MCQ Questions for Class 10 Physics Electricity Quiz 8

An electricity bulb of 100 watt is connected to supply of electricity of 220V. Resistance of filament is

0%
$\displaystyle 484\Omega$
0%
$\displaystyle 100\Omega$
0%
$\displaystyle 22000\Omega$
0%
$\displaystyle 242\Omega$

Explanation

Given :

$V =220$ volts

$P =100$ W

$\therefore$ Resistance of the filament

$R = \dfrac{V^2}{R}$

$\implies$

$R = \dfrac{220^2}{100} = 484\Omega$

When two resistances

$\displaystyle { R }_{ 1 }$ and

$\displaystyle { R }_{ 2 }$ are connected in series, they consume 12W power. When they are connected in parallel, they consume 50W power. What is the ratio of the powers of

$\displaystyle { R }_{ 1 }$ and

$\displaystyle { R }_{ 2 }$

0%
$1/4$
0%
$4$
0%
$3/2$
0%
$3$

Explanation

Let the powers of the resistances be

$P_1$ and

$P_2$ .

Parallel connection :

$P_{eq} = P_1+P_2$

$\therefore$

$50 = P_1+P_2$

$\implies P_2 = 50-P_1$

Series connection :

$P_{eq} = \dfrac{P_1P_2}{P_1+P_2}$

$\therefore$

$12 = \dfrac{P_1(50 - P_1)}{50}$

$P_1^2 - 50P_1 + 600 = 0$

$\implies$

$P_1 = 30$ W

$\implies$

$P_2 = 50 - 30 = 20$ W

Thus ratio of powers

$\dfrac{P_1}{P_2} = \dfrac{3}{2}$

Watt hour represents:

0%
Electric energy
0%
Current
0%
Voltage
0%
Power

Explanation

The watt-hour (Wh) is a unit of electric energy equivalent to one watt (1 W) of power expended for one hour (1 h) of time. Watt-hour is commonly used in electrical applications.

$1\ Wh = 1\times 60\times 60 = 3600\ J$

Unit used in selling electrical energy to consumer is:

0%
$Volt-Ampere$
0%
$Kilowatt-hour$
0%
$Volt/second$
0%
None of these

Explanation

The commercial unit of electrical energy is

$Kilowatt-hour\ (kWh)$ .

$1\ kWh$ is equal to

$1\ unit$ of energy.

Number of KWh in 1 Joule.

0%
$\displaystyle 3.6\times { 10 }^{ 6 }\ KWh$
0%
$\displaystyle 2.77\times { 10 }^{ -7 }\ KWh$
0%
$\displaystyle 600\ KWh$
0%
$\displaystyle 1.6\times { 10 }^{ -19 }\ KWh$

Explanation

We know that

$KWh$ is kilo watt-hour-

In SI unit-

$K = 10^3$

$hour = 3600\ s$

$1\ kWh = 3.6\times 10^6\ J$

$\therefore$

$1$

$J = \dfrac{1}{3.6\times 10^6} = 2.77\times 10^{-7}$

$kWh$

In the circuit,

$\displaystyle 5\Omega$ resistor develops 45J/s due to current flowing through it. The power developed per second across

$\displaystyle 12\Omega$ resistor is:

0%
16W
0%
192W
0%
36W
0%
64W

Explanation

Voltage across

$5\Omega$

$V = \sqrt{PR} = \sqrt{45\times 5} = 15$ volts

$\therefore$ Current

$I_1 = \dfrac{15}{5} = 3$ A

Equivalent resistance of series combination of

$3\Omega$ and

$6\Omega$

$R' = 9+6 = 15\Omega$

$\therefore$ Current

$I_2 = \dfrac{15}{15} = 1$ A

Thus total current flowing through

$12\Omega$

$I = I_1+I_2 = 3+1 = 4$ A

$\therefore$ Power developed across

$12\Omega$

$P = I^2 (12) = 4^2\times 12 = 192$ W

The equivalent resistance of parallel combination is:

0%
Lower than the value of lowest resistance in the combination
0%
Higher than the value of lowest resistance in the combination
0%
Both A and B
0%
None of these

Explanation

Consider two resistors having resistances

$R_1$ and

$R_2$ .

