Electricity Test

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Electricity Test - 42

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Question 1
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Two wires of same metal have the same length but their cross-sections are in the ratio 3:1. They are joined in series. The resistance of the thicker wire is $$10\Omega$$. The total resistance of the combination is:

A
$$\displaystyle \frac {5}{2} \Omega$$

B
$$\displaystyle \frac {40}{3} \Omega$$

C
$$40\Omega$$

D
$$100\Omega$$

Solution

Resistance of a wire $$R=\dfrac{\rho l}{A}$$ where $$\rho=$$ resistivity, $$l=$$ length, $$A=$$ cross section of the wire.
As both have same material and length so $$R \propto \dfrac{1}{A}$$
Thus, $$\dfrac{R_2}{R_1}=\dfrac{A_1}{A_2}=\dfrac{3}{1} \Rightarrow R_2=3R_1$$.
here $$R_1$$ is the resistance of thicker wire so its resistance $$R_1=10 \Omega$$ (given)
so, $$R_2=3(10)=30 \Omega$$
As they are connected in series so the equivalent resistance is
$$R_{eq}=R_1+R_2=10+30=40 \Omega$$
Question 2
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Directions For Questions

Two resistors A and B of resistance 4 $$\Omega$$ and 6 $$\Omega$$ respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance.

...view full instructions

The power supplied by the battery will be :

A
15 W

B
24 W

C
3.6 W

D
20 W

Solution

In this case, the resistances 4 ohms and 6 ohms are connected in parallel.
The total resistance is $$\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 6 } =\dfrac { 1 }{ 0.416 } =2.4\quad ohms$$. The voltage is given as 6 V.
The power is calculated as follows.
$$P=VI\quad =\quad \dfrac { { V }^{ 2 } }{ R } (as\quad I=\dfrac { V }{ R } )$$.
Therefore, $$P=\quad \dfrac { { 6 }^{ 2 } }{ 2.4 } \quad =\quad 15W$$.
Hence, the power supplied by the battery is 15 W.
Question 3
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Directions For Questions

Two resistors A and B of resistance 4 $$\Omega$$ and 6 $$\Omega$$ respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance.

...view full instructions

The power dissipated in each resistor will be :

A
$$P_A= 6 W, P_B=9 W$$

B
$$P_A= 9 W, P_B=6 W$$

C
$$P_A= 9 W, P_B=3 W$$

D
$$P_A= 3 W, P_B=6 W$$

Solution

The voltage across the circuit is given as 6 V.
So the power dissipated by the 4 ohms resistor is given as $$P=\dfrac { { V }^{ 2 } }{ R } =\dfrac { { 6 }^{ 2 } }{ 4 } =\quad 9W$$. And,
the power dissipated by the 6 ohms resistor is given as $$P=\dfrac { { V }^{ 2 } }{ R } =\dfrac { { 6 }^{ 2 } }{ 6 } =\quad 6W$$.
Hence, the power dissipated by the 4 ohms and 6 ohms resistors are 9 W and 6 W.
Question 4
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In an electric circuit containing a battery, the charge (assumed positive) inside the battery

A
always goes from the positive terminal to the negative terminal.

B
may go from the positive terminal to the negative terminal.

C
always goes from the negative terminal to the positive terminal.

D
does not move.

Solution

Inside the battery or cell, the positive charge move from negative terminal (lower potential) to positive terminal (higher potential) of the battery. While in external circuit the positive charge move from positive terminal to negative terminal of the battery.Hence correct option is C
Question 5
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A technician has $$10$$ resistors each of resistance $$0.1\ \Omega$$. The largest and smallest resistance that he can obtain by combining these resistors are:

A
$$10\ \Omega \ \text{and}\ 1\ \Omega$$

B
$$1\ \Omega \ \text{and}\ 0.1\ \Omega$$

C
$$1\ \Omega \ \text{and}\ 0.01\ \Omega$$

D
$$0.1\ \Omega \ \text{and}\ 0.01\ \Omega$$

Solution

Given,
Number of resistors, $$n=10$$
Resistance of each resistors, $$R=0.1\ \Omega$$

For the largest resistance, all ten resistors should be connected in series.
The effective resistance of $$n$$ resistors having equal resistances $$R$$ in series,
$$R_s=n\times R$$
$$R_s=10\times 0.1$$
$$R_s=1\ \Omega$$

For the smallest resistance, all ten resistors should be connected in parallel.
The effective resistance of $$n$$ resistors having equal resistances $$R$$ in parallel,
$$R_p=\dfrac{R}{n}$$

$$R_p=\dfrac{0.1}{10}$$

$$R_p=0.01\ \Omega$$
Question 6
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An electric toaster draws $$8\ A$$ current in a $$220\ V$$ circuit. It is used for $$2\ hr$$. What is the cost of operating the toaster if the cost of electrical energy is $$Rs.\ 4.5 / kWh$$?

