Exe-8B and 8C Current Electricity ICSE Class-10 Concise Physics Selina Solutions

Exe-8B and 8C Current Electricity ICSE Class-10 Concise Physics Selina Solutionss Chapter-8 . We Provide Step by Step Answer of Exercise-8(A), MCQs-8(A), Numericals-8(A), Exercise-8(B), MCQ-8(B), Numericals-8(B) Exercise-8(C), MCQ-8(C) and Numericals -8(C)  Questions of Exercise-8 Current Electricity ICSE Class-10. Visit official Website CISCE  for detail information about ICSE Board Class-10.

Board	ICSE
Publications	Selina Publication
Subject	Physics
Class	10th
Chapter-8	Current Electricity  (Exe-8B and 8C)
Book Name	Concise
Topics	Solution of Exercise-8(A), MCQs-8(A), Numericals-8(A),, Exercise-8(B), MCQ-8(B), Numericals-8(B) Exercise-8(C), MCQ-8(C) and Numericals -8(C)
Academic Session	2021-2022

Exe-8B and 8C Current Electricity ICSE Class-10 Concise Physics Selina Solutionss Chapter-8

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-: Select Exercise :-

Exercise-8(A), MCQs-8(A), Numericals-8(A),

Exe-8(B), MCQ-8(B), Num-8(B) Exe-8(C), MCQ-8(C) and Num-8(C)

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How to Solve Numericals of Current Electricity ICSE Class-10

Current Electricity deals with ohm’s law, concepts of E.M.F., Potential Difference, resistance, resistances in series and parallel; internal resistance also deals in Current Electricity. Concepts of P.D. Current Electricity in Physics of Chapter 8 and resistance and charge. Simple Numerical Problems using the above relations in Current Electricity.

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Excercise 8 (B) ICSE Class-10 Selina Solutions Physics Current Electricity

Page 200

Question 1

Explain the meaning of the terms e.m.f., terminal voltage and internal resistance of cell.

Answer 1

e.m.f.: When no current is drawn from a cell, the potential difference between the terminals of the cell is called its electro-motive force (or e.m.f.).

Terminal voltage: When current is drawn from a cell, the potential difference between the electrodes of the cell is called its terminal voltage.

Internal Resistance: The resistance offered by the electrolyte inside the cell to the flow of electric current through it is called the internal resistance of the cell.

Question 2

State two differences between the e.m.f. and terminal voltage of a cell.

Answer 2

e.m.f. of cell	Terminal voltage of cell
1.It is measured by the amount of  work done in moving a unit positive charge in the complete circuit inside and outside the cell.	1. It is measured by the amount of work done in moving a unit positive charge in the circuit outside the cell.
2.It is the characteristic of the cell i.e., it does not depend on the amount of current drawn from the cell	2. It depends on the amount of current drawn from the cell. More the current is drawn from the cell, less is the terminal voltage.
3.It is equal to the terminal voltage when cell is not in use, while greater than the terminal voltage when cell is in use.	3. It is equal to the emf of cell when cell is not in use, while less than the emf when cell is in use.

 

Question 3  (Exe-8B and 8C Current Electricity ICSE )

Name two factors on which the internal resistance of a cell depends and state how does it depend on the factors stated by you.

Answer 3

Internal resistance of a cell depends upon the following factors:

(i) The surface area of the electrodes: Larger the surface area of the electrodes, less is the internal resistance.

(ii) The distance between the electrodes: More the distance between the electrodes, greater is the internal resistance.

Question 4

A cell of emf ε and internal resistance r is used to send current to an external resistance R. Write expressions for (a) the total resistance of the circuit (b) the current drawn from the cell and (c) the p.d. across the cell (d) voltage drop inside the cell.

Answer 4

(a) Total resistance = R + r

(b) Current drawn from the circuit :

As we know that,

=V+v

= IR + Ir

=I(R+r)

I = e / (R + r)

(c) p.d. across the cell :

(d) voltage drop inside the cell:

Question 5

A cell is used to send current to an external circuit.

(a) How does the voltage across its terminal compare with its emf?

(b) Under what condition is the emf of a cell equal to its terminal voltage?