The equivalent resistance of

$R_1$ and

$R_2$ in parallel is,

$\dfrac{1}{R_{eq}}= \dfrac{1}{R_1}+\dfrac{1}{R_2}$

$\implies R_{eq}=\dfrac{R_1R_2}{R_1+R_2}$

For any values of

$R_1$ and

$R_2$ ,

$R_{eq}$ will be less than

$R_1$ and

$R_2$ .

For example,

Let

$R_1=2\ \Omega$

$R_2=6\ \Omega$

Then,

Equivalent resistance of

$R_1$ and

$R_2$ in parallel,

$R_{eq}=\dfrac{R_1R_2}{R_1+R_2}$

$R_{eq}=\dfrac{2\times 6}{2+8}$

$R_{eq}=\dfrac{12}{8}$

$R_{eq}=1.5\ \Omega$

$\implies R_{eq} \lt R_1\ \text{and}\ R_{eq} \lt R_2$

How much current is drawn by the motor of 1 H.P. from 220 volt supply.

0%
3.4A
0%
2A
0%
6A
0%
22A

Explanation

We know that

$1$ H.P

$= 746$ watt

Current

$I = \dfrac{P}{V} = \dfrac{746}{220} = 3.4$ A

A bulb of

$22\ \Omega$ is producing light when connected to a

$220\ V$ supply. What is the electric power of the bulb?

0%
$2200\ W$
0%
$1000\ W$
0%
$22\ W$
0%
$220\ W$

Explanation

Given,

Resistance of the bulb,

$R = 22\Omega$

Voltage,

$V = 220$ volts

We know,

Power of the bulb,

$P = \dfrac{V^2}{R}$

$P = \dfrac{220^2}{22}$

$P =2200\ W$

Calculate the monthly bill if a heater of 100 watt is used at the rate of rs. 1 per unit for 1 hour daily.

0%
Rs. 200
0%
Rs. 6000
0%
Rs. 100
0%
Rs. 3

Explanation

Given :

$P=0.1$ kW

$t = 1$ hour per day

Energy consumed per day

$E = Pt = 0.1\times 1 = 0.1$ kWh per day

Total energy consumed per month

$E_T = 30\times 0.1 = 3$ kWh per month

Cost of electricity

$=Re$

$1$ per unit i.e per kWh

$\therefore$ Monthly bill

$= Re.$

$1\times 3 =$ Rs.

$3$ per month

An electric bulb of resistance of

$1000 \Omega$ draws a current of

$0.05 A$ . Calculate the power of the bulb :

0%
$500 \ W$
0%
$25 \ kW$
0%
$250 \ W$
0%
$650 \ W$

Explanation

Given:

$Resistance(R) = 1000 \Omega, Current(I) = 0.5 A$
We know that, Power

$P = {I}^{2} \cdot R = {\left(0.5 A\right)}^{2} \times \left(1000\right) \Omega = 250 W$ .

An electric bulb of resistance

$20 \Omega$ draws a current of

$0.04 A$ . Calculate the potential difference at the ends.

0%
$0.8 V$
0%
$2 V$
0%
$0.032 V$
0%
$5 V$

Explanation

Given,

Resistance of resistor

$R = 20 \Omega$

Current draws

$I = 0.04 A$

According to ohm's law

Potential difference,

$V = IR = 0.04 \times 20 = 0.8 V$ .

Hence Option A

The attached circuit diagram shows connection of 3 resistors. All are connected in parallel . Thus what is the equivalent resistance.

