Question

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:

• A. \( \frac{R}{32} \)
• B. \( \frac{R}{4} \)
• C. \( \frac{R}{8} \)
• D. \( \frac{R}{6} \)

Hint:

For \(n\) equal resistors \(r\) in parallel, the equivalent resistance is \(r/n\). For resistors in series, the equivalent resistance is the sum of individual resistances.

Answer 1

The Correct Option is B

To solve this problem, we start with a wire of total resistance \( R \) that is cut into 8 equal pieces. Each piece will thus have a resistance of:

\[ R_{\text{piece}} = \frac{R}{8} \]

Next, four of these pieces are connected in parallel to form one equivalent resistance. The formula for calculating the equivalent resistance \( R_{\text{parallel}} \) of resistors in parallel (with equal resistance) is:

\[ \frac{1}{R_{\text{parallel}}} = \frac{1}{R_{\text{piece}}} + \frac{1}{R_{\text{piece}}} + \frac{1}{R_{\text{piece}}} + \frac{1}{R_{\text{piece}}} = 4 \times \frac{1}{R_{\text{piece}}} \]

Simplifying, we find:

\[ R_{\text{parallel}} = \frac{R_{\text{piece}}}{4} = \frac{R/8}{4} = \frac{R}{32} \]

Since we create two equivalent resistances using the same method (four pieces in parallel each), both will have the same resistance \( \frac{R}{32} \).

These two are then added in series, and the total resistance \( R_{\text{total}} \) for resistors in series is simply the sum:

\[ R_{\text{total}} = R_{\text{parallel}} + R_{\text{parallel}} = \frac{R}{32} + \frac{R}{32} = \frac{2R}{32} = \frac{R}{16} = \frac{R}{4} \]

Therefore, the net effective resistance of the combination is \( \frac{R}{4} \).