Answer 5

(a) Terminal voltage is less than the emf : Terminal Voltage < e.m.f.

(b) e.m.f. is equal to the terminal voltage when no current is drawn.

Question 6

Explain why the p.d. across the terminals of a cell is more in an open circuit and reduced in a closed circuit.

Answer 6

When the electric cell is in a closed circuit the current flows through the circuit. There is a fall of potential across the internal resistance of the cell. So, the p.d. across the terminals in a closed circuit is less than the p.d. across the terminals in an open circuit by an amount equal to the potential drop across the internal resistance of the cell.

Question 7  (Exe-8B and 8C Current Electricity ICSE )

Write the expressions for the equivalent resistance R of three resistors R₁, R₂ and R₃ joined in (a) parallel, (b) series.

Answer 7

(a) Total Resistance in series:

(b) Total Resistance in parallel:

Question 8

How would you connect two resistors in series? Draw a diagram. Calculate the total equivalent resistance.

Answer 8

If current I is drawn from the battery, the current through eac resistor will also be I.

On applying Ohm’s law to the two resistors separately, we further

have

V₁ = I R₁

V₂ = I R₂

V=V₁ + V₂

IR = I R₁+ I R₂

R= R₁+ R₂

Total Resistance in series R :

Question 9

Show by a diagram how two resistors R₁ and R₂ are joined in parallel. Obtain an expression for the total resistance of the combination.

Answer 9

On applying Ohm’s law to the two resistors separately, we further

Have

I₁ = V / R₁

I₂ = V / R₂

I = I₁ + I₂

Question 10

State how are the two resistors are joined with a battery in each of the following cases when:

(a) same current flows in each resistor

(b) potential difference is same across each resistor.

(c) equivalent resistance is less than either of the two resistances.

(d) equivalent resistance is more than either of the two resistances.

Answer 10

(a) series

(b) parallel

(c) parallel

(d) series

Question 11  (Exe-8B and 8C Current Electricity ICSE )

The V-I graph for a series combination and for a parallel combination of two resistors is shown in fig. Which of the two, A or B, represents the parallel combination? Give a reason for your answer.

Answer 11

For the same change in I, change in V is less for the straight line A than for the straight line B (i.e., the straight line A is less steeper than B), so the straight line A represents small resistance, while the straight line B represents more resistance. In parallel combination, the resistance decreases while in series combination, the resistance increases. So A represents the parallel combination.

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MULTIPLE CHOICE TYPE – 8 (B) Selina Solutions Physics Current Electricity

Page 200

Question 1

In series combination of resistances:
(a) p.d is same across each resistance
(b) total resistance is reduced
(c) current is same in each resistance
(d) all above are true

Answer 1

(c) In series combination of resistances, current is same in each resistance.

Question 2

In parallel combination of resistances:
(a) p.d is same across each resistance
(b) total resistance is increased
(c) current is same in each resistance
(d) all above are true

Answer 2 

(a) In parallel combination of resistances, P.D. is same across each resistance. 

Question 3

Which of the following combinations have the same equivalent resistance between X and Y?

Answer 3 

(a) and (d)

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NUMERICALS – 8 (B) Current Electricity Concise Physics Solution

Page 201

Question 1

The diagram in Fig. 8.42 shows a cell of e.m.f. ε = 2 volt and internal resistance r = 1 ohm to an external resistance R = 4 ohm. The ammeter A measures the current in the circuit and the voltmeter V measures the terminal voltage across the cell. What will be the readings of the ammeter and voltmeter when (i) the key K is open, (ii) the key K is closed.

(a) ε = 1.8 V

Total Resistance = 2 + 4.5 + 0.7 = 7.2 W

I=?

I= / R (total resistance)

I = 1.8 / 7.2 = 0.25 A

(b) Current (calculated in (a) part) I = 0.25 A

Now, total resistance excluding internal resistance = 4.5 + 0.7 = 5.2 ohm

V = IR = 0.25 x 5.2 = 1.3 V

Question 4

The music system draws a current of 400mA when connected to a 12V battery.

