Question

ᴛᴡᴏ ʀᴇsɪsᴛᴏʀ ᴏғ ʀᴇsɪsᴛᴀɴᴄᴇ ᴏғ 10 ᴏʜᴍs ᴀɴᴅ 20 ᴏʜᴍs ᴀʀᴇ ᴄᴏɴɴᴇᴄᴛᴇᴅ ɪɴ ᴘᴀʀᴀʟʟᴇʟ. ᴀ ʙᴀᴛᴛᴇʀʏ sᴜᴘᴘʟɪᴇs ᴀ ᴄᴜʀʀᴇɴᴛ ᴏғ 6ᴀ ᴛᴏ ᴛʜᴇ ᴄᴏᴍʙɪɴᴀᴛɪᴏɴ. ᴄᴀʟᴄᴜʟᴀᴛᴇ ᴛʜᴇ ᴄᴜʀʀᴇɴᴛ ɪɴ ᴇᴀᴄʜ ᴏғ ᴛʜᴇ ʀᴇsɪsᴛᴏʀ ᴀɴᴅ ᴘᴏᴡᴇʀ ᴏғ ᴇᴀᴄʜ ʀᴇsɪsᴛᴏʀ​

Answer 1

Answer :-

→ Current in 10 Ω resistor = 4 A

→ Current in 20 Ω resistor = 2 A

→ Power of the 10 Ω resistor = 160 W

→ Power of the 20 Ω resistor = 80 W

Explanation :-

We have :-

→ 1st resistance (R₁) = 10 Ω

→ 2nd resistance (R₂) = 20 Ω

→ Total current (I) = 6 A

______________________________

The current in the 10 Ω resistor is :-

I₁ = (R₂/R₁ + R₂) × I

⇒ I₁ = (20/10 + 20) × 6

⇒ I₁ = 20/30 × 6

⇒ I₁ = 4 A

The current in the 20 Ω resistor :-

I₂ = (R₁/R₁ + R₂) × I

⇒ I₂ = (10/10 + 20) × 6

⇒ I₂ = 10/30 × 6

⇒ I₂ = 2 A

______________________________

Power of 10 Ω resistor :-

P₁ = I²₁ × R₁

⇒ P₁ = (4)² × 10

⇒ P₁ = 16 × 10

⇒ P₁ = 160 W

Power of 20 Ω resistor :-

P₂ = I²₂ × R₂

⇒ P₂ = (2)² × 20

⇒ P₂ = 4 × 20

⇒ P₂ = 80 W

Answer 2

Answer:

Current in 10 ohm resistor = 4A

→ Current in 20Omega resistor= 2A

→ Power of the 10Omega resistor = 160W

→ Power of the 20 Omega resistor = 80W

Explanation :

We have :

→ 1st resistance (R_{1}) = 1O * Omega → 2nd resistance (R_{2}) = 20Omega

→ Total current (I) = 6A

The current in the 10 Q resistor is :

1₁ = (R₂/R₁ + R₂) × I

→ I_{1} = (20/10 + 20) * 6

→ I_{1} = 20/30 * 6

→ 1₁=4 A

The current in the 20 Q resistor :

12 = (R₁/R₁ + R₂) × I

12 (10/10 + 20) × 6

I_{2} = 10/30 * 6

→ I_{2} = 2A

Power of 10 Q resistor :

P_{1} = I_{1} ^ 2 * R_{1}

→ P_{1} = (4) ^ 2 * 10

→ P_{1} = 16 * 1C

P_{1} = 160W

Power of 20 Q resistor :

P 2 =I^ 2 2 * R 2

P_{2} = (2) ^ 2 * 20

→ P_{2} = 4 * 20

P_{2} = 80W