0%
1 ohm
0%
4 ohm
0%
3ohm
0%
2 ohm

Explanation

Equivalent resistance of resistors connected in parallel

$\dfrac{1}{R_{eq}} =\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$

$\therefore$

$\dfrac{1}{R_{eq}} =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}$

$\implies R_{eq} = 1$ ohm

A certain household has consumed

$250$ units of energy during a month. How much energy is this in joules?

0%
$9 \times 10^{10}\ J$
0%
$9 \times 10^{8}\ J$
0%
$9 \times 10^{9}\ J$
0%
$9 \times 10^{-8}\ J$

Explanation

Hint: Conversion of KWh to joule.

Step 1: Commercial unit of energy is KWh.

$1\; KWh = 10^3\; W = 10^3\; JS^{-1}$

$1\;hr = 60\times60 = 3600\;sec$

Therefore,

$1\;KWh = 1\; KW \times1\;hr$

$=10^3 \times 3600$

$=3.6 \times 10^6 \;J$

Step 2: Calculation according to question

Energy consumption

$=250\;KWh$

$= 250\times3.6\times10^6\;J$

$=900\times10^6\;J$

$=9\times10^8\;J$ or

$900\;MJ$

Hence, Option (B) is correct.

Find energy in kWh consumed in 10 hours by four devices of power

$500\ W$ each?

0%
$10\ kWh$
0%
$20\ kWh$
0%
$30\ kWh$
0%
$40\ kWh$

Explanation

Hint:

Power is the ratio of the work and the time. Energy consumed in four devices is four times the energy consumed in one device.

Step 1: Given data

Power consumed by each device, $P=500\ W=0.50\ kW$

Time, $t=10\ h$

Expression for the energy consumed is written as,

$W=P\times t$

Step 2: Calculations,

Energy consumed by each device, $W=0.50 \times 10 = 5kWh$

Energy consumed by four devices, $= 4 \times 5 = 20\ kWh$

Smallest commercial unit of energy is

0%
Kilowatt hour
0%
Watt second
0%
Watt hour
0%
Watt minutes

Explanation

Smallest commercial unit of energy is called Watt hour

$(Wh)$ . One watt hour is the energy consumed when

$1 \ watt$ of power is used for

$1 \ hour.$

Convert

$28.8 kJ$ into kilo watt hour.

0%
$8 \times {10}^{-3} kwh$
0%
$8 kwh$
0%
$8 \times {10}^{3} kwh$
0%
None of these

Explanation

We know that

$1 kWh=3.6 \times 10^6$

$Joules$

$E=28.8 \times 10^3$

$Joules$

$E = \dfrac{28.8 \times {10}^{3}}{3.6 \times {10}^{6}}$

$= 8 \times {10}^{-3} kwh$

Calculate the equivalent resistance when two resistances of

$3\ \Omega$ and

$6\ \Omega$ are connected in parallel.

0%
$1 \ \Omega$
0%
$3 \ \Omega$
0%
$2 \ \Omega$
0%
$6 \ \Omega$

Explanation

We know that in parallel combination, equivalent resistance is given by-

$\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}$

$\implies \dfrac{1}{R}=\dfrac{1}{3}+\dfrac{1}{6}$

$\implies \dfrac{1}{R}=\dfrac{2+1}{6}=\dfrac{1}{2}$

$\implies R=2\ \Omega$

Answer-(C)

If n resistance (each R) are connected in parallel then the resultant will be

0%
$nR$
0%
$R/n$
0%
$\displaystyle { n }^{ 2 }R$
0%
$\text{None}$

Explanation

Equivalent resistance of parallel combination

$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+.........$

Given, Resistanvce of all resistors

$=R$

$\therefore$

$\dfrac{1}{R_{eq}} = \dfrac{1}{R}+\dfrac{1}{R}+\dfrac{1}{R}+.........n$ terms

$\dfrac{1}{R_{eq}} = \dfrac{n}{R}$

$\implies$

$R_{eq} = \dfrac{R}{n}$

Commercial unit of energy is .......................

