A wire of resistance 20 \(\Omega\) is divided into 10 equal parts. A combination of two parts are connected in parallel and so on. Now resulting pairs of parallel combination are connected in series. The equivalent resistance of final combination is _______ \(\Omega\).
Correct Answer: 5
Approach Solution - 1
Given:
- Total resistance of the wire: \( 20 \, \Omega \)
- The wire is divided into 10 equal parts.
Step 1: Resistance of each part
The resistance of each part is:
\[ R_{\text{part}} = \frac{\text{Total resistance}}{\text{Number of parts}} = \frac{20 \, \Omega}{10} = 2 \, \Omega. \]
Step 2: Parallel combination of two parts
Each pair of two parts is connected in parallel. The equivalent resistance of a parallel combination is:
\[ R_{\text{parallel}} = \frac{R_{\text{part}}}{2} = \frac{2 \, \Omega}{2} = 1 \, \Omega. \]
Step 3: Total number of parallel combinations
Since there are 10 parts and they are paired in groups of 2, the total number of parallel combinations is:
\[ \text{Number of parallel combinations} = \frac{10}{2} = 5. \]
Step 4: Series combination of parallel pairs
The 5 parallel combinations (each of \( 1 \, \Omega \)) are connected in series. The equivalent resistance of a series combination is:
\[ R_{\text{eq}} = 5 \times R_{\text{parallel}} = 5 \times 1 \, \Omega = 5 \, \Omega. \]
Thus, the equivalent resistance of the final combination is \( R_{\text{eq}} = 5 \, \Omega \).
Approach Solution -2
Total resistance of the wire, R = 20 Ω
The wire is divided into 10 equal parts.
Hence, resistance of each part:
\[ R_1 = \frac{20}{10} = 2 \, \Omega \]
Step 2: Forming pairs in parallel.
Two parts are connected in parallel to form one pair.
Resistance of each pair (Rp):
\[ R_p = \frac{R_1}{2} = \frac{2}{2} = 1 \, \Omega \]
Since there are 10 parts, the number of such pairs = 10 / 2 = 5.
Step 3: Connecting all pairs in series.
Now, 5 such pairs (each of resistance 1 Ω) are connected in series.
Total resistance (Req):
\[ R_{eq} = 5 \times 1 = 5 \, \Omega \]
Step 4: Final Answer.
The equivalent resistance of the final combination is:
\[ \boxed{5 \, \Omega} \]
Final Answer: 5 Ω
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