Answer:
R₁ = 23.77 ohms and R₂ = 7.92 ohms
Explanation:
When connected in series, the equivalence resistance, R(eq) is given as
R(eq) = (R₁ + R₂)
When connected in parallel, the equivalence resistance, R(eq) is given as
[1/R(eq)] = [(1/R₁) + (1/R₂)]
R(eq) = (R₁R₂)/(R₁ + R₂)
The parallel and series combination are connected to a battery of emf 39.0 V with negligible internal resistance. And the power supplied is measured.
But power supplied is given as
P = IV = (V/R) V = (V²/R)
When connected in series, the power supplied is given as
P = 48.0 W,
V = 39.0 V,
R = R(eq) = (R₁ + R₂)
48 = (39²/R)
R = (39²/48)
R = 31.6875 ohms
R = (R₁ + R₂) = 31.6875
(R₁ + R₂) = 31.6875 (eqn 1)
When connected in series, the power supplied is given as
P = 256.0 W,
V = 39.0 V,
R = R(eq) = (R₁R₂)/(R₁ + R₂)
256 = (39²/R)
R = (39²/256)
R = 5.9414 ohms
R = R(eq) = (R₁R₂)/(R₁ + R₂) = 5.9414
(R₁R₂)/(R₁ + R₂) = 5.9414
But, recall eqn 1
(R₁ + R₂) = 31.6875
(R₁R₂)/(R₁ + R₂) = 5.9414
Substituting for (R₁ + R₂)
(R₁R₂)/(R₁ + R₂) = (R₁R₂)/31.6875 = 5.9414
(R₁R₂) = 31.6875 × 5.9414 = 188.2683
R₁ = (188.2683/R₂)
(R₁ + R₂) = 31.6875
Substituting for R₁
(188.2683/R₂) + R₂ = 31.6875
multiply through by R₂
188.2683 + R₂² = 31.6875R₂
R₂² - 31.6875R₂ + 188.2683 = 0
Solving the quadratic equation
R₂ = 23.77 ohms or 7.92 ohms
If R₂ = 23.77 ohms, R₁ = 7.92 ohms
If R₂ = 7.92 ohms, R₁ = 23.77 ohms
Since the question explains that R₁ > R₂
R₁ = 23.77 ohms and R₂ = 7.92 ohms
Hope this Helps!!!
