Question Set C

cbse 10th Standard Science Electricity Important Questions And Answers Study Material - QB365

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Questions:

Generic 2

8 x 2 = 16

Why is it not advisable to connect electric bulb and electric heater in series?

Calculate the electric energy consumed by 120 W toaster in 20 minutes.

Why is the current constant in series connection of circuit?

An electric iron has a rating of 750 W, 220 V. Calculate
(i) current passing through it, and
(ii) its resistance, when in use

A bulb is rated at 5.0 volt, 100 mA. Calculate its
(i) power and
(ii) resistance

(i) Draw a diagram to show how two resistor R₁ and R₂ are connected in series.
(ii) In a circuit, if the two resistors of 5 ohm and 10 ohm are connected in series, how does the current passing through the two resistors compare?

An electric lamp is marked 100 W, 220 V. It i used for 5 hours daily. Calculate.
(i) its reistance while glowing
(ii) energy consumed in kWh per day

An electric heater is used on 220 V supply and takes a current of 3.4 A. Calculate (i) its power and (ii) its power and (ii) its resistance when it is in use.

Generic 3

2 x 3 = 6

If a 12 V battery is connected to the arrangement of resistances given below, calculate
(i) the total effective resistance of the arrangement and
(ii) the total current flowing in the circuit

10.

(i) Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 Ohm resistor, a 10 Ohm resistor and a 15 Ohm resistor, and a plug key, all connected in series.
(ii) Calculate the electric current passing through the above circuit when the key is closed.

Multiple Choice Question

19 x 1 = 19

11.

The proper representation of series combination of cells obtaining maximum potential is

(a)

(b)

(c)

(d)

12.

Identify the circuit in which the electrical components have been properly connected.

(a)

(b)

(c)

(d)

13.

In the following circuits, heat produced in the resistor or combination of resistors connected to a 12 V battery will be

(a)

same in all the cases

(b)

maximum in case

(c)

maximum in case

(d)

minimum in case

14.

A cell, a resistor, a key and ammeter are arranged as shown in the circuit diagrams. The current recorded in the ammeter will be

(a)

maximum in

(b)

maximum in

(c)

maximum in

(d)

the same in all the cases

15.

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combination would be

(a)

1:2

(b)

2:1

(c)

1:4

(d)

4:1

16.

Which of the following terms does not represent electrical power in a circuit?

(a)

I²R

(b)

IR²

(c)

(d)

V²/R

17.

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 R/R' is

(a)

1/25

(b)

1/5

(c)

(d)

18.

Unit of electric power may also be expressed as

(a)

volt ampere

(b)

kilowatt hour

(c)

watt second

(d)

joule second

19.

Two resistors of resistance 2\(\Omega \) and 4\(\Omega \) when connected to a battery will have

(a)

same current flowing through them when connected in parallel

(b)

same current flowing through them when connected in series

(c)

same potential difference across them when connected in series

(d)

different potential difference across them when connected in parallel

20.

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)

(b)

(c)

(d)

21.

In an electrical circuit two resistors of 2\(\Omega \) and 4\(\Omega \) respectively are connected in series to a 6 V battery. The heat dissipated by the 4\(\Omega \) resistor in 5s will be

(a)

5 J

(b)

10 J

(c)

20 J

(d)

30 J

22.

In an electrical circuit three incandescent bulbs A, B and C of rating 40 W, 60 W and 100 W respectively are connected in parallel to an electric source. Which of the following is likely to happen regarding their brightness?

(a)

Brightness of all the bulbs will be the same

(b)

Brightness of bulb A will be the maximum

(c)

Brightness of bulb B will be more than that of A

(d)

Brightness of bulb B will be less than that of B

23.

The resistivity does not charge if

(a)

the material is charged

(b)

the temperature is charged

(c)

the shape of the resistor is charge

(d)

both material and temperature are changed

24.

If the current through a resistor is increased by 100% (assume that temperature remains unchanged), the increase in power dissipated will be

(a)

100%

(b)

200%

(c)

300%

(d)

400%

25.

A cylindrical conductor of length l and uniform area of cross section A has resistance R, another conductor of length 2l and resistance R of the same material has area of cross section?

(a)

A/2

(b)

3A/2

(c)

(d)

26.

Which of the following represents voltage?

(a)

\(\frac { Work\ done }{ current\ \times \ time } \)

(b)

Work done \(\times\) charge

(c)

\(\frac { Work\ done \ \times \ time }{ current } \)

(d)

Work done \(\times\) charge \(\times\) time

27.

What is the maximum resistance while can be made using five resistors each of 1/5\(\Omega \).

(a)

1/5\(\Omega \)

(b)

10\(\Omega \)

(c)

1/10\(\Omega \)

(d)

25\(\Omega \)

28.

A current of 1 A is drawn by a filament of an electric bulb. Number of electrons passing through a cross section of the filament in 16 seconds would be roughly

(a)

10²⁰

(b)

10¹⁶

(c)

10¹⁸

(d)

10²³

29.

Electrical resistivity of a given metallic wire depends upon

(a)

its length

(b)

its thickness

(c)

its shape

(d)

nature of the material

Generic 5

1 x 5 = 5

30.

Find out the following in the electric circuit given in Figure
(a) Effective resistance of two 8 \(\Omega\) resistors in the combination
(b)Current flowing through 4 \(\Omega\) resistor
(c) Potential difference across 8 \(\Omega\) resistance
(d) Power dissipated in 4 \(\Omega\) resistor
(e) Difference in ammeter readings, if any.

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Answers Key:

Generic 2

8 x 2 = 16

It is not advisable to connect electric bulb and electric heater in series because both the devices offer different resistance and the requirement of current for functioning is different. Moreover, the devices may damage and not function properly.
If one of the device stops working or fails then the current will not flow through the circuit.

