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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
When a 4 Ω resistor is connected across the terminals of a 12 V battery, the number of coulombs passing through the resistor per second is:
(a) 0.3
(b) 3
(c) 4
(d) 12
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