Book Store

Download books and chapters from book store.
Currently only available for.
CBSE Gujarat Board Haryana Board

Previous Year Papers

Download the PDF Question Papers Free for off line practice and view the Solutions online.
Currently only available for.
Class 10 Class 12

Electricity

What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?

The advantages of connecting electrical devices in parallel with the battery are:

1. Each device gets the full battery voltage.
2. The parallel circuit divides the current through the electrical devices. Each device gets proper current depending on its resistance.
3. Other devices do not get affected if one device is switched OFF/ON.

498 Views

What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?

(a) Highest resistance can be obtained by connecting the four coils in series. Then
R = R₁ + R₂ + R₃ +R₄
R = 4 + 8 + 12 + 24 = 48 Ω.

(b) Lowest resistance can be obtained by connecting the four coils in parallel.
Then

$1 over straight R space equals space 1 fourth plus 1 over 8 plus 1 over 12 plus 1 over 24 space equals 12 over 24 space equals space 1 half$

or $straight R space equals space 2 space straight capital omega. space space space space space Ans. space$

170 Views

How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?

Given three resistors of resistances 2 Ω, 3 Ω and 6 Ω.

(a) A total resistance of 4 Ω can be obtained by connecting the 2 Ω resistance in series with the parallel combination of 3 Ω and 6 Ω.

That is,

Equivalent resiastance, R = R_{1} + \frac{R_{2} R_{3}}{R_{2} + R_{3}} = 2 + \frac{3 \times 6}{3 + 6} = 4 Ω .

(b) We can obtain a total resistance of 1 Ω by connecting resistances of 2 Ω, 3 Ω and 6 Ω in parallel with each other.

\frac{1}{R} = \frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{1}{1} R = 1 Ω .

236 Views

Why does the cord of an electric heater not glow while the heating element does?

Current carried by the cord and the heating element of an electric heater is same. But, due to the high resistance of the heating element it becomes hot and starts glowing.
We have,
( H = I²RT)

And, the cord remains cold due to its low resistance and does not glow.

157 Views

Compute the heat generated while transferring 96,000 coulomb of charge in one hour through a potential difference of 50 V.

Given,

Charge transferred, Q = 96, 000 C
Time, t = 1 hour = 3600 s
Voltage, V = 50 V
Heat generated, H = V $\times$ Q
= 50 V x 96,000 C
= 48,00,000 J

299 Views

Chapter Chosen

Book Chosen

Subject Chosen

Book Store

Previous Year Papers

Electricity

How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?