Page 37 - CHAPTER-1 (Electricity)

CHAPTER 1

ELECTRICITY

This law implies that the amount of heat produced in a resistor always:

a. Increases in direct proportion with the square of current for a given

resistance.

b. Increases in direct proportion with the resistance for a given

current.

c. Increases proportionally with time for which the current will flow

through the resistor.

Example 12:

An electric iron of resistance 25Ω takes a current of 7A. Calculate the

heat developed in 0.5min.

Solution:

Given, resistance, = 25Ω; current = 7 ; time = 0.5 = 0.5 × 60 =

30 ; heat = ?

We know that, heat, = 7 × 25 × 30 = 36750
2
So, the heat developed is 3.68 × 10 .
4
Example 13:

200J of heat is produced 10s in a 5Ω resistance. Find the potential

difference across the resistor.

Solution:

Given, heat = 200 ; Resistance = 5Ω; Time = 10 ; Potential

difference = ? We know that,

Heat, =
2

H 200
⇒Current, I=√ =√ =2A
Rt 5×10

So, the potential difference across the resistor is, V=IR

=2×5=10V

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