110 volt is :

A
50 W

B
85 W

C
90 W

D
45 W

Solution

Answer is A.

The power dissipated by an electric bulb is given as P = VI.
In this case, Power P = 100 W and Voltage V = 220 V.
So, current I is I = P/V = 100/220 = 0.45 A.
When the supplied voltage is 110 V, the power consumed will be P=VI = 110 * 0.45 = 50 W.
Hence, the power consumed when operated at 110 V is 50 W.

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Electricity Test - 45

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Electricity Test - 45

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Question 1

$$800W$$

$$50W$$

$$106W$$

$$112W$$

Solution

Question 2

60V

120V

40V

150V

Solution

Question 3

$$0.5 \Omega$$

$$1.5 \Omega$$

$$2\Omega$$

$$4\Omega$$

Solution

Question 4

$$6\;R$$

$$\displaystyle\frac{4\;R}{11}$$

$$\displaystyle\frac{11\;R}{4}$$

$$\displaystyle\frac{R}{6}$$

Solution

Question 5

$$\displaystyle \frac{4}{5} \Omega$$

$$\displaystyle \frac{4} \Omega$$

$$20 \Omega$$

$$0.05 \Omega$$

Solution

Question 6

$$1 : 4$$

$$1 : 2$$

$$1 : 1$$

$$2 : 1$$

Solution

Question 7

24 ohm

12 ohm

6 ohm

3 ohm

Solution

Question 8

collision of conduction electrons with the atoms of metal gives them energy which appears as heat

.when electrons fall from higher energy level to lower energy level, heat energy is released

collisions of metal atoms with each other releases heat energy.

collisions of conduction electrons with each other releases heat energy.

Solution

Question 9

$$\frac{1}{10} \Omega$$

$$\frac{1}{5} \Omega$$

$$ \frac{1}{50} \Omega$$

$$\frac{1}{25} \Omega$$

Solution

Question 10

50 W

85 W

90 W

45 W

Solution

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