Question: Two resistances, a 60 ohm and an unknown one are connected to a power source in a series arrangement...

Question

Two resistances, a 60 ohm and an unknown one are connected to a power source in a series arrangement. This way the power of the unknown resistance is 60 watt. What is the least voltage of the power source?

Answer 1

Solution

$\frac{V_{1}}{V_{2}}$ = $\frac{60}{R}$ ........(1)

V₁ + V₂ = V .........(2)

from (1) & (2) Ž V₂ = $\frac{VR}{60 + R}$ ...........(3)

$\frac{V_{2}^{2}}{R}$ = 60 watt ........(4)

from (3) & (4) V²R = 60 (60 + R)²

V²R = 60(3600 + R² + 60R)

0 = 60R² + 3600 R – V²R + 60 × 3600

(3600 – V² – 60 × 60 × 2)

(3600 – V² + 60 × 60 × 2) > 0

(3 × 3600 – V²) (–3600 – V²) > 0

3 × 3600 – V² < 0

60 × 3600 < V²

60 × $\sqrt{3}$ < V