Question

calculate thew energy transferred when 2A current flows through a 10 ohm
resistor for 30 mins.

Answer 1

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➧ P = 1² × R

➙ Cᴜʀʀᴇɴᴛ => 2 ᴀᴍᴘᴇʀᴇꜱ
➙ Rᴇꜱɪꜱᴛᴀɴᴄᴇ => 2 ᴏʜᴍꜱ
➙ Pᴏᴡᴇʀ => (5)² × 2

➾ 25 × 2
➾ 30 ᴡᴀᴛᴛꜱ

➾ 50 / 1000 ᴋW
➾ 0.05 ᴋᴡ

➧ Tɪᴍᴇ=> 30 ᴍɪɴ

➾ 30 / 60
➾ ½ ʜʀꜱ
➾ 0.5 ʜʀꜱ

➧ Eɴᴇʀɢʏ = P × ᴛ

➾ 0.05 × 0.5
➾0.025 Kᴡʜ...✔

Answer 2

Given,  

Current flowing through resistors = 2 A

Resistance of resistors = 10 Ω

Time taken to flow the current = 30 mins = 30×60 = 1800 sec

To find,

The energy transferred

Solution,  

We can simply solve this numerical problem by using the subsequent mechanism  

Now,  

Current flowing through resistors (I) = 2 A

Resistance of resistors (R) = 10 Ω

Time taken to flow the current (T) = 1800 sec

Energy transferred (E) = ?

As we know that,

E = I²RT  

Where,

I =  Current flown through device

R = Resistance of device

T = Time for which current is flown

E = 2²×10×1800

E = 72000 J

E = 72 KJ

Hence, 72 KJ is the energy transferred when 2 A current flows through a resistor of 10 Ω for 30 minutes