Question

when two resistor are connected in series the equivalent resistance is 7 ohm and when connected in parallel the equivalent resistance 12/7 ohm. what are the value of resistors.​

Answer 1

R1 + R2 = 7

1/R1 + 1/R2 = 7/12

By substituting, say, 10-R1 for R2 in the second equation you get a simple quadratic equation in R1, which you can solve yourself. Good luck.

You’ll get two answers, because the values of R1 and R2 could be swapped, but basically its one answer. I don’t have a pen and paper available but doing it in my head gives me 6 and 4 ohms. But you should check it yourself

Answer 2

Answer:

3Ω and 4Ω

Explanation:

let the 2 resistors be 'xΩ' and 'yΩ'

When two resistors are in series they have values 7Ω

⇒ x + y = 7 ---(1)

⇒ y = 7 - x ---(2)

When two resistors are in parallel they have values 12/7Ω

⇒ $\frac{7}{12} =\frac{1}{x} +\frac{1}{y}$

⇒ $\frac{7}{12} =\frac{x+y}{xy}$

⇒ $\frac{7}{12} =\frac{7}{xy}$  (from equation 1)

⇒ xy = 12

⇒ x(7 - x) = 12 (from equation 2)

⇒ 7x - x² = 12

⇒ x² - 7x + 12 = 0

⇒ x² - 4x - 3x + 12 = 0

⇒ x(x - 4) - 3(x - 4) = 0

⇒ (x - 3)(x - 4) = 0

⇒ x = 3 (or) 4

⇒ y = 4 (or) 3   (from equation 2)

∴ the resistors are 3Ω and 4Ω

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VERIFICATION:

2 resistors 3Ω and 4Ω

(i) when they are connected in series

    Effective resistance = 3 + 4 = 7Ω

(ii) when they are connected in parallel

     1/(Effective resistance) = $\frac{1}{3} + \frac{1}{4}$

    ⇒  1/(Effective resistance) = $\frac{3+4}{12}$

    ⇒  1/(Effective resistance) = $\frac{7}{12}$

  ⇒  Effective resistance = 12/7 Ω

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