0%
Joule
0%
Volt
0%
Kilowatt hour
0%
Watt second

Explanation

Commercial unit of electrical energy is kilowatt hour

$\left(kWh\right)$

$1 kwh = 3.6 \times {10}^{6} J$

Whenever an electric current is passed through a conductor, it gets heated up. this means that a part of electrical energy given to the conductor gets converted to

0%
potential energy
0%
gravitational energy
0%
heat energy
0%
none of these

Imagine that you have a 100

$\Omega$ resistor. You want to add a resistor in series with this 100

$\Omega$ resistor in order to limit the current to 0.5A when

$110$ Volts is placed across two resistors in series. How much resistance should you use?

0%
$100\Omega$
0%
$120 \Omega$
0%
$250 \Omega$
0%
$300 \Omega$

Explanation

Given :

$V = 110$ volts

$I = 0.5$ A

$R_1 = 100\Omega$

Let a resistance

$R$ is connected in series with

$R_1$ .

Using Ohm's law,

$V = I(R_1+R)$

$\therefore$

$110 = 0.5(100+R)$

$\implies R = 120\Omega$

What is the highest equivalent resistance that can be secured by combinations of four coils of resistance

$4\ \Omega$ ,

$8\ \Omega$ ,

$12\ \Omega$ ,

$24\ \Omega$ ?

0%
$24\ \Omega$
0%
$48\ \Omega$
0%
$12\ \Omega$
0%
$4\ \Omega$

Explanation

Connecting resistances in series with each other increases the equivalent resistance of the combination.

Let

$4\ \Omega$ ,

$8\ \Omega$ ,

$12\ \Omega$ , and

$24\ \Omega$ resistors are connected in series.

$\therefore$ Equivalent resistance of the circuit

$R_{eq} = 4+8+12+24$

$\Rightarrow R_{eq} = 48$ ohm

A wire of resistance R is elongated n-fold to make a new uniform wire. The resistance of new wire

0%
$nR$
0%
$n^2R$
0%
$2nR$
0%
$2n^2R$

Explanation

We know that resistance

$R=\rho \dfrac{l}{A}$ , where

$\rho$ is the resistivity of the material,

$l$ is the length of the wire, and

$A$ is the cross-sectional area of the wire.

The wire is elongated keeping the volume fixed.

$\text{Old Volume} = \text{New volume}$

$\Rightarrow l\times A = l'\times A'$

$\Rightarrow l\times A=nl\times A'$

$\Rightarrow A'=\dfrac{A}{n}$

Thus, the area is decreased by n times.

So, the altered resistance would be

$R'=\rho \dfrac{nl}{\dfrac{A}{n}}=n^2 \rho \dfrac{l}{A}=n^2R$

A piece of wire of resistance

$R$ is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is

$R'$ , then the ratio

$\dfrac{R}{R'}$ is -

0%
$\dfrac{1}{25}$
0%
$\dfrac{1}{5}$
0%
$5$
0%
$25$

Explanation

$\textbf{Step 1: Resistance of each part after cutting}$

We know that, Resistance

$R = \rho \dfrac{\ell}{A}$

Since,

$R$ is directly proportional to length of the wire.

Therefore the resistance of every part after cutting the wire in

$5$ parts is

$R_{1} = R_{2} = R_{3} = R_{4} = R_{5} = \dfrac{R}{5}$

$\textbf{Step 2: Equivalent Resistance}$

When all the resistances are connected in parallel, let R' be the equivalent resistance

So,

$\dfrac{1}{R'} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3} + \dfrac{1}{R_4} + \dfrac{1}{R_5}$

$\Rightarrow \dfrac{1}{R'} = \dfrac{5}{R} + \dfrac{5}{R} + \dfrac{5}{R} + \dfrac{5}{R} + \dfrac{5}{R}$

$\Rightarrow R' = \dfrac{R}{25}$

$\Rightarrow$

$\dfrac{R}{R'} = 25$

Hence, Option (D) is correct.