(i) What is the resistance of the music system

(ii) The music system if left playing for several hours and finally the battery voltage drops and the music systems stops playing when the current drops to 320mA. At what battery voltage does the music system stop playing?

Answer 4

Question 5

A Cell of e.m.f. ε and internal resistance ? sends current 1.0 A when it is connected to an external resistance 1.9Ω. But it sends current 0.5 A when it is connected to an external resistance 3.9 Ω.Calculate the values of ε and ?.

Answer 5

In first case
I = 1 A, R = 1.9 ohm
ε = I(R + r) = 1(1.9+r)
ε = 1.9 + r————(1)
In second case
I = 0.5 A, R = 3.9 ohm
ε = I(R + r) = 0.5 (3.9 + r)
ε = 1.95 + 0.5r —————-(2)
From eq. (1) and (2),
1.9 + r = 1.95 + 0.5r
r = 0.05/0.5 = 0.1 ohm
Substituting value of r
ε = 1.9 + r = 1.9 + 0.1 = 2 V

Question 6  (Exe-8B and 8C Current Electricity ICSE )

Question 14

A uniform wire with a resistance of 27 ohm is divided into three equal pieces and then they are joined in parallel. Find the equivalent resistance of the parallel combination.

Answer 14

Wire cut into three pieces means new resistance = 27/3 = 9

Now three resistance connected in parallel :

Question 15

A circuit consists of a 1 ohm resistor in series with a parallel arrangement of 6 ohm and 3 ohm resistors. Calculate the total resistance of the circuit. Draw a diagram of the arrangement.

Answer 15

Question 16  (Exe-8B and 8C Current Electricity ICSE )

Calculate the effective resistance between the points A and B in the network shown below in figure.

Answer 16

Question 17

Calculate the equivalent resistance between the points A and B in the adjacent diagram in the figure.

Answer 17

R₁=3+2 = 5 ohm

R₂=30 W

R₃=6+4 = 10 ohm

R₁ , R₂ and R₃ are connected in parallel

Question 18

In the network shown in adjacent figure 8.49, calculate the equivalent resistance between the points (a) A and B, and (b) A and C

Answer 18

(a) R₁=2+2+2 = 6ohm

R₂=2ohm

R₁ and R₂ are connected in parallel

(b) R₁=2+2 = 4 ohm

R₂=2+2=4 W

R₁ and R₂ are connected in parallel

Question 19

Five resistors, each 3 ohm, are connected as shown in figure 8.50. Calculate the resistance (a) between points P and Q (b) between points X and Y.

Answer 19

(a) R₁=3+3=6 W

R₂=3 W

R₁ and R₂ are connected in parallel

(b) As calculated above R=2 ohm

R₃ = 3 ohm

R₄ = 3 ohm

R’=R+R₃+R₄=2+3+3 = 8 ohm

Question 20  (Exe-8B and 8C Current Electricity ICSE )

Two resistors of 2.0 ohm and 3.0 ohm are connected (a) in series, (b) in parallel, with a battery of 6.0 V and negligible internal resistance. For each case draw a circuit diagram and calculate the current through battery.

Answer 20

(a)

R₁=2 ohm

R₂=3 ohm

R = R₁ + R₂ = 2 + 3 = 5 ohm

V=6 V

(b)

R₁ and R₂ are connected in parallel

Question 21

A resistor of 6 ohm is connected in series with another resistor of 4 ohm. A potential difference of 20 V is applied across the combination. (a) Calculate the current in the circuit and (b) potential difference across the 6 ohm resistor.

Answer 21

(a) R₁=6 ohm

   R₂=4 ohm

   R = R₁ + R₂ = 6 + 4 = 10 ohm

   V=20 V

   I=V/R = 20/10 = 2 A

(b) R = 6 W

I = 2 A

V = ?

V = IR = 6 x 2 = 12 V

Question 22

Two resistors of resistance 4 Ω and 6 Ω are connected in parallel to a cell to draw 0.5 A current from the cell.

(a) Draw a labeled diagram of the arrangement

(b) Calculate current in each resistor.

Answer 22

(a) Circuit diagram

(b) Equivalent resistance of the circuit:

Hence, the emf of the cell is

Therefore, current through each resistor is

Question 23

Calculate current flowing through each of the resistors A and B in the circuit shown in figure 8.51?

