Chapter 1:Chemical Reaction And Equations

Chapter 2:Acids, Bases and Salts

Chapter 3:Metals and Non-metals

Chapter 4:Carbon and Its Compounds

Chapter 5: Periodic Classification of Elements

Chapter 10: Light Reflection and Refraction

Chapter 11:Human Eye and Colourful World

(a) 1/25

(b) 1/5

(c) 5

(d) 25

R
/
5

For equivalent resistance :

1
/
R'

=
5
/
R

+
5
/
R

+
5
/
R

+
5
/
R

+
5
/
R

=
5 + 5 + 5 + 5 + 5
/
R

=
25
/
R

R'
/
R

= 25
Answer: d) 25

(a) I^{2}R

(b) IR^{2}

(c) VI

(d) V^{2}/R

Answer: b) IR^{2}

Explanation:

We know P = VI. ............(i)

According to Ohm’s law,

V = IR

Puting Value of V in eq (i)

P = (IR) × I

P = I^{2}R

Again Ohm’s law,

I =
V
/
R

..............(ii)
From eq (i) and (ii)

P = V ×
V
/
R

= VSo IR^{2} does not represent electrical power

(a) 100 W

(b) 75 W

(c) 50 W

(d) 25 W

Answer: 25 W

We know P = VI = V^{2}/R

R = V^{2}/P

R = (220)^{2}/100 = 484 Ω

Again for V = 110

P = V^{2}/R

P = (110^{)2}/484 Ω = 25 W

(a) 1:2

(b) 2:1

(c) 1:4

(d) 4:1

Let R_{s} and R_{p} be the equivalent resistance of the wires when connected in series and parallel respectively.

R

1
/
R_{p}

=
1
/
R

+
1
/
R

=
2
/
R

R

R
/
2

We know for V constant

H

V^{2}
/
R_{s}

or H
1
/
R_{s}

Similarly H

V^{2}
/
R_{p}

or H
1
/
R_{p}

So
H_{p}
/
H_{s}

=
R_{p}
/
R_{s}

=
R
/
2R x 2

=
1
/
4

the voltmeter should be connected parallel in circuit to measure the potential difference

Here D = 0.5mm = 5 x 10

r =

5 x 10 ^{-4}
/
2

= 2.5 x 10 ρ =1.6 × 10

R = 10Ω

We know R = ρ

l
/
A

l =

RA
/
ρ

=
10 x 3.14 x (2.5 x 10 ^{-4})^{2}
/
(1.6 × 10^{–8} )^{2}

=
10 x 3.14 x 25
/
4 x 1.6

= 122.7m

Again R' = ρ

l
/
A

R' = ρ

l x (2)^{2}
/
π (D')^{2}

............[r =
D
/
2

] If D' =2D R = ρ

l x (2)^{2}
/
π (2D)^{2}

R' = ρ

R
/
4

=R = ρ
10
/
4

= 2.5 Ω
The length of the wire is 122.72 m and the new resistance is 2.5 Ω.

I (Ampere) | 0.5 | 1.0 | 2.0 | 3.0 | 4.0 |

V (Volts) | 1.6 | 3.4 | 6.7 | 10.2 | 13.2 |

Plot a graph between V and I and calculate the resistance of that resistor.

The slope graph give resistance.

Slope =
1
/
R

=
2
/
10.2-3.4

R = 6.8/2 = 3.4 Ω

The resistance is 3.4 Ω.

We know V= IR

R = V/I

R =
12
/
2.5 x 10^{-3}

= 4.8 x 10 For R_{s} = R_{1} + R_{2} + R_{3} +.........R_{n}

So R = 0.2 Ω + 0.3 Ω + 0.4 Ω + 0.5 Ω + 12 Ω = 13.4 Ω

Now, using Ohm’s law,

V =IR

I = V/R = 9 /31.4 = 0.67 AThe current flowing across the 12 Ω is 0.671 A.

Let number of resistors is n

We know
1
/
R _{p}

= n x
1
/
176

=
n
/
176

R

176
/
n

Now, using Ohm’s law,

V= IR
176
/
n

=
V
/
I

n =

176 x 5
/
220

= 4
The number of resistors is 4.

