Question

Determine the current flowing through a
potential difference across the 9ohm resistor.find
the power dissipated in the 11ohm resistor.
​

Answer 1

Answer:

Make the diagram in your answer sheet and then continue with the following answer.

This is a series connection.

∴ Total Resistance = Rₛ = R₁ + R₂ + R₃

= (4 + 9 + 11) Ω

= 24 Ω

Total potential difference = V = 12 V

∴ By Ohm's law,

Total current = Vₛ/Rₛ                                          [∵ V= IR]

⇒ I = Vₛ/Rₛ

⇒ I = 12/24

⇒ I = 1/2

⇒ I = 0.5 A

• As current remains constant in a series connection, thus the current flowing across the 9 Ω resistor = I = 0.5 A

Now,

Current passing through R₂ = 0.5 A

Resistance of R₂= 9 Ω

∴ Potential difference across R₂ = V₂ = ?

By Ohm's law,

V= IR

⇒ V₂ = 0.5 × 9

⇒ V₂ = 4.5 V

• Hence the potential difference across R₂, 9 Ω resistor = V₂ = 4.5 V

Now,

P = I²R

Resistance of R₃ = 11 Ω

Current across R₃ =I =  0.5 A

⇒ P = [(0.5)² × 11 ]W

⇒ P = (0.25 × 11) W

⇒ P = 2.75 W

• Hence, the power dissipated in the 11 Ω resistor = P = 2.75 W

Ans:

• Current across 9 Ω resistor = I = 0.5 A
• Potential difference across 9 Ω resistor = 4.5 V
• Power dissipated in the 11 Ω resistor = P = 2.75 W

Hope you understood.

Thank You.