Answer 23

For resistor A :

R=1 ohm

V=2 V

I=V/R = 2/1 = 2 A

For resistor B :

R = 2 ohm

V = 2 V

I=V/R = 2/2 = 1 A

Question 24

In figure 8.52, calculate :

(a) the total resistance of the circuit.

(b) the value of R, and

(c) the current flowing in R

Answer 24

(a)

V=4 V

I=0.4 A

Total Resistance R’=?

R’ = V/I = 0.4/4 = 10 ohm

(b)

R₁= 20 ohm

R’ = 10 ohm

(c)

R=20 ohm

V=4 V

I=V/R = 4/20 = 0.2 A

Question 25  (Exe-8B and 8C Current Electricity ICSE )

A particular resistance wire has a resistance of 3 ohm per meter. Find :

(a) The total resistance of three lengths of this wire each 1.5 m long, in parallel.

(b) The potential difference of the battery which gives a current of 2 A in each of the 1.5 m length when connected in the parallel to the battery (assume that resistance of the battery is negligible).

(c) The resistance of 5 m length of a wire of the same material, but with twice the area of cross section.

Answer 25

(a) Resistance of 1m of wire = 3 ohm

Resistance of 1.5 m of wire = 3 x 1.5 = 4.5 W

(b) I=2 A

V=IR = 2 x 4.5 = 9 V

(c) R=3 ohm for 1 m

For 5 m : R=3 x 5 = 15 ohm

But Area A is double i.e. 2 A and Resistance is inversely proportional to area so Resistance will be half.

R=15/2 = 7.5 ohm

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Page 203

Question 26

A cell supplies a current of 1.2 A through two 2 ohm resistors connected in parallel. When resistors are connected in series, it supplies a current of 0.4 A. Calculate: (i) the internal resistance and (ii) e.m.f. of the cell.

Answer 26

In parallel R = ½ + ½ = 1 ohm

I = 1.2 A

=I(R+r) = 1.2(1+r)= 1.2 + 1.2 r

In series R = 2+2=4 ohm

I=0.4 A

=I(R+r) = 0.4(4+r) = 1.6 + 0.4 r

It means :

1.2 + 1.2 r = 1.6 + 0.4 r

0.8 r = 0.4

r = 0.4 / 0.8 = ½ = 0.5 ohm

(i) Internal resistance r = 0.5 ohm

(ii) =I(R+r) = 1.2(1+0.5) = 1.8 V

Question 27

A battery of emf 15 V and internal resistance 3 ohm is connected to two resistors 3 ohm and 6 ohm connected in parallel. Find (a) the current through the battery (b) p.d. between the terminals of the battery (c) the current in 3 ohm resistor (d) the current in 6 ohm resistor.

Answer 27

(a)In parallel 1/R = 1/3 + 1/6 = 1/2

So R = 2 ohm

r = 3 W

=15 V

=I(R+r)

15=I(2+3)

I=15/5 = 3 A

(b)V=?

R=2 ohm

V=IR = 3 x 2 = 6 V

(c)V=6 V

R = 3 ohm

I=V/R = 6/3 = 2 A

(d) R=6 ohm

V= 6 V

I=V/R = 6/6 = 1 A

Question 28

The circuit diagram in Fig. 8.53 shows three resistors  2 Ω, 4 Ω and R Ω connected to a battery of e.m.f. 2 V and internal resistance 3 ohm . If main current of 0.25 A flows through the circuit, Find :

(a)  the p.d. across the 4 ohm resistor?

(b)  the p.d. across the internal resistance of the cell,

(c) the p.d. across the R ohm or 2 ohm resistor, and

(d) the value of R.