P = 120 W
t = 20 minutes E = P\(\times\)t
E = ? = 120 \(\times\) 20 \(\times\) 60 = 120 \(\times\) 1200
E = 144 \(\times\) 10³J

The number of electrons flowing through the circuit will remain constant. Therefore the current flowing will also be constant. The number of electrons have to travel in a fixed path. Therefore, the value of current is same at each and every point.

(i) Electric power is related to voltage and electric current as:
Power= Voltage x Current
It is given that the rating of the electric iron is 750W, 220V.
Let the current passing through the bulb be I.
Therefore,
750=220 x I
I=\(\frac{750}{220}\)=3.41A
Therefore, 3.41A current passes through the electric iron.
It is given that the rating of the electric bulb is 750W,220 V
Therefore, P=\(\frac { { V }^{ 2 } }{ P } \)=\(\frac { { 220 }^{ 2 } }{ 750 } \)=64.53\(\Omega \) when in use.
(ii) Electric power is related to voltage and resistance as:
\(\frac { { \left( Voltage \right) }^{ 2 } }{ Resistance } \)
That is,
P=\(\frac { { V }^{ 2 } }{ R } \)
=64.53\(\Omega \)

Voltage (V) = 5.0 V
Current (I) = 100 mA = 0.1 A
(i) Power = VI = 5\(\times\)0.1= 0.5W
(ii) Resistance = \(\frac { V }{ I } =\frac { 5 }{ 0.1 } =50\Omega \)

(i) Resistance of a glowing lamp is related to its power and voltage as
Power=\(\frac { { \left( voltage \right) }^{ 2 } }{ Resistance } \)
or, P=\(\frac { { V }^{ 2 } }{ R } \)
Therefore, R=\(\frac { { V }^{ 2 } }{ P } \)=\(\frac { { 220 }^{ 2 } }{ 100 } \)=484\(\Omega \)
Therefore, the resistance of the bulb when glowing is 484\(\Omega \)
(ii) Power=100 W=0.1 kW
Energy= Power x time
=0.1 kW x 5h
=0.5 kWh
0.5 kWh is the amount of energy is consumed by the bulb per day.

(i) Power = Voltage Current
= 220\(\times\)3.4
= 748 Watt
(ii) Power = Voltage / Resistance
P = V²/ R
R = V²/ P=220² / 748 = 64.71 \(\Omega \)

Generic 3

2 x 3 = 6

(i) The 5 Ω resistors are connected in series. Therefore, their effective resistance = (5 + 5) = 10 Ω
The 10 Ω resistors are connected in series. Therefore, their effective resistance = (10 + 10) = 20 Ω
Now these 10 Ω equivalent and 20 Ω equivalent are connected in parallel. Therefore, the equivalent resistance   (R_eq) will be:
   \(\frac { 1 }{ { R }_{ eq } } =\frac { 1 }{ { R }_{ 1 } } +\frac { 1 }{ { R }_{ 2 } } \)
     \(\frac { 1 }{ { R }_{ eq } } =\frac { 1 }{ 10 } +\frac { 1 }{ 20 } \)
     \(\frac { 1 }{ { R }_{ eq } } =\frac { 2+1 }{ 20 } \)
     \(\frac { 1 }{ { R }_{ eq } } =\frac { 20 }{ 3 } =6.76\Omega \)
(ii) Total Current = Total voltage / Total equivalent resistor
   =12 / 6.67
   = 1.8 A.

10.

(ii) When the key is closed, total voltage applied across the circuit will be = 2 ✕ 5V = 10 V
Total resistance (R) applied across the circuit in Ohms = 5+10+15= \(30\Omega \)
   By Ohm's Law,
   Current (I) = \(\frac { V }{ R } \)
     \(I=\frac { 10 }{ 30 } =0.33A\)

Multiple Choice Question

19 x 1 = 19

11.

12.

13.

14.

(d)
the same in all the cases

15.

16.

(b)
IR²

17.

(d)
25

18.

(a)
volt ampere

19.

(b)
same current flowing through them when connected in series

20.

(d)
5A

21.

22.

23.

24.

25.

26.

(a)
\(\frac { Work\ done }{ current\ \times \ time } \)

27.

(b)
10\(\Omega \)

28.

(a)
10²⁰

29.

(d)
nature of the material

Generic 5

1 x 5 = 5

30.

(a) Since two \(8\Omega \) resistors are in parallel, their effective resistance (\({ R }_{ P }\)) is given by,
\(\frac { 1 }{ { R }_{ P } } =\frac { 1 }{ 8 } +\frac { 1 }{ 8 } +\frac { 1 }{ 4 } \) or \({ R }_{ P }=4\Omega \)
(b) Total resistance in the circuit
\(R=4\Omega +{ R }_{ P }=4\Omega +4\Omega =84\Omega \)
Current through the electric circuit
\(I=\frac { V }{ R } =\frac { 8V }{ 8\Omega } =1A\)
(c) The potential difference \(4\Omega \) resistor V
I \(=IR=1\times 4=4V\)
(d) Power dissipated in \(4\Omega \) resistor P=\({ (I) }^{ 2 }\)
\(R={ (1) }^{ 2 }(4)=4W\)
(e) There is no difference in the readings of ammeters \({ A }_{ 1 }\) and \({ A }_{ 2 }\) as same current flows through all elements in a series circuit.

cbse 10th Standard Science Electricity Important Questions And Answers Study Material - QB365 Set C

QB365 Student Learing Managemnet System

Std: 10th Standard CBSE - Science

chapter: Electricity - Important Questions And Answers

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Questions:

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