The resistance across

$A$ and

$B$ in the given figure will be:

0%
$3R$
0%
$R$
0%
$\cfrac{R}{3}$
0%
None of the above

Explanation

Given:

$R_1 = R_2 = R_3 = R$

The three resistances are connected in parallel between the points A and B.

Equivalent resistance for parallel combination:

$\Rightarrow \dfrac{1}{R_{AB}} =\dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}$

$\Rightarrow \dfrac{1}{R_{AB}} =\dfrac{1}{R} + \dfrac{1}{R} + \dfrac{1}{R} = \dfrac{3}{R}$

$\Rightarrow$

$R_{AB} = \dfrac{R}{3}$

An electric bulb is rated

$220\ V$ and

$100\ W%$ . When it is operated on

$110\ V$ , the power consumed will be:

0%
$100\ W$
0%
$75\ W$
0%
$50\ W$
0%
$25\ W$

Explanation

$P=V^2/R$

$220\ V,$

$100=220^2/R \quad ..........(i)$

$110\ V,$

$P=110^2/R\quad ...........(ii)$

Dividing (i) from (ii),

$P/100=110^2/220^2$

$P=25\ W$

In shown diagram, if voltage of battery is doubled and resistance kept constant then what is the new Power dissipated at resistor(Consider

$P$ was the initial power dissipation) ?

0%
$\dfrac{P}{4}$
0%
$\dfrac{P}{2}$
0%
P
0%
2P
0%
4P

Explanation

Power,

$P=\dfrac{V^2}{R}$

As R is kept constant so

$P \propto V^2$

$P'$ is new power dissipation.

So,

$\dfrac{P'}{P}=\dfrac{(2V)^2}{V^2}=\dfrac{4V^2}{V^2}$ or

$P'=4P$

Thus, option E is correct.

Five resistors each of

$1\ ohm$ are connected in parallel. The resultant resistance of the combination is

0%
$5\Omega$
0%
$1\Omega$
0%
$0.2\Omega$
0%
$2\Omega$

Explanation

All resistors are connected in parallel

$\dfrac{1}{R_e}=\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}+\dfrac{1}{R_4}+\dfrac{1}{R_5}$

$\dfrac{1}{R_e}=\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1} = \dfrac{5}{1}$

$R_e=\dfrac{1}{5}=0.2\Omega$

If a third identical resistor is added in parallel in a circuit of two identical parallel resistors. Calculate the ratio of the new equivalent resistance to the old?

0%
$\dfrac{4}{9}$
0%
$\dfrac{2}{3}$
0%
$1$
0%
$\dfrac{3}{2}$
0%
$\dfrac{9}{4}$

Explanation

Let the resistance of each resistor be

$R$ .

Equivalent resistance of two resistors in parallel,

$\dfrac{1}{R_{eq}} = \dfrac{1}{R} + \dfrac{1}{R}$

$\implies R_{eq} = \dfrac{R}{2}$

Equivalent resistance of three resstors in parallel,

$\dfrac{1}{R'_{eq}} = \dfrac{1}{R} + \dfrac{1}{R}+ \dfrac{1}{R}$

$\implies R'_{eq} = \dfrac{R}{3}$

$\therefore\ \dfrac{R'_{eq}}{R_{eq}} = \dfrac{R/3}{R/2} = \dfrac{2}{3}$

Three resistor of

$10 \Omega$ ,

$20 \Omega$ , and

$30 \Omega$ are connected in parallel in a circuit. Calculate the equivalent resistance.