Answer 28

(a)R = 4

I = 0.25 A

V=IR = 0.25 x 4 = 1 V

(b)Internal Resistance r=3 ohm

I = 0.25 A

V=IR = 0.25 x 3 = 0.75 V

(c) Effective resistance of parallel combination of two 2 ohm resistances = 1 ohm

V= I/R = 0.25/1 = 0.25 V

(d) I=0.25 A

=2 V, r = 3 ohm

=I(R’+r)

2=0.25(R’+3)

R’=5 W

Question 29  (Exe-8B and 8C Current Electricity ICSE )

Three resistors of 6.0 ohm, 2.0 ohm and 4.0 ohm are joined to ammeter A and a cell of emf 6.0 V as shown in figure. Calculate :

(a) the effective resistance of the circuit.

(b) the reading of ammeter

Answer 29

(a) R₁=6 W

R’=R₂+R₃=2+4=6 W

R₁ and R’ in parallel :

(b) R= 3 ohm

V=6 V

I=?

I=V/R=6/3=2 A

Question 30

The diagram below in Fig. 8.55 shows the arrangement of five different resistances connected to a battery of e.m.f. 1.8 V. Calculate :

(a) The total resistance of the circuit, and

(b) The reading of ammeter A.

Answer 30

Question 31

A cell of e.m.f. 2 V and internal resistance 1.2 Ω is connected to an ammeter of resistance 0.8 Ω and two resistors of 4.5 Ω and 9 Ω as shown in fig.

Find:

(a) The reading of the ammeter,

(b) The potential difference across the terminals of the cells, and

(c) The potential difference across the 4.5 Ω resistor.

Answer 31

The total resistance of the circuit is

(a) Therefore, the current through the ammeter is

(b) The potential difference across the ends of the cells is

(c) The potential difference across the 4.5Ωresistor is

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Chapter -8, “Current Electricity” Selina Physics Solution  Excercise – 8 (C)

Page 211

Question 1

Write an expression for the electrical energy spent in flow of current through an electrical appliance in terms of current, resistance and time.

Answer 1

Question 2

Write an expression for the electrical power spent in flow of current through a conductor in terms of (a) resistance and potential difference, (b) current and resistance.

Answer 2

 

Question 3  (Exe-8B and 8C Current Electricity ICSE )

Electrical power P is given by the expression P= (Q x V)/time

(a)What do the symbols Q and V represent?

(b)Express the power P in terms of current and resistance explaining the meanings of symbols used there in.

Answer 3

 (a) Q represents Charge and V represents Voltage.

 (b)  

Question 4

Name the S.I. unit of electrical energy. How is it related to Wh ?

Answer 4

The S.I. unit of electrical energy is joule.

1 Wh = 3600 J

Question 5

Explain the meaning of the statement ‘the power of an appliance is 100 W’.

Answer 5

The power of an appliance is 100 W. It means that 100 J of electrical energy is consumed by the appliance in 1 second

Question 6

State the S.I. units of electrical power.

Answer 6

 The S.I. unit of electrical power is Watt.

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Page 212

Question 7

(i)State and define the household unit of electricity.

(ii)What is the voltage of the electricity that is generally supplied to a house?

(iii) What is consumed while using different electrical appliances, for which electricity bills are paid?

Answer 7

(i)The household unit of electricity is kilowatt-hour (kWh).

One kilowatt-hour (kWh) is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for one hour.

(ii)The voltage of the electricity that is generally supplied to a house is 220 Volt.

(iii) electrical energy is consumed.

Question 8

Name the physical quantity which is measured in

(i) kW, (ii) kWh. (iii) Wh

Answer 8

(i)Electrical power is measured in kW and

(ii)Electrical energy is measured in kWh.

(iii)Energy consume is measured in Wh.

Question 9

Define the term kilowatt – hour and state its value in S.I. unit.

Answer 9

One kilowatt-hour (kWh) is the electrical energy consumed by an electrical appliance of power 1 kW when it is used for one hour.

Its value in SI unit is 

Question 10

How do kilowatt and kilowatt-hour differ ?

Answer 10

Kilowatt is the unit of electrical power whereas kilowatt-hour is the unit of electrical energy.

Question 11  (Exe-8B and 8C Current Electricity ICSE )

Complete the following:

(a)……………………..

(b)……………………..

1 kWh= ________ J

Answer 11

(a)

(b) 3.6 x 10⁶ J

Question 12

What do you mean by power rating of an electrical appliance? How do you use it to calculate (a) the resistance of the appliance and (b) the safe limit of the current in it, while in use?

