Electricity Test

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Electricity Test - 34

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Question 1
1 / -0

A house is fitted with ten lamps of each $$60\ W$$. Each lamp burns for $$5\ hrs$$ a day on an average. The cost of consumption in a month of $$30\ days$$ at $$2. 80$$ rupees per unit is:

A
$$Rs. 152$$

B
$$Rs. 252$$

C
$$Rs. 352$$

D
$$Rs. 452$$

Solution

Total energy spent is given by: $$E = nE'$$
where $$n:$$ number of bulbs
$$E':$$ Energy spent by each bulb
$$E' = Pt$$
$$E' = 60 \times 5 \times 30\ Wh$$

$$ E = 60 \times 5 \times 30 \times 10$$
      $$ = 9000 \times 10$$
      $$ = 90 KWh$$

Cost $$= 90 \times 2.8$$
           $$ = 28 \times 9$$
          $$ = 252 Rs$$
Question 2
1 / -0

An electric kettle consumes 1 kW of electric power when operated at 220 V. A fuse wire of what rating must be used for it?

A
1A

B
2A

C
4A

D
5A

Solution

Hint : A fuse wire should be able pass that amount of current which the appliance requires

Step 1 : Find the maximum current that the appliance requires
Given, Power $$ = 1KW= 1000W$$ & Voltage $$= 220\ V$$
$$P= V \times I$$
So, $$I=\dfrac{P}{V}$$
$$I=\dfrac{1000}{220}= 4.545 A$$
The fuse wire should be able to pass at least $$4.545\ A$$ current.
So, the rating should be $$5A$$.
Question 3
1 / -0

Observe the figure given below.
Find the current passing through $$6\Omega$$ resistor.

A
0.72 A

B
0.80 A

C
0.48 A

D
Cannot be said

Solution

Let the equivalent resistance between $$E$$ and $$F$$ consisting of $$R_1=6 \Omega$$ and $$R_2=4 \Omega$$ in parallel be $$R_{eq}.$$
$$\dfrac{1}{R_{eq}}=\dfrac{1}{R_1}+\dfrac{1}{R_2}=\dfrac{R_1+R_2}{R_1R_2}$$

$$\Rightarrow R_{eq}=\dfrac{R_1R_2}{R_1+R_2}=\dfrac{(6)(4)}{6+4}=2.4 \Omega$$

Let the voltage across $$E$$ and $$F$$ be $$V_{EF}.$$
Applying Ohm's law for $$R_{eq}$$
$$V_{EF}=IR_{eq}$$
$$V_{EF}=(1.2 \ A) \times (2.4 \ \Omega)=2.88 \ V$$

The voltage $$V_{EF}=2.88 \ V$$ also appears across $$R_1$$ and $$R_2$$.
$$\therefore V_{EF}=V_{BC}=V_{AD}=2.88 \ V$$

Let the current through $$6\Omega$$ resistance be $$I_{BC}.$$
Applying Ohm's Law for resistance $$R_1=6 \Omega $$
$$V_{BC}=I_{BC}R_1$$
$$\Rightarrow I_{BC}=\dfrac{V_{BC}}{R_1}=\dfrac{2.88}{6}=0.48 \ A$$
Question 4
1 / -0

Find power dissipated in $$4 \ \Omega$$ resistor :

A
12 W

B
16 W

C
4 W

D
6 W

Solution

Power dissipated $$P = I^2 R = 1^2 \times 4 = 4 W$$
Question 5
1 / -0

Name the effect of current responsible for the glow of the bulb in an electric circuit.

A
Chemical

B
Heating

C
Magnetic

D
Heating and Magnetic

Solution

The electric bulb has a filament called tungsten when electricity passes through this filament, it heats up and glows.
This heat is generated due to the passage of electric current, the drift of electrons due to the current and the resistance it has.
Hence, the heating effect of electric current is responsible for the glow of the bulb in an electric circuit.
Question 6
1 / -0

You have a large supply of light bulbs and a battery. You start with one light bulb connected to the battery and notice its brightness. You then add one light bulb at a time, each new bulb being added in series to the previous bulbs then

A
Brightness of the bulbs will increase

B
Current through the bulbs will increase

C
Power transferred from the battery will decrease

D
Lifetime of the battery will decrease
Question 7
1 / -0

Answer question numbers 72 and 73 based on the following information.
Solar surface radiates energy uniformly at a rate of $$4 \times 10^{26}W$$. This energy spreads or distributes uniformly and normally outwards.

72. Considering Earth and the Sun to be spherical object, the amount of radiant energy received by the Earth per second is nearly

A
$$1360 \:W.$$

B
$$1.75 \times 10^{17}\:W$$

C
$$3.5 \times 10^{17}\:W$$

D
$$7 \times 10^{17}\:W$$
Question 8
1 / -0

The resistance between A and B in the given figure will be(in ohm):

A
20

B
30

C
90

D
more than 10 but less than 20

Solution

In the given circuit, suppose, $$r=10\Omega$$
$$r_1=10\Omega$$
$$r_2=20\Omega$$
and, $$r_3=50\Omega$$
(Please refer the diagram.)
Now, $$r_1, r_2, r_3$$ are connected in parallel. Suppose, resultant of these three resistances is $$R_e$$.
Therefore, $$\dfrac{1}{r_e}=\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3} = \dfrac{1}{10}+\dfrac{1}{20}+\dfrac{1}{50} = \dfrac{17}{100}$$
Hence, $$r_e = \dfrac{100}{17} \Omega$$
Now, resistances $$r$$ and $$r_e$$ are connected in series.
Therefore, equivalent resistance in the given circuit, $$R=r+r_e$$
$$R=10+\dfrac{100}{17} = \dfrac{270}{17} \Omega$$, which is more than $$10\Omega$$, but less than $$20 \Omega$$.
Question 9
1 / -0

In the electric circuit shown in the figure, the effective resistance of two 8 $$\Omega$$ resistors in the combination is :

A
$$4 \Omega$$

B
$$16 \Omega$$

C
$$2 \Omega$$

D
$$ 12 \Omega$$

Solution

In this case, the 8 ohms resistances are connected in parallel.
So the effective resistance ($$R_{eff}$$) is given as:
$$\dfrac{1}{R_{eff}}=\dfrac { 1 }{ 8 } +\dfrac { 1 }{ 8 } =\dfrac { 1 }{ 4 }$$.
Hence, the effective resistance of two 8 ohms resistors in the combination is given as $$ 4 $$ ohm.
Question 10
1 / -0

Which effect of current is responsible for a compass placed in an electric field to be deflected?

A
Heating effects of current.

B
Magnetic effects of current.

C
Conducting effects of heat.

D
Chemical effects of current

Solution