0%
$6000 \ \Omega$
0%
$0.017 \ \Omega$
0%
$1 \ \Omega$
0%
$60 \ \Omega$
0%
$5.45 \ \Omega$

Explanation

Given :

$R_1 = 10\Omega$

$R_2= 20\Omega$

$R_3 = 30\Omega$

Equivalent resistance of parallel combination

$\dfrac{1}{R_{eq}} = \dfrac{1}{R_1} +\dfrac{1}{R_2} + \dfrac{1}{R_3}$

$\therefore$

$\dfrac{1}{R_{eq}} = \dfrac{1}{10} +\dfrac{1}{20} + \dfrac{1}{30}$

$\implies R_{eq} = 5.45$

$\Omega$

Three resistors

$R_1, R_2$ , and

$R_3$ are connected as shown below.
The resistance of

$R_3$ is

$1 \Omega$ and it is known that the value of

$R_3$ is smaller than the values of

$R_1$ or

$R_2$ . The resistances of

$R_1$ and

$R_2$ are unknown.
Select the best statement describing the equivalent resistance of this combination:

0%
The equivalent resistance of this combination must be lower than $1\Omega$ .
0%
The equivalent resistance of this combination may be lower than $1\Omega$ .
0%
The equivalent resistance of this combination must be greater than $1\Omega$ .
0%
The equivalent resistance of this combination may be greater than $1\Omega$ .
0%
The equivalent resistance of this combination cannot be estimated without additional information.

Explanation

The resistances in given diagram are in parallel combination , therefore equivalent resistance of the given combination ,

$1/R=1/R_{1}+1/R_{2}+1/R_{3}$ ,

it is clear from the equation that value of

$R$ is less than

$R_{1}$ or

$R_{2}$ or

$R_{3} =1$ , therefore equivalent resistance of this combination is less than 1 ohm .

In a DC circuit when a switch is closed to operate the circuit, the electrons that cause the current:

0%
travel from the negative plate of the battery to the positive plate
0%
travel from the positive plate of the battery to the negative plate
0%
travel from one resistance to the next resistance
0%
travel from the negative plate of the battery to the positive plate inside the battery
0%
travel from one atom in the circuit to the next atom

A current is flowing in a circuit consisting of a resistor and a battery.
What happens to the power dissipated in the resistor when the resistance is quadrupled and the voltage remains constant?

0%
It is doubled
0%
It is quadrupled
0%
It is halved
0%
It is quartered
0%
It remains the same

Explanation

Let the voltage of the battery be

$V$ and initial resistance be

$R$ .

Thus, the initial power dissipated is

$P= \dfrac{V^2}{R}$

Now the resistance is quadrupled keeping the voltage constant i.e

$R' = 4R$

$\therefore$ New power dissipattion is

$P' = \dfrac{V^2}{R'} = \dfrac{V^2}{4R} = \dfrac{P}{4}$

Hence the power dissipated is quartered.

Calculate the unknown resistance R of the circuit as shown in figure, all resistance are connected in the series. The current is flowing through circuit is 2A and battery is of 20 voltage.

0%
$1 \Omega$
0%
$2 \Omega$
0%
$4 \Omega$
0%
$6 \Omega$
0%
$12 \Omega$

Explanation

Given :

$V =20$ volts

$I = 2$ A

Total resistance of the circuit

$R_{eq} =R + 4+4 = R+8$

Using

$V = IR_{eq}$

$\therefore$

$20 =2 (R+8)$

$\implies R = 2\Omega$

A 100 and a 50 ohm resistor are wired to a 12 V battery as shown in the diagram below. From the list of statements listed below please select the correct statements.

0%
These resistors are wired in parallel.
0%
These resistors are wired in series.
0%
The potential difference across these resistors is the same.
0%
The potential difference across the 100 resistor is higher.
0%
The electrical current passing through the 50 resistor is half the current through the 100 resistor.

Explanation

The resistors are wired in parallel as their initial and final points are attatched to each other , bieng in parallel the potential difference across both resistors is same .

Now equivalent resistance in circuit ,

$R=\frac{50\times100}{50+100}=\frac{5000}{150}=33.3\Omega$

therefore total current ,

$I=E/R=12/33.3=0.36A$

current in the

$50\Omega$ resistor ,

$I_{1}=E/50=12/50=0.24A$

current in the

$100\Omega$ resistor ,

$I_{1}=E/100=12/100=0.12A$ ,

it is clear that current in 50 resistor is double the current through the 100 resistor .