When two resistance connect in parallel and third resistance with series to get total resistance 9 Ω

So
1
/
R_{p}

=
1
/
R_{1}

+
1
/
R_{1}

=
1
/
6

+
1
/
6

=
2
/
6

=
1
/
3

Now R

R

= 3 + 6 = 9 Ω

ii)

When two resistance connect in series and third resistance with parallel to get total resistance 4 Ω

R

= 6 + 6 =12

Now R

1
/
R_{e}

=
1
/
R_{s}

+
1
/
6

=
1
/
12

+
1
/
6

=
2+1
/
12

=
1
/
4

R

Given

P = 10 W and V = 220 V

Let the total number of blub is 'n'

Current I = 5 A

We know

P =
V^{2}
/
R

R =

V^{2}
/
P

=
(220)^{2}
/
10

= 4840 Ω We also know

V = IR

R =
V
/
R

=
220
/
5

= 44 ΩSo 'n' number of resistance each 4840 Ω connect in parrallel have equivalent resistance 44 Ω

1
/
R

=
1
/
R

+
1
/
R

+
1
/
R

+...........+ 'n'times
1
/
R

=
1
/
R

x n
1
/
44

=
n
/
4840

n =

4840
/
44

= 110So, 110 lamps can be connected in parallel.

Case (i) When coils are used separately

Given

V = 220 V and R = 24 Ω

Using Ohm’s law

I =
V
/
R

=
220
/
24

=9.18 Awhen they resistor used separately,9.166 A of current flows through each resistor

Case (ii) When coils connected in series

The total resistance = 24 Ω + 24 Ω = 48 Ω

I =
V
/
R

=
220
/
48

= 4.58 ASo, 4.58 A flows through the series circuit.

Case (iii) When coils connected in parallel

Total resistance =
1
/
R_{1}

+
1
/
R_{2}

=
1
/
24

+
1
/
24

=

24 + 24
/
24 x 24

=
48
/
576

=

576
/
48

=12 ΩI =

V
/
R

=
220
/
12

=18.33 A18.33 A current flow in parallel circuit .

(i) Here R_{1} =1 Ω

R_{2} = 2 Ω

V = 6V

Total resistance R_{e}= R_{1} + R_{2}

= 1+ 2= 3 Ω

Using ohm's law

I =
V
/
R

=
6
/
3

= 2 ANow Power

P =I

= 2

The power consumed by the 2 Ω is 8 W.

(ii) In Parallel combination Voltage remains same So

P =
V^{2}
/
R

=
4^{2}
/
2

=
16
/
2

= 8 AThe power consumed by the 2 Ω resistor is 8 W.

Here bulbs are connected in parallel So the voltage remains same.

For P = 100 W

P = V × I

I_{1} = P/V

I = 100 W/220 V = 100/220 = 0.455A

For P = 60 W

I_{2} = 60 W/220 V = 60/220 =0.273A

Given W = 250 W and time = 1hr = 3600 seconds

We know

H = Pt

H = 250 × 3600

= 9 × 10^{5} J

Similarly For 1200 W and time =10 min = 600 seconds

H = 1200 W × 600 s = 7.2 × 10^{5} J

Energy consumed by the TV is more than the toaster.

We know

P = I^{2} R

R = 8 Ω , I = 15 A

time = 2 hour

P = (15A) ^{2} × 8 Ω = 1800 J/s

a. Why is the tungsten used almost exclusively for filament of electric lamps?

b. Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal?

c. Why is the series arrangement not used for domestic circuits?

d. How does the resistance of a wire vary with its area of cross-section?

e. Why copper and aluminium wires are usually employed for electricity transmission?

1
/
A

As We see resistance is inversely proportional to the area of cross section. When the area of cross section increases the resistance decreases and vice versa

Chapter 1:Chemical Reaction And Equations

Chapter 2:Acids, Bases and Salts

Chapter 3:Metals and Non-metals

Chapter 4:Carbon and Its Compounds

Chapter 5: Periodic Classification of Elements

Chapter 10: Light Reflection and Refraction

Chapter 11:Human Eye and Colourful World

Chapter 13:Magnetic Effects of Electric Current

Chapter 16:Sustainable Management of Natural Resources