The question asks us to find the equivalent resistance when an electric wire is cut into equal parts and then connected in parallel. Let's break down the steps.
Initially, we have an electric wire with a resistance of 50 ohm. This wire is cut into five equal pieces. When a wire of uniform material and thickness is cut into equal lengths, the resistance of each piece is proportional to its length. Since the wire is cut into five equal pieces, the resistance of each individual piece will be one-fifth of the original resistance.
Let the original resistance be \(R = 50 \, \Omega\). The wire is cut into \(n = 5\) equal parts. The resistance of each part, let's call it \(r\), is given by:
\(r = \frac{R}{n}\)
Substituting the given values:
\(r = \frac{50 \, \Omega}{5} = 10 \, \Omega\)
So, each of the five pieces of wire has a resistance of 10 ohm.
These five wires, each having a resistance of 10 ohm, are then connected in parallel. When resistances are connected in parallel, the reciprocal of the equivalent resistance is the sum of the reciprocals of the individual resistances.
For \(n\) resistances connected in parallel, the equivalent resistance \(R_{eq}\) is given by:
\(\frac{1}{R_{eq}} = \frac{1}{r_1} + \frac{1}{r_2} + \dots + \frac{1}{r_n}\)
In this case, we have \(n = 5\) equal resistances, where each \(r_i = r = 10 \, \Omega\). So the formula simplifies to:
\(\frac{1}{R_{eq}} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} = \frac{5}{r}\)
Therefore, the equivalent resistance is:
\(R_{eq} = \frac{r}{5}\)
Using the resistance of each wire \(r = 10 \, \Omega\) and the number of wires \(n = 5\) connected in parallel:
\(R_{eq} = \frac{10 \, \Omega}{5}\)
\(R_{eq} = 2 \, \Omega\)
Thus, the equivalent resistance of the combination is 2 ohm.
| Parameter | Value |
|---|---|
| Initial Resistance (R) | 50 ohm |
| Number of parts (n) | 5 |
| Resistance of each part (r) | 10 ohm |
| Connection Type | Parallel |
| Equivalent Resistance (Req) | 2 ohm |
| Combination Type | Formula for \(n\) Resistors (\(R_1, R_2, \dots, R_n\)) | Formula for \(n\) Equal Resistors (\(r\)) |
|---|---|---|
| Series | \(R_{eq} = R_1 + R_2 + \dots + R_n\) | \(R_{eq} = nr\) |
| Parallel | \(\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \dots + \frac{1}{R_n}\) | \(R_{eq} = \frac{r}{n}\) |
The electrical resistance \(R\) of a uniform conductor is directly proportional to its length \(L\) and inversely proportional to its cross-sectional area \(A\). This relationship is given by the formula:
\(R = \rho \frac{L}{A}\)
where \(\rho\) (rho) is the resistivity of the material, a constant that depends on the material properties.
When a wire is cut into equal parts, say into \(n\) parts, the length of each part becomes \(L/n\), while the material (\(\rho\)) and the cross-sectional area (\(A\)) remain the same for each part. Therefore, the resistance of each part \(r\) is:
\(r = \rho \frac{L/n}{A} = \frac{1}{n} \left(\rho \frac{L}{A}\right) = \frac{R}{n}\)
This confirms why cutting a wire into \(n\) equal parts results in each part having \(1/n\) times the original resistance.
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