Three resistors (

$2, 5$ and

$7$ ohm) are wires as shown in the diagram. The equivalent resistance of this combination (in ohm) is:

0%
$\dfrac{59}{70}$
0%
$\dfrac{70}{59}$
0%
$\dfrac{1}{14}$
0%
$14$
0%
Cannot be determined without additional information.

Explanation

Given :

$R_1 = 2\Omega$

$R_2= 5\Omega$

$R_3 = 7\Omega$

Equivalent resistance of parallel combination

$\dfrac{1}{R_{eq}} =\dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}$

$\therefore$

$\dfrac{1}{R_{eq}} =\dfrac{1}{2}+\dfrac{1}{5}+\dfrac{1}{7}$

$\Rightarrow \dfrac{1}{R_{eq}}=\dfrac{35+14+10}{70}$

$\Rightarrow R_{eq} =\dfrac{70}{59}$

$\Omega$

This question relates to the DC circuit shown below.
The values for current, voltage, and resistance in the circuit are graphed from point A through point G in the graphs directly below.
Which of the graphs shows the resistance from point A to point G?

0%
A
0%
B
0%
C
0%
D
0%
E

An electric circuit contains an operating heating element and a lit lamp. Which statement best explains why the lamp remains lit when the heating element is removed from the circuit?

0%
The lamp and heating element were connected in parallel.
0%
The lamp has less resistance than the heating element.
0%
The lamp has more resistance than the heating element.
0%
The lamp and heating element were connected in series.

All resistors in the circuit have the same resistance,

$R$ , and the battery is having a constant voltage of

$V$ .
Find out power dissipated by resistor

$f$ . Given that the power dissipated by resistor

$e$ is

$P$ :

0%
$\dfrac{P}{6}$
0%
$\dfrac{P}{3}$
0%
$\dfrac{P}{2}$
0%
$P$
0%
$2P$

Explanation

The power dissipated by a resistor is

$i^2R$

where

$i$ is the current through the resistor of resistance

$R$ .

Since both

$i$ and

$R$ for the resistors

$e$ and

$f$ are equal (they are of equal resistance and connected in series), the power dissipated by both of them is the same and equal to

$P.$

Two electrical bulbs have tungsten filament of same length. If one of them gives 60 watts and other 100 watts, then

0%
100 watts bulb has thicker filament
0%
60 watt bulb has thicker filament
0%
Both filaments are of same thickness
0%
It is impossible to get different wattage unless lengths are different

Explanation

power

$P=\dfrac{V^2}{R}$ , where

$V$ is the voltage and

$R$ is the resistance.

So for higher power, the resistance should be less.

Therefore, the resistance of

$100\ W$ bulb should be less.

We know that resistance

$R=\dfrac{\rho l}{A}$ , where

$\rho$ is the resistivity,

$l$ is the length, and

$A$ is the area.

when the resistivity and the length are fixed, the area should be more for less resistance.

Thus, the

$100\ W$ bulb should have a thicker filament.

What is the power dissipated by each lamp of resistance

$60 \Omega$ ?

0%
1 W
0%
2 W
0%
6 W
0%
12 W

Explanation

Resistance are in series

So total resistance

$R_T=R+R=6+6=12$

current,

$I=\dfrac{V}{R_T}=\dfrac{12}{12}=1A$

Power dissipated across each reistance

$P=i^2R=1^2\times 6=6W$

Metals are good conductor because

0%
outer electrons are strongly bound to the atom.
0%
outer electrons are loosely bound to the atom.
0%
inner electrons are loosely bound to the atom.
0%
protons can detach from the nucleus and conduct electricity.

If four resistances, each of value

$1\ ohm$ , are connected in series, what will be the resultant resistance?