Answer 12

An electrical appliance such as electric bulb, geyser etc. is rated with power (P) and voltage (V) which is known as its power rating. For example: If an electric bulb is rated as 50W-220V, it means that when the bulb is lighted on a 220 V supply, it consumes 50 W electrical power.

(a) To calculate the resistance of the appliance, the expression is:

(b) The safe limit of current I is: 

Question 13

An electric bulb is marked ‘100 W, 250 V’. What information does this convey?

Answer 13

It means that if the bulb is lighted on a 250 V supply, it consumes 100 W electrical power (which means 100 J of electrical energy is converted in the filament of bulb into the light and heat energy in 1 second).

Question 14

List the names of three electrical gadgets used in your house. Write their power, voltage rating and approximate time for which each one is used in a day. Hence find the electrical energy consumed by each in day in a month of 30 days.

Answer 14

Appliance	Power (in Watt)	Voltage (in Volts)	Time (hours)	Electrical energy (E=Pxt)
Fluorescent tube	40	220	12 x 30 = 360hrs	14.4kwh
Television set	120	220	4 x 30 = 120hrs	14.4kwh
Refrigerator	150	220	24 x 30 = 720hrs	108kwh

Question 15

Two lamps, one rated 220 V, 50 W and the other rated 220 V, 100 W, are connected in series with mains of 220 V. Explain why does the 50 W lamp consume more power.

Answer 15

   

Question 16  (Exe-8B and 8C Current Electricity ICSE )

Name the factors on which the heat produced in a wire depends when current is passed in it, and state how does it depend on the factors stated by you.

Answer 16

When current is passed in a wire, the heat produced in it depends on the three factors:

(i) on the amount of current passing through the wire,

(ii) on the resistance of wire and (iii) on the time for which current is passed in the wire.

(iii)Dependence of heat produced on the current in wire: The amount of heat H produced in the wire is directly proportional to the square of current I passing through the wire.

(iv)Dependence of heat produced on the resistance of wire: The amount of heat H produced in the wire is directly proportional to the resistance R of the wire.

(v)Dependence of heat produced on the time: The amount of heat H produced in the wire is directly proportional to the time t for which current is passed in the wire.

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Chapter :- 8, “Current Electricity” Selina Physics Solution MULTIPLE CHOICE TYPE – 8 (C)

Page 212

Question 1.

When a current I flows through a resistance R for time t, the electrical energy spent is:

(a) IRt

(b) I²Rt

(c) IR²t

(d) I²R/t

Answer 1

(b) I²Rt

Question 2

An electrical appliances has a rating 100 W, 120 V. The resistance of element of appliance when in use is:

(a) 1.2 ohm

(b) 144 ohm

(c) 120 ohm

(d) 100 ohm

Answer 2

(b) 144 ohm

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NUMERICALS – 8 (C) Chapter :- 8, “Current Electricity” Selina Physics Solution 

Page 212

Question 1

An electric bulb of resistance 500 ohm draws current 0.4 A from the source. Calculate: (a) the power of bulb and (b) the potential difference at its end.

Answer 1

Resistance of electric bulb (R) = 500Ω

Current drawn from the source (I) = 0.4 A

Power of the bulb (P) = VI

V = I x R

V = 0.4 x 500 = 200 V

The potential difference at its end is 200 V.

Hence,

Power (P) = VI

P = 200 x 0.4 = 80 W

The power of the bulb is 80 Watt.

Question 2

A current of 2 A is passed through a coil of resistance 75 ohm for 2 minutes. (a) How much heat energy is produced ? (b) How much charge is passed through the resistance?

Question 3

Calculate the current through a 60 W lamp rated for 250 V. If the line voltage falls to 200 V, how is the power consumed by the lamp affected?

Answer 3

Question 4

An electric bulb is rated ‘100 W, 250 V’. How much current will the bulb draw if connected to a 250 V supply?

Answer 4

Question 5   (Exe-8B and 8C Current Electricity ICSE )

An electric bulb is rated at 220 V, 100 W. (a) What is its resistance? (b) What safe current can be passed through it?

