Physics, asked by adwaidps123, 10 months ago

[0.1A current is flowing through a 100 ohm resistor for 2 minutes. a. Calculate the heat released. b.Calculate the power.]

Answers

Answered by ashishkushwaha16
9

Answer:

a) heat released =120 J

b)power =1 watt

Explanation:

a) I= 0.1A ,. r=100 ohm,. t=2mins=120sec

according to Joule law of heating

w=I^2Rt

=0.1*0.1*100*120

=120J

b) power is the rate at which work is done.

p=w/t

= I^2Rt/t

=120/120=1

therefore,power=1 watt

Answered by nirman95
0

POWER IS 1 WATT & HEAT RELEASED IS 120 JOULES.

Given:

  • Resistance = 100 ohm
  • Time = 2 min = 2 × 60 = 120 sec
  • Current = 0.1 Ampere

To find:

  • Heat released & Power ?

Calculation:

The general expression of heat released in a circuit is given as :

H =  {i}^{2}  \times r \times t

  • i is current, r is resistance and t is time.

 \implies H =  {(0.1)}^{2}  \times 100 \times 120

 \implies H =   {10}^{ - 2}   \times 100 \times 120

 \implies H =   120 \: joule

Now, power dissipated in circuit is :

P =  {i}^{2}  \times r

 \implies \: P =  {(0.1)}^{2}  \times 100

 \implies \: P = 1 \: watt

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