In an electrical circuit, 'n' number of 1 Ω resistors are connected in parallel. When one of the resistors is eliminated, the overall resistance of the circuit is ____
1/(n-1)
The solution to this problem involves understanding the concept of parallel resistance. When resistors are connected in parallel, the reciprocal of the total resistance is equal to the sum of the reciprocals of the individual resistances.
Initially, we have 'n' resistors, each with a resistance of 1 Ω, connected in parallel. The total resistance \(R_n\) is given by:
\(\frac{1}{R_n} = \frac{1}{1} + \frac{1}{1} + ... + \frac{1}{1}\) (n times)
\(\frac{1}{R_n} = n\)
\(R_n = \frac{1}{n}\)
When one resistor is removed, we are left with (n-1) resistors in parallel. The new total resistance \(R_{n-1}\) is:
\(\frac{1}{R_{n-1}} = \frac{1}{1} + \frac{1}{1} + ... + \frac{1}{1}\) (n-1 times)
\(\frac{1}{R_{n-1}} = n - 1\)
\(R_{n-1} = \frac{1}{n-1}\)
Therefore, when one resistor is eliminated, the overall resistance of the circuit becomes \(\frac{1}{n-1}\) Ω.
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