Question

how much energy in joule is dissipated when an electronic iron draws a current of 9amp for 2min at 220 volts

Answer 1

Solution:

An iron box dissipates energy in the form of heat energy.

According to Joule's Heating Law,

Heat Energy Dissipated in an electric circuit is given as;

→ Heat = VIt

where, I is the current flowing in the circuit, V is the potential drop offered by the circuit and t is the time of flow of current in the circuit.

According to the question,

• I = 9 A
• V =  220 V
• t = 2 minutes = 120 seconds

Substituting in the formula we get:

→ Heat = 220 × 9 × 120

→ Heat = 237600 J or 237.6 KJ

Hence heat energy dissipated in the circuit is 237600 J.

Answer 2

$\huge \mathtt{ \fbox{Solution :)}}$

Given ,

• Current (I) = 9 amp
• Time (t) = 2 min or 120 sec
• Potential difference (V) = 220 volt

We know that , the product ot potential difference and electric current is electric power

$\mathtt{\large \fbox{Electric \: power = VI}}$

Thus ,

Electric power = 220 × 9

Electric power = 1980 watt

Hence , the electric power is 1980 watt

And the rate of work done of unit charge is also termed as electric power

$\large \mathtt{ \fbox{Electric \: power = \frac{work \: done}{time} }}$

Thus ,

1980 = work/120

work = 237600 joule

Since , work done = energy

Hence , the total energy is 237600 joule

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