Electricity Test

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Electricity Test - 38

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Weekly Quiz Competition

Question 1
1 / -0

If a man has five resistors each of value $$\dfrac { 1 }{ 5 } \Omega $$ then is the maximum resistance he can obtain by connecting them, is :

A
$$1\Omega $$

B
$$5\Omega $$

C
$$\dfrac { 1 }{ 2 } \Omega $$

D
$$\dfrac { 2 }{ 5 } \Omega $$

Solution

Given,
$$R=\dfrac15\ Omega$$ total number of resistance, n=5
For maximum effective resistance, all resistors should connect in series.
For series combination
$$R_{eq}=R_1+R_2+ ..........+R_n\\R_{eq}=nR$$ (All resistor have same resistance)

On putting value of $$n$$ and $$R$$
$$R_{eq}=5\times \dfrac15=1\ \Omega$$

Hence maximum resistance is 1 ohm.
Question 2
1 / -0

Conventionally, the direction of the current is taken as the direction of flow

A
of negative charges

B
the direction of flow of atoms

C
of molecules

D
of positive charges

Solution

In metals, there are plenty of free electrons which flow and make electric current. The direction of the current is taken opposite to the flow of electrons. Since electrons are negative charges, the direction opposite to the flow of electrons is the same as the direction of the flow of positive charges.
So conventionally, the direction of current is the same as the direction of the flow of positive charges.
Question 3
1 / -0

Three fuse wires A, B and C tinned copper having identical cross section and length 8 cm, 12 cm and 16 cm are used as fuses. For the same value of current

A
fuse A will melt first

B
fuse B will melt first

C
fuse C will melt first

D
all fuse will melt for same current

Solution

Heat lost per second per unit surface area of fuse wire is
$$H = \dfrac{I^2 \rho }{2 \pi ^2 r^3}$$

$$\Rightarrow I^2 \propto r^3$$

$$\Rightarrow I \propto r^{\dfrac{3}{2}}$$

Hence, all fuse will melt for same current as they are independent of length of fuse wire.
Question 4
1 / -0

An electric heater kept in a vacuum is heated continuously by passing an electric current. Its temperature:

A
It will go on rising with time

B
It will stop after some time as it will lose heat to the surroundings by conduction

C
It will rise for sometime and thereafter will starts falling

D
It will become constant after sometime because of loss of heat due to radiation

Solution

Treating initially that there is no radiation and on applying a voltage to the heater, the temperature of the heater rises. As the temperature rises, the radiation also increases.
As the input power($$i^2r$$) is constant, the output power should also be constant.
Initially, the temperature will be increasing and radiation as well. When the heater reaches a certain temperature, the radiation power equals the input power and the temperature of the heater stops rising.
Thus temperature becomes constant after some time.
Question 5
1 / -0

A uniform wire of resistance $$50\Omega$$ is cut into 5 equal parts. These parts are now connected in parallel. The equivalent resistance of the combination is

A
$$2\Omega$$

B
$$10\Omega$$

C
$$250\Omega$$

D
$$6250\Omega$$

Solution

Here, a uniform wire of resistance $$50 \Omega$$ is cut into $$5$$ equal parts. Hence, resistance of each part is
$$r = \dfrac{R}{n} $$
$$ \Rightarrow r = \dfrac{50}{5} $$
$$ \Rightarrow r= 10\ \Omega$$
Now, the parts are connected in parallel, the equivalent resistance of the combination is
$$\dfrac{1}{R_{eq}} = \dfrac{1}{10} + \dfrac{1}{10} + \dfrac{1}{10} + \dfrac{1}{10} + \dfrac{1}{10}$$
$$\dfrac{1}{R_{eq}} = \dfrac{5}{10} = \dfrac{1}{2}$$
$$R_{eq} = 2\Omega$$
Question 6
1 / -0

Three resistors each of $$10 \Omega$$ are connected in series to a battery of potential difference 150 v. The current flowing through it is ..... A.

A
45

B
5

C
15

D
20

Solution

$$V=IR$$
Current is given by,
$$V=I\times{R_{eq}}$$
$$I=\frac{V}{R_{eq}}$$

The three resistance are in series. So, the equivalent resistance will be:
$$R_{eq}=R_1+R_2+R_3$$
$$R_{eq}=10+10+10$$
$$=30\ \Omega$$

Substitute the values in the given formula:
$$I=150/30$$
$$I=5$$ $$A$$
Question 7
1 / -0

An electric bulb rated 220 V - 100 W will fuse if it consumes 131 W. What voltage fluctuations can it withstand?

A
up to 230 V

B
up to 241 V

C
up to 225 V

D
up to 252 V

Solution

We know that: $$P = \dfrac{V^{2}}{R}$$
Using the fact that the bulb is $$220V - 100W$$
$$ 100 = \dfrac{220^{2}}{R}$$

$$R = \dfrac{48400}{100} = 484\ ohm$$

Given that it will fuse on use of power of $$131W$$.
Substituting this:
$$131 = \dfrac{V^{2}}{484}$$

$$V^{2} = 131 \times 484 = 63,404$$

$$\Rightarrow V = 251.8V$$ Which is the highest voltage it can experience without fusing.
Question 8
1 / -0

Fuse should be connected to ..... wire of the circuit

A
Earth

B
Neutral

C
Phase

D
Any of the above

Solution

Fuse should be connected to live / phase wire of circuit. If it is connected to neutral wire, the fuse will melt when excess current flow, but the appliance will still be connected to high potential through live wire. Thus, if a person touches the appliance, he receives shock. Hence, choice is (C).
Question 9
1 / -0

How much energy in kilowatt hour is consumed in operating ten 50 watt bulbs for 10 hours per day in a month (30 days)

A
1500

B
5000

C
15

D
150

Solution

Energy consumed $$=10\times 50\times 10\times 30\times 3600 J$$
$$[1 Kwh=3600\times 1000J]$$
$$=\frac {10\times 50\times 10\times 30\times 3600}{3600\times 1000}kWh=150$$
Question 10
1 / -0

An 1800 W toaster, a 1.3 kW frying pan are plugged into the same $$20 A-120V$$ line then

A
fuse will not blow

B
fuse will blow

C
supply will spark

D
fuse is just saved

Solution

$$I_{toaster}=\dfrac{1800}{120}= 15 A,$$

$$I_{frying pan}=\dfrac{1300}{120}= 10.83 A$$

$$I_{net}= 25.83 A > 20 A$$
Because the applied current is more than the possible value of current which is given in the question, the fuse will blow.

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Electricity Test - 38

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Electricity Test - 38

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Question 1

$$1\Omega $$

$$5\Omega $$

$$\dfrac { 1 }{ 2 } \Omega $$

$$\dfrac { 2 }{ 5 } \Omega $$

Solution

Question 2

of negative charges

the direction of flow of atoms

of molecules

of positive charges

Solution

Question 3

fuse A will melt first

fuse B will melt first

fuse C will melt first

all fuse will melt for same current

Solution

Question 4

It will go on rising with time

It will stop after some time as it will lose heat to the surroundings by conduction

It will rise for sometime and thereafter will starts falling

It will become constant after sometime because of loss of heat due to radiation

Solution

Question 5

$$2\Omega$$

$$10\Omega$$

$$250\Omega$$

$$6250\Omega$$

Solution

Question 6

45

5

15

20

Solution

Question 7

up to 230 V

up to 241 V

up to 225 V

up to 252 V

Solution

Question 8

Earth

Neutral

Phase

Any of the above

Solution

Question 9

1500

5000

15

150

Solution

Question 10

fuse will not blow

fuse will blow

supply will spark

fuse is just saved

Solution

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