A wire of resistance \(R\) is cut into 8 equal pieces. From these pieces, two equivalent resistances are made by adding four of these together in parallel. Then these two are added in series. The net effective resistance of the combination is:
Show Hint
- \( \frac{R}{32} \)
- \( \frac{R}{4} \)
- \( \frac{R}{8} \)
- \( \frac{R}{6} \)
The Correct Option is B
Solution and Explanation
The problem begins with a wire of total resistance \( R \) divided into 8 equal segments. The resistance of each segment is calculated as:
\[ R_{\text{piece}} = \frac{R}{8} \]
Subsequently, four of these segments are connected in parallel. The equivalent resistance \( R_{\text{parallel}} \) for these parallel resistors is determined by:
\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_{\text{piece}}} + \frac{1}{R_{\text{piece}}} + \frac{1}{R_{\text{piece}}} + \frac{1}{R_{\text{piece}}} = 4 \times \frac{1}{R_{\text{piece}}} \]
This simplifies to:
\[ R_{\text{parallel}} = \frac{R_{\text{piece}}}{4} = \frac{R/8}{4} = \frac{R}{32} \]
Two such parallel combinations are formed, each yielding a resistance of \( \frac{R}{32} \).
These two equivalent resistances are then connected in series. The total resistance \( R_{\text{total}} \) for series connections is the sum of individual resistances:
\[ R_{\text{total}} = R_{\text{parallel}} + R_{\text{parallel}} = \frac{R}{32} + \frac{R}{32} = \frac{2R}{32} = \frac{R}{16} = \frac{R}{4} \]
Consequently, the net effective resistance of the entire combination is \( \frac{R}{4} \).
