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प्रश्न
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उत्तर
The resistors of values 20 Ω, 10 Ω and 20 Ω are connected in series. Therefore, their net resistance is:
R = 20 Ω + 10 Ω + 20 Ω
R = 50 Ω
This combination of resistors are connected in parallel with that of value 30 Ω. Therefore, the resistance R' is:
1/R' = (1/50) + (1/30)
1/R' = (3 + 5)/150
R' = 150/8
R' = 18.75 Ω
The resistance, R' is connected in series with the two resistors of values 10 Ω each. Hence, the total resistance between A and B is:
18.75 + 10 + 10 = 38.75 Ω
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संबंधित प्रश्न
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