Question

Four resistors of equal resistance R each are connected in various possible ways both in series and/or parallel to get different combined resistances. Which of the following values of resistance cannot be achieved by using all the four resistors?

• A. 0.25 R,
• B. 1.33 R
• C. 2 R
• D. 2.5 R

Answer 1

Answer:

C. 2 R

Combinations of Resistors

Understanding Series and Parallel Combinations

• When resistors are connected in series, the total resistance is the sum of individual resistances: R_total = R₁ + R₂ + ... + R_n.
• When resistors are connected in parallel, the reciprocal of the total resistance is the sum of the reciprocals of individual resistances: 1/R_total = 1/R₁ + 1/R₂ + ... + 1/R_n.

Analyzing Possible Combinations with Four Equal Resistors (R)

Given four resistors, each with resistance R, we can explore various configurations to find achievable combined resistances.

1. All Four in Series

• Resistance = R + R + R + R = 4R

2. All Four in Parallel

• 1/R_total = 1/R + 1/R + 1/R + 1/R = 4/R
• R_total = R/4

3. Three in Series, One in Parallel

• Resistance of the series part = R + R + R = 3R
• This 3R is in parallel with the fourth R.
• 1/R_total = 1/(3R) + 1/R = (1 + 3)/(3R) = 4/(3R)
• R_total = 3R/4

4. One in Series, Three in Parallel

• Resistance of the parallel part: 1/R_parallel = 1/R + 1/R + 1/R = 3/R => R_parallel = R/3
• This R/3 is in series with the first R.
• R_total = R + R/3 = (3R + R)/3 = 4R/3

5. Two in Series, Two in Parallel

• Resistance of one series pair = R + R = 2R
• Resistance of the other series pair = R + R = 2R
• These two 2R resistances are in parallel.
• 1/R_total = 1/(2R) + 1/(2R) = 2/(2R) = 1/R
• R_total = R

6. Two in Parallel, Two in Series

• Resistance of one parallel pair = (R * R) / (R + R) = R²/2R = R/2
• Resistance of the other parallel pair = (R * R) / (R + R) = R²/2R = R/2
• These two R/2 resistances are in series.
• R_total = R/2 + R/2 = R

7. Three in Parallel, One in Series

• Resistance of the parallel group = R/3 (as calculated in point 4)
• This R/3 is in series with the fourth R.
• R_total = R + R/3 = 4R/3

8. One in Parallel, Three in Series

• Resistance of the series group = 3R (as calculated in point 3)
• This 3R is in parallel with the fourth R.
• R_total = 3R/4 (as calculated in point 3)

9. One in Series, One in Parallel, One in Series

• Let's connect R1 in series with R2, then R3 in parallel with the combination, and then R4 in series with that.
• R1+R2 = 2R
• (2R) || R3 = (2R * R) / (2R + R) = 2R²/3R = 2R/3
• (2R/3) + R4 = 2R/3 + R = 5R/3
• R_total = 5R/3

10. Other Complex Combinations

• By arranging the resistors in different patterns (e.g., bridging combinations), other resistance values might be obtained. For instance, a Wheatstone bridge configuration with equal arms would result in an open circuit if balanced, but slight imbalances can lead to complex outcomes.

Achievable Resistances

The following resistance values can be achieved:

• 4R (all in series)
• R/4 (all in parallel)
• 3R/4
• 4R/3
• R (two pairs in series, or two pairs in parallel)
• 5R/3
• Other values are also possible depending on the precise arrangement.

Unachievable Resistance Value

The value 2R cannot be achieved by connecting four resistors of equal resistance R in any series and/or parallel combination. This is because any combination of series and parallel connections of equal resistors will result in a total resistance that is a rational multiple of R, where the numerator and denominator are derived from the number of resistors and their arrangement. For 2R to be achieved, it would typically require different ratios or numbers of resistors in series and parallel groups that cannot be formed with exactly four equal resistors.