0%
3 ohm
0%
4 ohm
0%
5 ohm
0%
2 ohm

Explanation

In series combination , equivalent resistance is given by-

$R=R_1+R_2+R_3+R_4+R_5+.....$

Here there are four resistors.

$R=1+1+1+1$

$\implies R=4\Omega$

Answer-(B)

The dry cell of E.M.F

$12\ V$ , is connected in series to a

$3 \Omega$ and a

$6 \Omega$ resistor. Calculate the total power supplied by the dry cell.

0%
$9\ W$
0%
$72\ W$
0%
$16\ W$
0%
$2\ W$

Explanation

Equivalent resistance for resistance connected in series is given by

$R_{eq} = R_1 +R_2$

Since, in the given case

$3 \ \Omega$ and

$6 \ \Omega$ are connected in series.

Hence, equivalent resistance,

$R_{eq} =3+6=9 \ \Omega$

Given, emf of cell

$V = 12 \ V$

Hence, the power supplied by the cell is given by,

$P = \dfrac{V^2}{R}$

$P = \dfrac{12 \times 12}{9} = 16 \ W$

2104896_523200_ans_ca3ff680ee5446e3b6272ad687d25bac.png

An air conditioner is rated 240 V, 1.5 kW. The air conditioner is switched on 8 hours each day. What is electrical energy consumed in 30 days?

0%
2.88 kWh
0%
360 kWh
0%
120 kWh
0%
240 kWh

Explanation

Given,

Power of air conditioner

$P=1.5kW$

Time to use per day

$t=8\ hr$

Time to use in 30 days

$T=8\times 30\ hr$

Energy consumed

$E=Power\times time =1.5\ kW\times 8\times 30\ hr=360kWh$

Option B

Electrical power P is given by the expression P =

$\dfrac{Q \times V}{t}$ . time. What does

$Q$ and

$V$ stand for?

0%
Charge, energy
0%
Current, voltage
0%
Current, power
0%
Charge, voltage

Explanation

$Q$ is the symbol for charge.

$V$ is the symbol for voltage.

Two resistance are connected in series as shown in diagram. What is the current through the

$5\ ohm$ resistance?

0%
$2 A$
0%
$3 A$
0%
$4 A$
0%
$5 A$

Explanation

Voltage across

$5\Omega$ resistor is

$V=10V$

Current,

$I=\dfrac{V}{R}=\dfrac{10}{5}$

$\implies I=2A$

Answer-(A)

Two resistance are connected in series as shown in diagram. What is the value of

$R$ ?

0%
$4 \Omega$
0%
$6 \Omega$
0%
$5 \Omega$
0%
$3 \Omega$

Explanation

Voltage across

$5\Omega$ resistor is

$V=10V$

Current,

$I=\dfrac{V}{R}=\dfrac{10}{5}$

$\implies I=2A$

Since resistance

$5\Omega$ and

$R$ are in series and hence current through both resistance is same.

Hence current through

$R$ is also

$2A$ .

Now for

$R$ , potential difference is

$V=6V$ and

$I=2A$ .

Hence ,

$R=\dfrac{V}{I}=\dfrac{6}{2}$

$\implies R=3\Omega$

Answer-(D)

When a number of resistances are connected in parallel then combined resistance is less than the smallest individual resistance.

0%
True
0%
False

Explanation

When two resistors

$R$ and

$r$ are connected in paralle.

Equivalent resistance

$=R_o=\dfrac{Rr}{R+r}=\dfrac{R}{1+\dfrac{R}{r}}$

Since,

$1+\dfrac{R}{r}>1$

$\implies R_o<R$

Similarly,

$R_o<r$

Hence, when resistors are connected in parallel, equivalent resistance is less than each individual resistance.

Given statement is

$True$ .

CBSE Questions for Class 10 Physics Electricity Quiz 8 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

CBSE Questions for Class 10 Physics Electricity Quiz 8 - MCQExams.com

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Practice Class 10 Physics Quiz Questions and Answers

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