Question

A current of 5A flows in a resistor of 2Ω. Calculate the energy dissipated in 300sec in the resistor? *

1 point

a) 15kJ

b) 15000kJ

c) 1500J

d) 15J

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Answer 1

Given :-

• Current (I) = 5 Ampere
• Resistance (R) = 2 ohm
• Time (t) = 300 seconds

To find :-

• The energy dissipated

Solution:-

According to joules law of heating

H= I² Rt

Where,

H= heat energy

I= current

R= resistance

t= time

• Now find

→H=I²Rt

→H=(5)² ×2×300

→ H= 25×600

→H= 15000 jolues

or ,

1 kilo jolues = 1000 joules

→H= 15 kJ

So the correct option is a).

Answer 2

Answer:

a) 15 KJ

Explanation:

Given :

Current flowing through the resistor = I = 5 Amperes

Resistance of the resistor = R = 2 ohms

Time = 300 seconds

To find :

Energy dissipated (heat generated)

Heat generated = I²RT

Substituting the values :

Heat generated = 5²×2×300

Heat generated = 25×2×300

Heat generated = 50×300

Heat generated = 15,000 Joules

1000 joules = 1 Kilo joule

15000 joules = 15 Kilo joules

The energy dissipated is 15 KJ

So option a) is correct