A
$$Rs.\ 9.50$$

B
$$Rs.\ 25.50$$

C
$$Rs.\ 14.84$$

D
$$Rs.\ 15.84$$

Solution

Given,
$$i=8\ A\\E=220\ v\\t=2\ hr$$

Energy dissipated is given by:
$$E = E\ i\ t$$
$$=220 \times 8 \times 2\ Wh$$
$$=3.52\ kWh$$

The cost per kWh is $$Rs.\ 4.50$$
Hence, the cost of running the kettle for two hours is:
$$3.52 \times 4.50 = Rs.\ 15.84$$
Question 7
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Fill in the blank.
The heat generated in calorie is equal to _________.

A
$$ VIt$$

B
$$(VIt)/4.2 $$

C
$$(V^2Rt)/4.2 $$

D
$$VI^2t$$

Solution

The right matches are given below.
Heat Generated$$\rightarrow $$Propotional to the square of current$$\rightarrow $$ $$\frac { VIt }{ 4.18 } cal$$
Resistance in parallel$$\rightarrow $$Is used to reduce effective resistance in a circuit$$\rightarrow $$ $$\frac { 1 }{ { R }_{ p } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } $$
Resistivity$$\rightarrow $$Depends on the material of the conductor$$\rightarrow $$ $$\rho =\frac { RA }{ l } $$
Ohm's law$$\rightarrow $$Gives relation between V and I$$\rightarrow $$V=IR
Question 8
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An electric heater of power $$3 \ kW$$ is used for $$10 \ h$$. How much energy does it consume? Express your answer in joule.

A
$$1.08\times 10^5$$

B
$$1.08\times 10^8$$

C
$$1.08\times 10^7$$

D
$$1.08\times 10^3$$

Solution

We know that
$$Power = work/time$$
$$Work \ or\ energy = Power\times time$$
$$Work = 3 kW \times 10 h$$
$$Energy = 30 KWh$$.
$$Energy = 30\times 10^{3}\times 60\times 60\times =1.08\times 10^{8}\ J$$.
Question 9
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A bulb is connected to a cell. How is the resistance of circuit affected if another identical bulb is connected in series ?

A
Resistance is doubled

B
Resistance is same.

C
Resistance is tripled

D
Resistance is one half.

Solution

Components connected in series are connected along a single path, so the same current flows through all of the components. The current through each of the components is the same, and the voltage across the circuit is the sum of the voltages across each component. The total resistance of resistors in series is equal to the sum of their individual resistances. That is,
$${ R }_{ total }={ R }_{ 1 }+{ R }_{ 2 }+{ R }_{ 3 }$$.
When a bulb of the same resistance is connected in the series, the total resistance in the circuit is given as sum of both resistances connected. Hence the resistance is doubled.
Question 10
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Substances which allow electricity to pass through them are known as

A
Electrical conductors

B
Electrical insulators

C
Both conductors & insulators

D
None of these

Solution

Substances which allow electricity to pass through them easily are called conductors of electricity or electrical conductors.For example:Silver,Copper,Aluminium,etc.

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Electricity Test - 42

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Electricity Test - 42

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Question 1

$$\displaystyle \frac {5}{2} \Omega$$

$$\displaystyle \frac {40}{3} \Omega$$

$$40\Omega$$

$$100\Omega$$

Solution

Question 2

15 W

24 W

3.6 W

20 W

Solution

Question 3

$$P_A= 6 W, P_B=9 W$$

$$P_A= 9 W, P_B=6 W$$

$$P_A= 9 W, P_B=3 W$$

$$P_A= 3 W, P_B=6 W$$

Solution

Question 4

always goes from the positive terminal to the negative terminal.

may go from the positive terminal to the negative terminal.

always goes from the negative terminal to the positive terminal.

does not move.

Solution

Question 5

$$10\ \Omega \ \text{and}\ 1\ \Omega$$

$$1\ \Omega \ \text{and}\ 0.1\ \Omega$$

$$1\ \Omega \ \text{and}\ 0.01\ \Omega$$

$$0.1\ \Omega \ \text{and}\ 0.01\ \Omega$$

Solution

Question 6

$$Rs.\ 9.50$$

$$Rs.\ 25.50$$

$$Rs.\ 14.84$$

$$Rs.\ 15.84$$

Solution

Question 7

$$ VIt$$

$$(VIt)/4.2 $$

$$(V^2Rt)/4.2 $$

$$VI^2t$$

Solution

Question 8

$$1.08\times 10^5$$

$$1.08\times 10^8$$

$$1.08\times 10^7$$

$$1.08\times 10^3$$

Solution

Question 9

Resistance is doubled

Resistance is same.

Resistance is tripled

Resistance is one half.

Solution

Question 10

Electrical conductors

Electrical insulators

Both conductors & insulators

None of these

Solution

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