Answer 5

Question 6

A bulb of 40 W is used for 12.5 h each day for 30 days. Calculate the electrical energy consumed.

Answer 6

Question 7

An electric press is rated 750 W, 230 V. Calculate the electrical energy consumed by the iron in 16 hours.

Answer 7

Question 8

An electrical appliance having a resistance of 200 ohm is operated at 220 V. calculate the energy consumed by the appliance in 5 minutes. (i) in joules (ii) in kWh.

Answer 8

Question 9

A bulb marked 12 V, 24 W operates on a 12 V battery for 20 minutes. Calculate:

(i)The current flowing through it, and

(ii)The energy consumed.

Answer 9

Question 10  (Exe-8A and 8C Current Electricity ICSE )

A current of 0.2 A flows through a wire whose ends are at a potential difference of 15 V. Calculate:

(i)The resistance of the wire, and

(ii)The heat energy produced in 1 minute.

Answer 10

Question 11

What is the resistance, under normal working conditions, of an electric lamp rated at ‘240 v’, 60 W? If two such lamps are connected in series across a 240 V mains supply, explain why each one appears less bright.

Answer 11

When one lamp is connected across the mains, it draws 0.25 A current, while if two lamps are connected in series across the mains, current through each bulb becomes

(i.e., current is halved), hence heating (  ) in each bulb becomes one-fourth, so each bulb appears less bright.

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Page 213

Question 12

Two bulbs are rated 60 W, 220 V and 60 W, 110 V respectively. Calculate the ratio of their resistances.

Answer 12

Question 13

An electric bulb is rated 250 W, 230 V. Calculate :

(i) the energy consumed in one hour, and

(ii) the time in which the bulb will consume 1.0 kWh energy when connected to 230 V mains?

Answer 13

Question 14  (Exe-8B and 8C Current Electricity ICSE )

Three heaters each rated 250 W, 100 V are connected in parallel to a 100 V supply. Calculate:

(i)The total current taken from the supply,

(ii)The resistance of each heater, and

(iii)The energy supplied in kWh to the three heaters in 5 hours.

Answer 14

Question 15

A bulb is connected to a battery of p.d. 4 V and internal resistance  . A steady current of 0.5 A flows through the circuit. Calculate:

(i)The total energy supplied by the battery in 10 minutes,

(ii)The resistance of the bulb, and

(iii)The energy dissipated in the bulb in 10 minutes.

Answer 15

Question 16

Two resistors A and B of 4 ohm and 6 ohm  respectively are connected in parallel. The combination is connected across a 6 volt battery of negligible resistance. Calculate: (i) the power supplied by the battery, (ii) the power dissipated in each resistor.

Answer 16

Question 17   (Exe-8B and 8C Current Electricity ICSE )

A battery of e.m.f. 15 V and internal resistance is connected to two resistors of 4 ohm and 6 ohm joined (a) in series. Find in each case the electrical energy spent per minute in 6 ohm resistor.

Answer 17

Question 18

Water in an electric kettle connected to a 220 V supply took 5 minutes to reach its boiling point. How long would it have taken if the supply had been of 200 V?

Answer 18

 

 

 

 

 

 

 

 

 

 

 

 

 Question 19

An electric toaster draws 8 A current in a 220 V circuit. It is used for 2 h. Find the cost of operating the toaster if the cost of electrical energy is Rs. 4.50 per kWh.

Answer 19

Question 20

An electric iron is rated 220V, 2kW.

If the iron is used for 3h daily find the cost of running it for one week if it costs Rs 4.25 per kwh.

Answer 20

For one day energy used will be

Energy =Power x time

Energy=2 x 3=6kwh

Energy used in one week

6 x 7=42kwH

Total cost= 42x 4.25=178.5

Rs 178.5

Question 21  (Exe-8B and 8C Current Electricity ICSE )

A geyser is rated 1500 W, 250 V. This geyser is connected to 250 V mains. Calculate:

(i)The current drawn,

(ii)The energy consumed in 50 hours, and

(iii)The cost of energy consumed at Rs 4.20 per kWh.

Answer 21

Return to Concise Selina ICSE Physics Class-10 

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