Electricity Test

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Electricity Test - 33

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Weekly Quiz Competition

Question 1
1 / -0

A copper $$(\rho =1.6\times 10^{-8}\Omega m)$$ wire and an Aluminium $$(\rho =2.5\times 10^{-8}\Omega m)$$ wire have equal cross sectional area and equal resistances. The ratio of their lengths is

A
$$5: 4$$

B
$$4:5$$

C
$$16:25$$

D
$$25:16$$

Solution

We know,
$$R= \dfrac{\rho \times L }{A} $$
Here,
$$ A{}_{Cu} = A{}_{Al} $$
$$ R{}_{Cu} = R{}_{Al} $$

Hence,
$$ {\rho}^{}_{Cu} \times {{L}^{}_{Cu}} = {\rho}^{}_{Al} \times {{L}^{}_{Al}} $$

or, $$ \dfrac{ {L}^{}_{Cu}} {{L}_{Al}} = \dfrac{ {\rho}_{Al}} {{\rho}_{Cu}} = \dfrac{2.5 \times 10^{-8}} {1.6 \times 10^{-8}} = \dfrac{25}{16}$$
Question 2
1 / -0

The equivalent conductance of two wires of resistance 10 ohm and 5 ohm joined together in parallel is:

A
$$\dfrac{50}{15}$$ mho

B
$$\dfrac{35}{50}$$ mho

C
$$0.3$$ mho

D
$$\dfrac{2}{3}$$ mho

Solution

Effective resistance of the parallel combination:
$$\dfrac{1}{R}=\dfrac{1}{5}+\dfrac{1}{10}=\dfrac{3}{10}$$
So, conductance $$= \dfrac{3}{10}=0.3$$ mho
Question 3
1 / -0

The current that passes through 20 ohm resistance when it is connected in parallel with a 30 ohm resistance and this set is connected to a battery of 2V is

A
0.2A

B
0.3A

C
0.1A

D
0.016A

Solution

Given,
$$R_1=20\Omega\\R_2=30\Omega\\E=2\ V$$
According to question, both resistance are in parallel and it is connected with $$2\ V$$ battery.
Potential difference across each resistance $$2\Omega$$

Current in $$20\Omega$$ resistance $$I_{20\Omega}=\dfrac{\Delta V}{R}=\dfrac{2}{20}=0.1\ A$$
Question 4
1 / -0

The resistance between A and B is:

A
$$9\Omega $$

B
$$2\Omega $$

C
$$12\Omega $$

D
$$8\Omega $$

Solution

$$ Req \ of \ AC = \dfrac{6\times6}{12} = 3$$

$$ Req \ of \ AD = \dfrac{6\times6}{12} = 3$$

$$ Req \ of \ AE = \dfrac{6\times6}{12} = 3$$

$$ R_{AB} = \dfrac{6\times3}{9} = 2\Omega$$
Question 5
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There are $$5$$ tube-lights each of $$40\ W$$ in a house. These are used on an average for $$5\ hrs/day$$. In addition, there is an immersion heater of $$1500\ W$$ used on an average for $$1\ hr/day$$. The number of units of electricity that are consumed in a month of $$30\ days$$ is:

A
$$25\ units$$

B
$$50\ units$$

C
$$75\ units$$

D
$$100\ units$$

Solution

$$E=n_1{ P }_{ 1 }{ t }_{ 1 }+n_2{ P }_{ 2 }{ t }_{ 2 }$$
$$E= 5 \times 40\times 5\times 30+1 \times 1500\times 30$$
$$=200\times 150+45000$$
$$E=75\ kWh$$
$$\Rightarrow E = 75\ units$$
Question 6
1 / -0

If an electric iron of $$1200 \ W$$ is used for $$30 \ minutes$$ everyday, find electric energy consumed in the month of April.

A
$$12\ kWh$$

B
$$18 \ kWh$$

C
$$36\ kWh$$

D
$$6\ kWh$$

Solution

$$Power \ \ P= 1200 W = \dfrac{1200}{1000}=1.2 kW$$

$$time \ \ t = 30 \ min =\dfrac{30}{60}=0.5 h$$

Number of days in the month of april = $$30$$
Hence, Total time for which iron is used is given by (in hrs) = $$ 0.5 \times 30$$

$$Energy \ \ E =Power\times Total \ time$$
$$= 1.2\times 0.5\times 30$$
$$= 18 kW h$$
Question 7
1 / -0

The resultant resistance of two resistors when connected in series is 48 ohm. The ratio of their resistances is 3 : 1. The value of each resistance is

A
20$$\Omega $$ ,28$$\Omega $$

B
32$$\Omega $$ ,16$$\Omega $$

C
36$$\Omega $$, 12$$\Omega $$

D
24$$\Omega $$, 24$$\Omega $$

Solution

$$ R_{1} + R_{2} = 48 $$ ohm

$$ \frac{R_{1}}{R_{2}} = \frac{3}{1}$$

$$ \Rightarrow R_{1} = 3 R_{2}$$

$$ \Rightarrow 4R_{2} = 48 \Rightarrow R_{2} = 12 $$ ohm

$$ R_{1} = 36 $$ ohm
Question 8
1 / -0

If the electric current in a lamp decreases by $$5\ \%$$, then the power output decreases by:

A
$$20\ \%$$

B
$$10\ \%$$

C
$$5\ \%$$

D
$$2.5\ \%$$

Solution

Hint: Power consumed by resistor is $$P = {i^2}R$$ ,where $$i$$ is the current flowing through the resistor and $$R$$ is the total resistance.

