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प्रश्न
The resistance of two resistors joined in series is 8Ω and in parallel is 1.5Ω. Find the value of the two resistances.
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उत्तर
In series, R1 + R2 = 8 Ω
In parallel `("R"_1"R"_2)/("R"_1 + "R"_2) = 1.5` Ω
∴ R1R2 = 8 × 1.5 = 12 Ω
Now (R1 - R2)2 = (R1 + R2)2 - 4R1R2
∴ (R1 - R2)2 = (8)2 - 4 × 12
or (R1 - R2)2 = 64 - 48 = 16
or R1 - R2 = 4 Ω
On solving equations (i) and (ii) R1 = 6 Ω and R2 = 2 Ω
संबंधित प्रश्न
(i) Two sets A and B, of three bulbs each, are glowing in two separate rooms. When one of the bulbs in set A is fused, the other two bulbs also cease to glow. But in set B, when one bulb fuses, the other two bulbs continue to glow. Explain why this phenomenon occurs.
(ii) Why do we prefer arrangements of Set B for house circuiting?
Find the equivalent resistance between A and B
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