Question 620: Current Electricity

Question

A wire of length '$$l$$' and resistance $$100 \Omega$$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

Accepted Answer

To solve this problem, we first need to understand how the resistance changes when we cut the wire into equal parts and how it behaves when connected in different configurations (series and parallel).
Given:
Original length of the wire, $ l $.
Total resistance of the wire $ R = 100 \Omega $.
The wire is divided into 10 equal parts.
Resistance of each part:
Since the wire is divided into 10 equal parts, the length of each part is $\frac{l}{10}$. Resistance is proportional to length (as long as the cross-sectional area and material of the wire remain constant). Therefore, the resistance of each part, denoted as $ r $, is $\frac{1}{10}$th of the total resistance:
$$ r = \frac{R}{10} = \frac{100 \Omega}{10} = 10 \Omega $$
First 5 parts in series:
When resistors are connected in series, the total resistance is the sum of the individual resistances:
$$ R_{	ext{series}} = 5 	imes 10 \Omega = 50 \Omega $$
Next 5 parts in parallel:
When resistors are connected in parallel, the total resistance $ R_{	ext{parallel}} $ can be calculated using the reciprocal formula:
$$ \frac{1}{R_{	ext{parallel}}} = \frac{1}{10 \Omega} + \frac{1}{10 \Omega} + \frac{1}{10 \Omega} + \frac{1}{10 \Omega} + \frac{1}{10 \Omega} = 5 	imes \frac{1}{10 \Omega} = \frac{5}{10 \Omega} = \frac{1}{2 \Omega} $$
Thus,
$$ R_{	ext{parallel}} = 2 \Omega $$
Final combination in series:
The total resistance of the combination, where the series and parallel groups are again connected in series, will be:
$$ R_{	ext{total}} = R_{	ext{series}} + R_{	ext{parallel}} = 50 \Omega + 2 \Omega = 52 \Omega $$
Thus, the correct answer to the resistance of the final combination is:
Option B $$52 \Omega$$

Answer

$$26 \Omega$$

Answer

$$52 \Omega$$

Answer

$$55 \Omega$$

Answer

$$60 \Omega$$

Current Electricity

A wire of length '$l' and resistance 100 \Omega$ is divided into 10 equal parts. The first 5 parts are connected in series while the next 5 parts are connected in parallel. The two combinations are again connected in series. The resistance of this final combination is:

Choose the correct answer: