Question

4.
0.1 A current is flowing through a 100 ohm resistor for 2 minutes.
a) calculate the heat released.
b) calculate the power.​

Answer 1

Answer:

• Heat released = 120 Joules
• Power = 1 Watt

Explanation:

Given

• Current flowing through the resistor, I = 0.1 A
• Resistance of resistor, R = 100 Ω
• Time for which current flowed, t = 2 min = 2 × 60 sec = 120 sec

To find

• Heat released, H =?
• Power, P =?

Formulae required

• Formula for Joule's law of heating  

       H = I² × R × t

• Formula to calculate power

       P = I² × R

[ Where H is heat produced, I is current, R is resistance, t is time taken and P is power ]

Solution

Using expression for Joule's heating law

→ H = I² × R × t

→ H = ( 0.1 )² × ( 100 ) × 120

→ H = 0.01 × 100 × 120

→ H = 120 J

Now, Using Formula to calculate power

→ P = I² × R

→ P = ( 0.1 )² × ( 100 )

→ P = 0.01 × 100

→ P = 1 W

Therefore,

• Heat released is equal to 120 Joules. and
• Power is equal to 1 Watt.

Answer 2

GiveN :

• Current Flowing in wire (I) = 0.1 A
• Time (t) = 2mins = 120s
• Resistance = 100 ohms

To FinD :

• Heat Raised
• Power

SolutioN :

⇒H = I²Rt

⇒H = (0.1)² * (100) * (120)

⇒H = 0.01 * 100 * 120

⇒H = 1 * 120

⇒Heat Energy Raised = 120 Joules

_______________________

Electric Power is given by the relation :

⇒P = VI

⇒P = (IR)*I

⇒P = I²R

⇒P = (0.1)² * 100

⇒P = 0.01 * 100

⇒Electric Power = 1 Watt