Correct option: B

Explanation for correct answer:

Step 1: Find the decrease in power of the lamp.

As, power of the lamp, $$P = {i^2}R$$

$$ \Rightarrow \dfrac{{dP}}{P} = 2\dfrac{{di}}{i}$$

$$ \Rightarrow \dfrac{{dP}}{P} = 2 \times 5\% $$

$$ \Rightarrow \dfrac{{\vartriangle P}}{P} = 10\% $$

So, the decrease in the power of the lamp is $$10\% $$ .
Question 9
1 / -0

Three unequal resistors in parallel are equivalent to a resistance $$1\ \Omega$$. If two of them are in the ratio $$1 : 2$$ and if no resistance value is fractional, the largest of the three resistances in $$\Omega$$ is:

A
$$4$$

B
$$6$$

C
$$8$$

D
$$12$$

Solution

Let $$R_1, R_2,$$ and $$R_3$$ be the three resistances connected in parallel.
$$R_{p}$$ is equivalent resistance in parallel combination.
Hence,
$$\dfrac {1}{R_{p}} = \dfrac {1}{R_{1}} + \dfrac {1}{R_{2}}+\dfrac {1}{R_{3}}$$
Let, $$R_{2} = 2R_{3}$$
Therefore,
$$\dfrac {1}{R_{p}} = \dfrac {1}{R_{1}} + \dfrac {1}{2R_{3}}+\dfrac {1}{R_{3}}$$

$$\dfrac {1}{R_{p}} = \dfrac {1}{R_{1}} + \dfrac {3}{2R_{3}}$$

$$R_p = 1\ \Omega$$ ; given

$$\dfrac {1}{R_{1}} = 1- \dfrac {3}{2R_{3}}$$

$$\therefore R_{1} = \dfrac {2R_{3}}{(2R_{3}-3)}$$

If $$R_{3} = 3\Omega$$, $$R_{1} = 2\Omega$$ and $$R_{2} = 6\Omega$$
$$\therefore$$ Largest resistance $$= 6\Omega $$
Question 10
1 / -0

The electrical energy in kilowatt hours consumed in operating ten $$50\ W$$ bulbs for $$10\ hrs/day$$ in a month of $$30\ days$$ is

A
$$1500$$

B
$$15000$$

C
$$15$$

D
$$150$$

Solution

Let
$$n$$ be the number of bulbs, $$n= 10$$
$$P$$ be the power of each bulb, $$P= 50\ W$$
total time to burn the bulb, $$t =10 \times 30 = 300\ hr$$

Now the energy $$E$$, consumed in operating the bulbs is:
Energy consume by each built in 30 days $$E=Pt=50\times 300$$
Energy consume by 10 bulbs in 30 days $$E=nPt=10 \times 50 \times 300\ Wh$$

$$E=150\ kWh$$ (Option D)

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Electricity Test - 33

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Electricity Test - 33

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SHARING IS CARING

Question 1

$$5: 4$$

$$4:5$$

$$16:25$$

$$25:16$$

Solution

Question 2

$$\dfrac{50}{15}$$ mho

$$\dfrac{35}{50}$$ mho

$$0.3$$ mho

$$\dfrac{2}{3}$$ mho

Solution

Question 3

0.2A

0.3A

0.1A

0.016A

Solution

Question 4

$$9\Omega $$

$$2\Omega $$

$$12\Omega $$

$$8\Omega $$

Solution

Question 5

$$25\ units$$

$$50\ units$$

$$75\ units$$

$$100\ units$$

Solution

Question 6

$$12\ kWh$$

$$18 \ kWh$$

$$36\ kWh$$

$$6\ kWh$$

Solution

Question 7

20$$\Omega $$ ,28$$\Omega $$

32$$\Omega $$ ,16$$\Omega $$

36$$\Omega $$, 12$$\Omega $$

24$$\Omega $$, 24$$\Omega $$

Solution

Question 8

$$20\ \%$$

$$10\ \%$$

$$5\ \%$$

$$2.5\ \%$$

Solution

Question 9

$$4$$

$$6$$

$$8$$

$$12$$

Solution

Question 10

$$1500$$

$$15000$$

$$15$$

$$150$$

Solution

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