Question

The total current drawn by a circuit consisting of three resistors connected in parallel is 12A. The voltage drop across the first resistor is 12V, the value of a second resistor is 3 Ohms and the power dissipation of the third resistor is 24W. What are the resistance of first and Third resistors?

(Please give a detailed answer and faster) ​

Answer 1

Answer:

All we can say for sure is that 24V is across each resistor and that the equivalent resistance of the three parallel resistors is 24V/12A = 2Ω.

This resistance can be the result of different combinations of three parallel resistors…

[5Ω, 4Ω, 20Ω],

[7Ω, 3Ω, 42Ω],

[8Ω, 3Ω, 24Ω],

[9Ω, 3Ω, 18Ω],

[10Ω, 3Ω, 15Ω],

[12Ω, 3Ω,12Ω]

Shuffling the order of the resistors within each bracketed group gives more results.

Answer 2

Given: Current = 12A,

           Voltage = 12V,

           Second resistor = 3Ω

            Power dissipation in the third resistor = 24W

To find: Resistance of first and third resistors.

Solution: Current in the circuit is 12A

The voltage across the first resistor is 12V

The voltage across the first resistor will be equal to the voltage across the second and third resistor.

i₁R₁ = i₂R₂ = i₃R₃

i₁+i₂+i₃ = 12A

Now, in third resistor, power = V²/R₃

24 = (12)²/R₃

R₃ = (12×12)/24

R₃ = 6Ω

Now, also power is also equal to i₃²R₃

24 = i₃²×6

i₃ = 2Ω

As, i₂R₂ = i₃R₃

R₂ = 3Ω, therefore, i₂×3 = 2×6

i₂ = 4A

As, i₁R₁ = i₂R₂

i₁ = 12-6 = 6A

6×R₁ = 4×3

R₁ = 2Ω

Therefore, the resistance of the first resistor is 2Ω, and the resistance of the third resistor is 6Ω.