An infinite combination of resistors, each having resistance R = 4 Ω, is given below. What is the net resistance between the points A and B ? (Each resistance is of equal value, R = 4)
Wait, re-reading the visual interpretation and the standard ladder problem formulation: The equivalent resistance R_eq of the infinite ladder from point A to point B (across the first vertical rung) can be determined by considering that adding one more segment at the beginning does not change the total resistance. The segment added at the beginning consists of a horizontal R, a vertical R, and another horizontal R.
Let R_eq be the resistance of the infinite ladder starting from one junction pair (like A and B) to infinity. The structure is: A — R — J1 — …inf… ; B — R — K1 — …inf… ; J1 — R — K1.
The resistance between J1 and K1 onwards is also R_eq.
So, the resistance between A and B is R (A to J1) in series with the parallel combination of R (J1 to K1) and R_eq (J1 onwards), and this whole combination in series with R (K1 to B). This assumes A and B are the start points on the two rails *before* the first vertical resistor. This interpretation doesn’t match the diagram precisely, where A and B are connected to the points *before* the first horizontal resistors on the top and bottom rails respectively.
Let’s use the standard ladder network definition where A and B are the input points. The network structure repeats. Let R_eq be the equivalent resistance of the infinite network from points A and B.
Adding the first segment (R horizontal top, R vertical, R horizontal bottom) such that the rest of the infinite ladder connected after the first vertical resistor has resistance R_eq.
The total resistance R_eq is the resistance of the first horizontal top R, plus the parallel combination of the first vertical R and the resistance of the rest of the ladder. The rest of the ladder, starting from the points where the first vertical R connects to the rails, itself looks like the infinite ladder.
Let’s assume A and B are the input terminals, and the network extends to infinity. The resistance seen from A and B is R_AB. The first segment is horizontal R on top, horizontal R on bottom, and vertical R between them. This interpretation still doesn’t match the usual ladder structure formula derived above.
Let’s re-examine the formula derivation that matches option B. The formula R_eq = R * (1 + sqrt(5)) / 2 corresponds to an infinite ladder where *each segment* consists of a series resistance R on one rail, a series resistance R on the other rail, and a shunt resistance R between the rails. The resistance is calculated between corresponding points on the two rails at the beginning.
A — R — R — …
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R R
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B — R — R — …
Let R_eq be the resistance between A and B. If we add one more segment (R top, R bottom, R vertical) at the beginning, the resistance between the new input points should still be R_eq.
This structure is slightly different from the image. The image shows horizontal R, vertical R, horizontal R, vertical R…
A — R — J1
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R
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K1 — R — J2
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R
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K2 — …
B is connected to K1. This is not A and B being the input points of the ladder. A and B are connected as shown in the diagram.
Let the resistance of the infinite network starting from J1 and K1 onwards be R_inf.
The resistance between J1 and B is the series combination of R (J1 to K1) and R (K1 to B). So R_J1B = R + R = 2R.
The resistance between A and J1 is R.
The resistance from A to B is R (A to J1) in series with the parallel combination of R (vertical from J1) and the resistance of the rest of the network starting from K1. This is still confusing based on the diagram.
Let’s assume the question *implies* the standard infinite ladder network structure where A and B are the input terminals across the first vertical resistor, and the formula R_eq = R * (1 + sqrt(5)) / 2 applies to this structure.
In that case, with R = 4 Ω:
R_eq = 4 * (1 + sqrt(5)) / 2 = 2 * (1 + sqrt(5)) = 2 + 2√5 Ω.
This matches option B. Given this is a standard physics/engineering problem, it’s highly likely the question intends this standard structure and formula, despite the potentially ambiguous labeling of A and B in the diagram relative to the text description (“between the points A and B”). Let’s proceed with this interpretation as it’s the only one yielding a matching option.
– The key to solving infinite networks is to recognize that adding or removing a finite number of repeating units does not change the overall equivalent resistance of the infinite part.
– Let the equivalent resistance of the infinite ladder from a typical junction pair onwards be R_eq.
– By considering the first segment and the remaining infinite part, a relationship can be set up to solve for R_eq.
– For a standard ladder with series resistors R on rails and shunt resistors R between rails, the equivalent resistance R_eq satisfies R_eq = R + (R * R_eq) / (R + R_eq). This leads to R_eq^2 – R * R_eq – R^2 = 0, with the positive solution R_eq = R * (1 + sqrt(5)) / 2.
– Applying R = 4 Ω, R_eq = 4 * (1 + sqrt(5)) / 2 = 2 * (1 + sqrt(5)) = 2 + 2√5 Ω.